sem5 em electomagnet ph217 note2to7 exam1n2nfinal

12/21/2015 Lecture Notes Chapter 2

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Chapter 2. Electrostatics2.1. The Electrostatic Field2.2. Divergence and Curl of Electrostatic Fields

2.2.1. The curl of E2.2.2. Applications of Gauss's law2.3. The Electric Potential2.4. Work and Energy in Electrostatics2.5. Metallic Conductors2.6. Capacitors

Chapter 2. Electrostatics

2.1. The Electrostatic Field

To calculate the force exerted by some electric charges, q1, q2, q3, ... (the source charges) on anothercharge Q (the test charge) we can use the principle of superposition. This principle states that theinteraction between any two charges is completely unaffected by the presence of other charges. Theforce exerted on Q by q1, q2, and q3 (see Figure 2.1) is therefore equal to the vector sum of the force exerted by q1 on Q, the force exerted by q2 on Q, and the force exerted by q3 on Q.

Figure 2.1. Superposition of forces. The force exerted by a charged particle on another charged particle depends on their separationdistance, on their velocities and on their accelerations. In this Chapter we will consider the specialcase in which the source charges are stationary. The electric field produced by stationary source charges is called and electrostatic field. The electricfield at a particular point is a vector whose magnitude is proportional to the total force acting on a testcharge located at that point, and whose direction is equal to the direction of the force acting on apositive test charge. The electric field , generated by a collection of source charges, is defined as

where is the total electric force exerted by the source charges on the test charge Q. It is assumedthat the test charge Q is small and therefore does not change the distribution of the source charges.The total force exerted by the source charges on the test charge is equal to



The electric field generated by the source charges is thus equal to

In most applications the source charges are not discrete, but are distributed continuously over someregion. The following three different distributions will be used in this course:

1. line charge λ: the charge per unit length.

2. surface charge σ: the charge per unit area.

3. volume charge ρ: the charge per unit volume.

To calculate the electric field at a point generated by these charge distributions we have to replacethe summation over the discrete charges with an integration over the continuous charge distribution:

1. for a line charge:

2. for a surface charge:

3. for a volume charge:

Here is the unit vector from a segment of the charge distribution to the point at which we areevaluating the electric field, and r is the distance between this segment and point .

Example: Problem 2.2a) Find the electric field (magnitude and direction) a distance z above the midpoint between two equalcharges q a distance d apart. Check that your result is consistent with what you would expect when z »d.b) Repeat part a), only this time make he right-hand charge -q instead of +q.

Figure 2.2. Problem 2.2a) Figure 2.2a shows that the x components of the electric fields generated by the two point chargescancel. The total electric field at P is equal to the sum of the z components of the electric fields



generated by the two point charges:

When z » d this equation becomes approximately equal to

which is the Coulomb field generated by a point charge with charge 2q.

b) For the electric fields generated by the point charges of the charge distribution shown in Figure2.2b the z components cancel. The net electric field is therefore equal to

Example: Problem 2.5 Find the electric field a distance z above the center of a circular loop of radius r which carries auniform line charge λ.

Figure 2.3. Problem 2.5. Each segment of the loop is located at the same distance from P (see Figure 2.3). The magnitude ofthe electric field at P due to a segment of the ring of length dl is equal to

When we integrate over the whole ring, the horizontal components of the electric field cancel. Wetherefore only need to consider the vertical component of the electric field generated by each segment:

The total electric field generated by the ring can be obtained by integrating dEz over the whole ring:



Example: Problem 2.7 Find the electric field a distance z from the center of a spherical surface of radius R, which carries auniform surface charge density σ. Treat the case z < R (inside) as well as z > R (outside). Express youranswer in terms of the total charge q on the surface.

Figure 2.4. Problem 2.7. Consider a slice of the shell centered on the z axis (see Figure 2.4). The polar angle of this slice is θand its width is dθ. The area dA of this ring is

The total charge on this ring is equal to

where q is the total charge on the shell. The electric field produced by this ring at P can be calculatedusing the solution of Problem 2.5:

The total field at P can be found by integrating dE with respect to θ:

This integral can be solved using the following relation:

Substituting this expression into the integral we obtain:



Outside the shell, z > r and consequently the electric field is equal to

Inside the shell, z < r and consequently the electric field is equal to

Thus the electric field of a charged shell is zero inside the shell. The electric field outside the shell isequal to the electric field of a point charge located at the center of the shell.

2.2. Divergence and Curl of Electrostatic Fields

The electric field can be graphically represented using field lines. The direction of the field linesindicates the direction in which a positive test charge moves when placed in this field. The density offield lines per unit area is proportional to the strength of the electric field. Field lines originate onpositive charges and terminate on negative charges. Field lines can never cross since if this wouldoccur, the direction of the electric field at that particular point would be undefined. Examples of fieldlines produced by positive point charges are shown in Figure 2.5.

Figure 2.5. a) Electric field lines generated by a positive point charge with charge q. b) Electricfield lines generated by a positive point charge with charge 2q.

The flux of electric field lines through any surface is proportional to the number of field lines passingthrough that surface. Consider for example a point charge q located at the origin. The electric flux through a sphere of radius r, centered on the origin, is equal to

Since the number of field lines generated by the charge q depends only on the magnitude of thecharge, any arbitrarily shaped surface that encloses q will intercept the same number of field lines.Therefore the electric flux through any surface that encloses the charge q is equal to . Using theprinciple of superposition we can extend our conclusion easily to systems containing more than onepoint charge:

We thus conclude that for an arbitrary surface and arbitrary charge distribution



where Qenclosed is the total charge enclosed by the surface. This is called Gauss's law. Since thisequation involves an integral it is also called Gauss's law in integral form. Using the divergence theorem the electric flux can be rewritten as

We can also rewrite the enclosed charge Qencl in terms of the charge density ρ:

Gauss's law can thus be rewritten as

Since we have not made any assumptions about the integration volume this equation must hold forany volume. This requires that the integrands are equal:

This equation is called Gauss's law in differential form. Gauss's law in differential form can also be obtained directly from Coulomb's law for a chargedistribution :

where . The divergence of is equal to

which is Gauss's law in differential form. Gauss's law in integral form can be obtained by integrating over the volume V:

Example: Problem 2.42 If the electric field in some region is given (in spherical coordinates) by the expression

where A and B are constants, what is the charge density ρ?

The charge density ρ can be obtained from the given electric field, using Gauss's law in differentialform:



2.2.1. The curl of E

Consider a charge distribution ρ(r). The electric field at a point P generated by this charge distributionis equal to

where . The curl of is equal to

However, for every vector and we thus conclude that

2.2.2. Applications of Gauss's law

Although Gauss's law is always true it is only a useful tool to calculate the electric field if the chargedistribution is symmetric:

1. If the charge distribution has spherical symmetry, then Gauss's law can be used with concentricspheres as Gaussian surfaces.

2. If the charge distribution has cylindrical symmetry, then Gauss's law can be used with coaxialcylinders as Gaussian surfaces.

3. If the charge distribution has plane symmetry, then Gauss's law can be used with pill boxes asGaussian surfaces.

Example: Problem 2.12 Use Gauss's law to find the electric field inside a uniformly charged sphere (charge density ρ) ofradius R.

The charge distribution has spherical symmetry and consequently the Gaussian surface used to obtainthe electric field will be a concentric sphere of radius r. The electric flux through this surface is equalto

The charge enclosed by this Gaussian surface is equal to



Applying Gauss's law we obtain for the electric field:

Example: Problem 2.14 Find the electric field inside a sphere which carries a charge density proportional to the distance fromthe origin: ρ = k r, for some constant k.

The charge distribution has spherical symmetry and we will therefore use a concentric sphere ofradius r as a Gaussian surface. Since the electric field depends only on the distance r, it is constant onthe Gaussian surface. The electric flux through this surface is therefore equal to

The charge enclosed by the Gaussian surface can be obtained by integrating the charge distributionbetween r' = 0 and r' = r:

Applying Gauss's law we obtain:

or

Example: Problem 2.16 A long coaxial cable carries a uniform (positive) volume charge density ρ on the inner cylinder (radiusa), and uniform surface charge density on the outer cylindrical shell (radius b). The surface charge isnegative and of just the right magnitude so that the cable as a whole is neutral. Find the electric fieldin each of the three regions: (1) inside the inner cylinder (r < a), (2) between the cylinders (a < r < b),(3) outside the cable (b < r).

The charge distribution has cylindrical symmetry and to apply Gauss's law we will use a cylindricalGaussian surface. Consider a cylinder of radius r and length L. The electric field generated by thecylindrical charge distribution will be radially directed. As a consequence, there will be no electricflux going through the end caps of the cylinder (since here ). The total electric flux through thecylinder is equal to

The enclosed charge must be calculated separately for each of the three regions:

1. r < a:



2. a < r < b:

3. b < r:

Applying Gauss's law we find

Substituting the calculated Qencl for the three regions we obtain

1. r < a: .

2. a < r < b:

3. b < r

Example: Problem 2.18 Two spheres, each of radius R and carrying uniform charge densities of +ρ and -ρ, respectively, areplaced so that they partially overlap (see Figure 2.6). Call the vector from the negative center to thepositive center . Show that the field in the region of overlap is constant and find its value.

To calculate the total field generated by this charge distribution we use the principle of superposition.The electric field generated by each sphere can be obtained using Gauss' law (see Problem 2.12).Consider an arbitrary point in the overlap region of the two spheres (see Figure 2.7). The distancebetween this point and the center of the negatively charged sphere is r-. The distance between thispoint and the center of the positively charged sphere is r+. Figure 2.7 shows that the vector sum of and is equal to . Therefore,

The total electric field at this point in the overlap region is the vector sum of the field due to thepositively charged sphere and the field due to the negatively charged sphere:

Figure 2.6. Problem 2.18.



Figure 2.7. Calculation of Etot.The minus sign in front of shows that the electric field generated by the negatively charged sphere isdirected opposite to . Using the relation between and obtained from Figure 2.7 we can rewrite as

which shows that the field in the overlap region is homogeneous and pointing in a direction oppositeto .

2.3. The Electric Potential

The requirement that the curl of the electric field is equal to zero limits the number of vector functionsthat can describe the electric field. In addition, a theorem discussed in Chapter 1 states that any vectorfunction whose curl is equal to zero is the gradient of a scalar function. The scalar function whosegradient is the electric field is called the electric potential V and it is defined as

Taking the line integral of between point a and point b we obtain

Taking a to be the reference point and defining the potential to be zero there, we obtain for V(b)

The choice of the reference point a of the potential is arbitrary. Changing the reference point of thepotential amounts to adding a constant to the potential:

where K is a constant, independent of b, and equal to

However, since the gradient of a constant is equal to zero

Thus, the electric field generated by V' is equal to the electric field generated by V. The physical



behavior of a system will depend only on the difference in electric potential and is thereforeindependent of the choice of the reference point. The most common choice of the reference point inelectrostatic problems is infinity and the corresponding value of the potential is usually taken to beequal to zero:

The unit of the electrical potential is the Volt (V, 1V = 1 Nm/C).

Example: Problem 2.20 One of these is an impossible electrostatic field. Which one?a) b) Here, k is a constant with the appropriate units. For the possible one, find the potential, using theorigin as your reference point. Check your answer by computing .

a) The curl of this vector function is equal to

Since the curl of this vector function is not equal to zero, this vector function can not describe anelectric field.

b) The curl of this vector function is equal to

Since the curl of this vector function is equal to zero it can describe an electric field. To calculate theelectric potential V at an arbitrary point (x, y, z), using (0, 0, 0) as a reference point, we have toevaluate the line integral of between (0, 0, 0) and (x, y, z). Since the line integral of is pathindependent we are free to choose the most convenient integration path. I will use the followingintegration path:

The first segment of the integration path is along the x axis:

and

since y = 0 along this path. Consequently, the line integral of along this segment of the integrationpath is equal to zero. The second segment of the path is parallel to the y axis:



and

since z = 0 along this path. The line integral of along this segment of the integration path is equal to

The third segment of the integration path is parallel to the z axis:

and

The line integral of along this segment of the integration path is equal to

The electric potential at (x, y, z) is thus equal to

The answer can be verified by calculating the gradient of V:

which is the opposite of the original electric field .

The advantage of using the electric potential V instead of the electric field is that V is a scalarfunction. The total electric potential generated by a charge distribution can be found using thesuperposition principle. This property follows immediately from the definition of V and the fact thatthe electric field satisfies the principle of superposition. Since

it follows that

This equation shows that the total potential at any point is the algebraic sum of the potentials at thatpoint due to all the source charges separately. This ordinary sum of scalars is in general easier toevaluate then a vector sum.

Example: Problem 2.46



Suppose the electric potential is given by the expression

for all r (A and λ are constants). Find the electric field , the charge density , and the totalcharge Q.

The electric field can be immediately obtained from the electric potential:

The charge density can be found using the electric field and the following relation:

This expression shows that

Substituting the expression for the electric field we obtain for the charge density :

The total charge Q can be found by volume integration of :

The integral can be solved easily:

The total charge is thus equal to

The charge distribution can be directly used to obtained from the electric potential



This equation can be rewritten as

and is known as Poisson's equation. In the regions where this equation reduces to Laplace'sequation:

The electric potential generated by a discrete charge distribution can be obtained using the principle ofsuperposition:

where is the electric potential generated by the point charge . A point charge located at theorigin will generate an electric potential equal to

In general, point charge will be located at position and the electric potential generated by thispoint charge at position is equal to

The total electric potential generated by the whole set of point charges is equal to

To calculate the electric potential generated by a continuous charge distribution we have to replace thesummation over point charges with an integration over the continuous charge distribution. For thethree charge distributions we will be using in this course we obtain:

1. line charge λ :

2. surface charge σ :

3. volume charge ρ :

Example: Problem 2.25 Using the general expression for V in terms of ρ find the potential at a distance z above the center ofthe charge distributions of Figure 2.8. In each case, compute . Suppose that we changed theright-hand charge in Figure 2.8a to -q. What is then the potential at P? What field does this suggest?Compare your answer to Problem 2.2b, and explain carefully any discrepancy.



Figure 2.8. Problem 2.35.a) The electric potential at P generated by the two point charges is equal to

The electric field generated by the two point charges can be obtained by taking the gradient of theelectric potential:

If we change the right-hand charge to -q then the total potential at P is equal to zero. However, thisdoes not imply that the electric field at P is equal to zero. In our calculation we have assumed rightfrom the start that x = 0 and y = 0. Obviously, the potential at P will therefore not show an x and ydependence. This however not necessarily indicates that the components of the electric field along thex and y direction are zero. This can be demonstrated by calculating the general expression for theelectric potential of this charge distribution at an arbitrary point (x,y,z):

The various components of the electric field can be obtained by taking the gradient of this expression:



The components of the electric field at P = (0, 0, z) can now be calculated easily:

b) Consider a small segment of the rod, centered at position x and with length dx. The charge on thissegment is equal to . The potential generated by this segment at P is equal to

The total potential generated by the rod at P can be obtained by integrating dV between x = - L and x =L

The z component of the electric field at P can be obtained from the potential V by calculating the zcomponent of the gradient of V. We obtain

c) Consider a ring of radius r and width dr. The charge on this ring is equal to

The electric potential dV at P generated by this ring is equal to

The total electric potential at P can be obtained by integrating dV between r = 0 and r = R:

The z component of the electric field generated by this charge distribution can be obtained by takingthe gradient of V:




Find the electric field a distance z above the center of a circular loop of radius r, which carries auniform line charge λ.

The total charge Q on the ring is equal to

The total electric potential V at P is equal to

The z component of the electric field at P can be obtained by calculating the gradient of V:

This is the same answer we obtained in the beginning of this Chapter by taking the vector sum of thesegments of the ring.

We have seen so far that there are three fundamental quantities of electrostatics:

1. The charge density ρ

2. The electric field

3. The electric potential V

If one of these quantities is known, the others can be calculated:

In general the charge density ρ and the electric field do not have to be continuous. Consider forexample an infinitesimal thin charge sheet with surface charge σ. The relation between the electricfield above and below the sheet can be obtained using Gauss's law. Consider a rectangular box ofheight ε and area A (see Figure 2.9). The electric flux through the surface of the box, in the limit ε →0, is equal to



Figure 2.9. Electric field near a charge sheet.where and are the perpendicular components of the electric field above and below thecharge sheet. Using Gauss's law and the rectangular box shown in Figure 2.9 as integration volume weobtain

This equation shows that the electric field perpendicular to the charge sheet is discontinuous at theboundary. The difference between the perpendicular component of the electric field above and belowthe charge sheet is equal to

The tangential component of the electric field is always continuous at any boundary. This can bedemonstrated by calculating the line integral of around a rectangular loop of length L and height ε(see Figure 2.10). The line integral of , in the limit ε → 0, is equal to

Figure 2.10. Parallel field close to charge sheet.Since the line integral of around any closed loop is zero we conclude that

or

These boundary conditions for can be combined into a single formula:

where is a unit vector perpendicular to the surface and pointing towards the above region. The electric potential is continuous across any boundary. This is a direct results of the definition of Vin terms of the line integral of :

If the path shrinks the line integral will approach zero, independent of whether is continuous ordiscontinuous. Thus




a) Check that the results of examples 4 and 5 of Griffiths are consistent with the boundary conditionsfor .b) Use Gauss's law to find the field inside and outside a long hollow cylindrical tube which carries auniform surface charge σ. Check that your results are consistent with the boundary conditions for .c) Check that the result of example 7 of Griffiths is consistent with the boundary conditions for V.

a) Example 4 (Griffiths): The electric field generated by an infinite plane carrying a uniform surfacecharge σ is directed perpendicular to the sheet and has a magnitude equal to

Therefore,

which is in agreement with the boundary conditions for . Example 5 (Griffiths): The electric field generated by the two charge sheets is directed perpendicularto the sheets and has a magnitude equal to

The change in the strength of the electric field at the left sheet is equal to

The change in the strength of the electric field at the right sheet is equal to

These relations show agreement with the boundary conditions for .

b) Consider a Gaussian surface of length L and radius r. As a result of the symmetry of the system, theelectric field will be directed radially. The electric flux through this Gaussian surface is thereforeequal to the electric flux through its curved surface which is equal to

The charge enclosed by the Gaussian surface is equal to zero when r < R. Therefore

when r < R. When r > R the charge enclosed by the Gaussian surface is equal to



The electric field for r > R, obtained using Gauss' law, is equal to

The magnitude of the electric field just outside the cylinder, directed radially, is equal to

The magnitude of the electric field just inside the cylinder is equal to

Therefore,

which is consistent with the boundary conditions for E.

c) Example 7 (Griffiths): the electric potential just outside the charged spherical shell is equal to

The electric potential just inside the charged spherical shell is equal to

These two equations show that the electric potential is continuous at the boundary.

2.4. Work and Energy in Electrostatics

Consider a point charge q1 located at the origin. A point charge q2 is moved from infinity to a point adistance r2 from the origin. We will assume that the point charge q1 remains fixed at the origin whenpoint charge q2 is moved. The force exerted by q1 on q2 is equal to

where is the electric field generated by q1. In order to move charge q2 we will have to exert a forceopposite to . Therefore, the total work that must be done to move q2 from infinity to r2 is equal to

where is the electric potential generated by q1 at position r2. Using the equation of V for a pointcharge, the work W can be rewritten as



This work W is the work necessary to assemble the system of two point charges and is also called theelectrostatic potential energy of the system. The energy of a system of more than two point chargescan be found in a similar manner using the superposition principle. For example, for a systemconsisting of three point charges (see Figure 2.11) the electrostatic potential energy is equal to

Figure 2.11. System of three point charges.In this equation we have added the electrostatic energies of each pair of point charges. The generalexpression of the electrostatic potential energy for n point charges is

The lower limit of j (= i + 1) insures that each pair of point charges is only counted once. Theelectrostatic potential energy can also be written as

where Vi is the electrostatic potential at the location of qi due to all other point charges. When the charge of the system is not distributed as point charges, but rather as a continuous chargedistribution ρ, then the electrostatic potential energy of the system must be rewritten as

For continuous surface and line charges the electrostatic potential energy is equal to

and

However, we have already seen in this Chapter that ρ, V, and carry the same equivalent information.The charge density ρ, for example, is related to the electric field :



Using this relation we can rewrite the electrostatic potential energy as

where we have used one of the product rules of vector derivatives and the definition of in terms ofV. In deriving this expression we have not made any assumptions about the volume considered. Thisexpression is therefore valid for any volume. If we consider all space, then the contribution of thesurface integral approaches zero since will approach zero faster than 1/r2. Thus the totalelectrostatic potential energy of the system is equal to

Example: Problem 2.45 A sphere of radius R carries a charge density (where k is a constant). Find the energy of theconfiguration. Check your answer by calculating it in at least two different ways.

Method 1: The first method we will use to calculate the electrostatic potential energy of the charged sphere usesthe volume integral of to calculate W. The electric field generated by the charged sphere can beobtained using Gauss's law. We will use a concentric sphere of radius r as the Gaussian surface. Firstconsider the case in which r < R. The charge enclosed by the Gaussian surface can be obtained byvolume integration of the charge distribution:

The electric flux through the Gaussian surface is equal to

Applying Gauss's law we find for the electric field inside the sphere (r < R):

The electric field outside the sphere (r > R) can also be obtained using Gauss's law:

The total electrostatic energy can be obtained from the electric field:



Method 2: An alternative way calculate the electrostatic potential energy is to use the following relation:

The electrostatic potential V can be obtained immediately from the electric field . To evaluate thevolume integral of we only need to know the electrostatic potential V inside the charged sphere:

The electrostatic potential energy of the system is thus equal to

which is equal to the energy calculated using method 1.

2.5. Metallic Conductors

In a metallic conductor one or more electrons per atom are free to move around through the material.Metallic conductors have the following electrostatic properties:

1. The electric field inside the conductor is equal to zero. If there would be an electric field inside the conductor, the free charges would move and produce anelectric field of their own opposite to the initial electric field. Free charges will continue to flow untilthe cancellation of the initial field is complete.

2. The charge density inside a conductor is equal to zero. This property is a direct result of property 1. If the electric field inside a conductor is equal to zero,then the electric flux through any arbitrary closed surface inside the conductor is equal to zero. Thisimmediately implies that the charge density inside the conductor is equal to zero everywhere (Gauss'slaw).

3. Any net charge of a conductor resides on the surface. Since the charge density inside a conductor is equal to zero, any net charge can only reside on thesurface.

4. The electrostatic potential V is constant throughout the conductor. Consider two arbitrary points a and b inside a conductor (see Figure 2.12). The potential differencebetween a and b is equal to



Figure 2.12. Potential inside metallic conductor. Since the electric field inside a conductor is equal to zero, the line integral of between a and b is

equal to zero. Thus

or

5. The electric field is perpendicular to the surface, just outside the conductor. If there would be a tangential component of the electric field at the surface, then the surface chargewould immediately flow around the surface until it cancels this tangential component.

Example: A spherical conducting shella) Suppose we place a point charge q at the center of a neutral spherical conducting shell (see Figure2.13). It will attract negative charge to the inner surface of the conductor. How much induced chargewill accumulate here?b) Find E and V as function of r in the three regions r < a, a < r < b, and r > b.

Figure 2.13. A spherical conducting shell.a) The electric field inside the conducting shell is equal to zero (property 1 of conductors). Therefore,the electric flux through any concentric spherical Gaussian surface of radius r (a<r<b) is equal tozero. However, according to Gauss's law this implies that the charge enclosed by this surface is equalto zero. This can only be achieved if the charge accumulated on the inside of the conducting shell isequal to -q. Since the conducting shell is neutral and any net charge must reside on the surface, thecharge on the outside of the conducting shell must be equal to +q.

b) The electric field generated by this system can be calculated using Gauss's law. In the threedifferent regions the electric field is equal to

for b < r

for a < r < b

for r < a

The electrostatic potential V(r) can be obtained by calculating the line integral of from infinity to apoint a distance r from the origin. Taking the reference point at infinity and setting the value of theelectrostatic potential to zero there we can calculate the electrostatic potential. The line integral of has to be evaluated for each of the three regions separately.For b < r:



For a < r < b:

For r < a:

Figure 2.14. Arbitrarily shaped conductor.In this example we have looked at a symmetric system but the general conclusions are also valid foran arbitrarily shaped conductor. For example, consider the conductor with a cavity shown in Figure2.14. Consider also a Gaussian surface that completely surrounds the cavity (see for example thedashed line in Figure 2.14). Since the electric field inside the conductor is equal to zero, the electricflux through the Gaussian surface is equal to zero. Gauss's law immediately implies that the chargeenclosed by the surface is equal to zero. Therefore, if there is a charge q inside the cavity there will bean induced charge equal to -q on the walls of the cavity. On the other hand, if there is no charge insidethe cavity then there will be no charge on the walls of the cavity. In this case, the electric field insidethe cavity will be equal to zero. This can be demonstrated by assuming that the electric field inside thecavity is not equal to zero. In this case, there must be at least one field line inside the cavity. Sincefield lines originate on a positive charge and terminate on a negative charge, and since there is nocharge inside the cavity, this field line must start and end on the cavity walls (see for example Figure2.15). Now consider a closed loop, which follows the field line inside the cavity and has an arbitraryshape inside the conductor (see Figure 2.15). The line integral of inside the cavity is definitely notequal to zero since the magnitude of is not equal to zero and since the path is defined such that and are parallel. Since the electric field inside the conductor is equal to zero, the path integral of inside the conductor is equal to zero. Therefore, the path integral of along the path indicated inFigure 2.15 is not equal to zero if the magnitude of is not equal to zero inside the cavity. However,the line integral of along any closed path must be equal to zero and consequently the electric fieldinside the cavity must be equal to zero.

Figure 2.15. Field line in cavity.

Example: Problem 2.35 A metal sphere of radius R, carrying charge q, is surrounded by a thick concentric metal shell (innerradius a, outer radius b, see Figure 2.16). The shell carries no net charge.a) Find the surface charge density σ at R, at a, and at b.



b) Find the potential at the center of the sphere, using infinity as reference.c) Now the outer surface is touched to a grounding wire, which lowers its potential to zero (same as atinfinity). How do your answers to a) and b) change?

a) Since the net charge of a conductor resides on its surface, the charge q of the metal sphere willreside its surface. The charge density on this surface will therefore be equal to

As a result of the charge on the metal sphere there will be a charge equal to -q induced on the innersurface of the metal shell. Its surface charge density will therefore be equal to

Since the metal shell is neutral there will be a charge equal to +q on the outside of the shell. Thesurface charge density on the outside of the shell will therefore be equal to

Figure 2.16. Problem 2.35.b) The potential at the center of the metal sphere can be found by calculating the line integral of between infinity and the center. The electric field in the regions outside the sphere and shell can befound using Gauss's law. The electric field inside the shell and sphere is equal to zero. Therefore,

c) When the outside of the shell is grounded, the charge density on the outside will become zero. Thecharge density on the inside of the shell and on the metal sphere will remain the same. The electricpotential at the center of the system will also change as a result of grounding the outer shell. Since theelectric potential of the outer shell is zero, we do not need to consider the line integral of in theregion outside the shell to determine the potential at the center of the sphere. Thus

Consider a conductor with surface charge σ, placed in an external electric field. Because the electricfield inside the conductor is zero, the boundary conditions for the electric field require that the fieldimmediately above the conductor is equal to



Figure 2.17. Patch of surface of conductor.This electric field will exert a force on the surface charge. Consider a small, infinitely thin, patch of the surface with surface area dA (see Figure 2.17). Theelectric field directly above and below the patch is equal to the vector sum of the electric fieldgenerated by the patch, the electric field generated by the rest of the conductor and the externalelectric field. The electric field generated by the patch is equal to

The remaining field, , is continuous across the patch, and consequently the total electric fieldabove and below the patch is equal to

These two equations show that is equal to

In this case the electric field below the surface is equal to zero and the electric field above the surfaceis directly determined by the boundary condition for the electric field at the surface. Thus

Since the patch cannot exert a force on itself, the electric force exerted on it is entirely due to theelectric field . The charge on the patch is equal to . Therefore, the force exerted on the patch isequal to

The force per unit area of the conductor is equal to



This equation can be rewritten in terms of the electric field just outside the conductor as

This force is directed outwards. It is called the radiation pressure.

2.6. Capacitors

Consider two conductors (see Figure 2.18), one with a charge equal to +Q and one with a charge equalto -Q. The potential difference between the two conductors is equal to

Since the electric field is proportional to the charge Q, the potential difference ∆V will also beproportional to Q. The constant of proportionality is called the capacitance C of the system and isdefined as

Figure 2.18. Two conductors.The capacitance C is determined by the size, the shape, and the separation distance of the twoconductors. The unit of capacitance is the farad (F). The capacitance of a system of conductors can ingeneral be calculated by carrying out the following steps:1. Place a charge +Q on one of the conductors. Place a charge of -Q on the other conductor (for a twoconductor system).2. Calculate the electric field in the region between the two conductors.3. Use the electric field calculated in step 2 to calculate the potential difference between the twoconductors.4. Apply the result of part 3 to calculate the capacitance:

We will now discuss two examples in which we follow these steps to calculate the capacitance.

Example: Example 2.11 (Griffiths) Find the capacitance of two concentric shells, with radii a and b.

Place a charge +Q on the inner shell and a charge -Q on the outer shell. The electric field between theshells can be found using Gauss's law and is equal to



The potential difference between the outer shell and the inner shell is equal to

The capacitance of this system is equal to

Figure 2.19. Example 2.11. A system does not have to have two conductors in order to have a capacitance. Consider for examplea single spherical shell of radius R. The capacitance of this system of conductors can be calculated byfollowing the same steps as in Example 12. First of all, put a charge Q on the conductor. Gauss's lawcan be used to calculate the electric field generated by this system with the following result:

Taking infinity as the reference point we can calculate the electrostatic potential on the surface of theshell:

Therefore, the capacitance of the shell is equal to

Let us now consider a parallel-plate capacitor. The work required to charge up the parallel-platecapacitor can be calculated in various ways:

Method 1: Since we are free to chose the reference point and reference value of the potential we willchose it such that the potential of the positively charges plate is and the potential of thenegatively charged plate is . The energy of this charge distribution is then equal to

Method 2: Let us look at the charging process in detail. Initially both conductors are uncharged and . At some intermediate step in the charging process the charge on the positively charged



conductor will be equal to q. The potential difference between the conductors will then be equal to

To increase the charge on the positively charged conductor by dq we have to move this charge dqacross this potential difference . The work required to do this is equal to

Therefore, the total work required to charge up the capacitor from q = 0 to q = Q is equal to

Example: Problem 2.40. Suppose the plates of a parallel-plate capacitor move closer together by an infinitesimal distance ε, asa result of their mutual attraction.a) Use equation (2.45) of Griffiths to express the amount of work done by electrostatic forces, interms of the field E and the area of the plates A.b) Use equation (2.40) of Griffiths to express the energy lost by the field in this process.

a) We will assume that the parallel-plate capacitor is an ideal capacitor with a homogeneous electricfield E between the plates and no electric field outside the plates. The electrostatic force per unitsurface area is equal to

The total force exerted on each plate is therefore equal to

As a result of this force, the plates of the parallel-plate capacitor move closer together by aninfinitesimal distance ε. The work done by the electrostatic forces during this movement is equal to

b) The total energy stored in the electric field is equal to

In an ideal capacitor the electric field is constant between the plates and consequently we can easilyevaluate the volume integral of :

where d is the distance between the plates. If the distance between the plates is reduced, then theenergy stored in the field will also be reduced. A reduction in d of ε will reduce the energy stored byan amount ∆W equal to



which is equal to the work done by the electrostatic forces on the capacitor plates (see part a).



Chapter 3. Special Techniques for Calculating Potentials3.1. Solutions of Laplace's Equation in One-, Two, and Three Dimensions

3.1.1. Laplace's Equation in One Dimension3.1.2. Laplace's Equation in Two Dimensions3.1.3. Laplace's Equation in Three Dimensions3.1.4. Uniqueness Theorems

3.2. Method of Images3.3. Separation of Variables

3.3.1. Separation of variables: Cartesian coordinates3.3.2. Separation of variables: spherical coordinates

3.4. Multipole Expansions3.4.1. The monopole term.3.4.2. The dipole term.

Chapter 3. Special Techniques for Calculating Potentials

Given a stationary charge distribution we can, in principle, calculate the electric field:

where . This integral involves a vector as an integrand and is, in general, difficult to calculate.In most cases it is easier to evaluate first the electrostatic potential V which is defined as

since the integrand of the integral is a scalar. The corresponding electric field can then be obtainedfrom the gradient of V since

The electrostatic potential V can only be evaluated analytically for the simplest charge configurations.In addition, in many electrostatic problems, conductors are involved and the charge distribution ρ isnot known in advance (only the total charge on each conductor is known). A better approach to determine the electrostatic potential is to start with Poisson's equation

Very often we only want to determine the potential in a region where ρ = 0. In this region Poisson'sequation reduces to Laplace's equation

There are an infinite number of functions that satisfy Laplace's equation and the appropriate solutionis selected by specifying the appropriate boundary conditions. This Chapter will concentrate on thevarious techniques that can be used to calculate the solutions of Laplace's equation and on theboundary conditions required to uniquely determine a solution.

3.1. Solutions of Laplace's Equation in One-, Two, and Three Dimensions



3.1.1. Laplace's Equation in One Dimension

In one dimension the electrostatic potential V depends on only one variable x. The electrostaticpotential V(x) is a solution of the one-dimensional Laplace equation

The general solution of this equation is

where s and b are arbitrary constants. These constants are fixed when the value of the potential isspecified at two different positions.

Example Consider a one-dimensional world with two point conductors located at x = 0 m and at x = 10 m. Theconductor at x = 0 m is grounded (V = 0 V) and the conductor at x = 10 m is kept at a constantpotential of 200 V. Determine V.

The boundary conditions for V are

and

The first boundary condition shows that b = 0 V. The second boundary condition shows that s = 20V/m. The electrostatic potential for this system of conductors is thus

The corresponding electric field can be obtained from the gradient of V

The boundary conditions used here, can be used to specify the electrostatic potential between x = 0 mand x = 10 m but not in the region x < 0 m and x > 10 m. If the solution obtained here was the generalsolution for all x, then V would approach infinity when x approaches infinity and V would approachminus infinity when x approaches minus infinity. The boundary conditions therefore provide theinformation necessary to uniquely define a solution to Laplace's equation, but they also define theboundary of the region where this solution is valid (in this example 0 m < x < 10 m).

The following properties are true for any solution of the one-dimensional Laplace equation:

Property 1: V(x) is the average of V(x + R) and V(x - R) for any R as long as x + R and x - R are located in theregion between the boundary points. This property is easy to proof:

This property immediately suggests a powerful analytical method to determine the solution of



Laplace's equation. If the boundary values of V are

and

then property 1 can be used to determine the value of the potential at (a + b)/2:

Next we can determine the value of the potential at x = (3 a + b)/4 and at x = (a + 3 b)/4 :

This process can be repeated and V can be calculated in this manner at any point between x = a and x= b (but not in the region x > b and x < a).

Property 2: The solution of Laplace's equation can not have local maxima or minima. Extreme values must occurat the end points (the boundaries). This is a direct consequence of property 1. Property 2 has an important consequence: a charged particle can not be held in stable equilibrium byelectrostatic forces alone (Earnshaw's Theorem). A particle is in a stable equilibrium if it is locatedat a position where the potential has a minimum value. A small displacement away from theequilibrium position will increase the electrostatic potential of the particle, and a restoring force willtry to move the particle back to its equilibrium position. However, since there can be no local maximaor minima in the electrostatic potential, the particle can not be held in stable equilibrium by justelectrostatic forces.

3.1.2. Laplace's Equation in Two Dimensions

In two dimensions the electrostatic potential depends on two variables x and y. Laplace's equation nowbecomes

This equation does not have a simple analytical solution as the one-dimensional Laplace equationdoes. However, the properties of solutions of the one-dimensional Laplace equation are also valid forsolutions of the two-dimensional Laplace equation:

Property 1: The value of V at a point (x, y) is equal to the average value of V around this point

where the path integral is along a circle of arbitrary radius, centered at (x, y) and with radius R.



Property 2: V has no local maxima or minima; all extremes occur at the boundaries.

3.1.3. Laplace's Equation in Three Dimensions

In three dimensions the electrostatic potential depends on three variables x, y, and z. Laplace'sequation now becomes

This equation does not have a simple analytical solution as the one-dimensional Laplace equationdoes. However, the properties of solutions of the one-dimensional Laplace equation are also valid forsolutions of the three-dimensional Laplace equation:

Property 1: The value of V at a point (x, y, z) is equal to the average value of V around this point

where the surface integral is across the surface of a sphere of arbitrary radius, centered at (x,y,z) andwith radius R.

Figure 3.1. Proof of property 1. To proof this property of V consider the electrostatic potential generated by a point charge q locatedon the z axis, a distance r away from the center of a sphere of radius R (see Figure 3.1). The potentialat P, generated by charge q, is equal to

where d is the distance between P and q. Using the cosine rule we can express d in terms of r, R and θ

The potential at P due to charge q is therefore equal to

The average potential on the surface of the sphere can be obtained by integrating across the surfaceof the sphere. The average potential is equal to



which is equal to the potential due to q at the center of the sphere. Applying the principle ofsuperposition it is easy to show that the average potential generated by a collection of point charges isequal to the net potential they produce at the center of the sphere.

Property 2: The electrostatic potential V has no local maxima or minima; all extremes occur at the boundaries.

Example: Problem 3.3 Find the general solution to Laplace's equation in spherical coordinates, for the case where V dependsonly on r. Then do the same for cylindrical coordinates.

Laplace's equation in spherical coordinates is given by

If V is only a function of r then

and

Therefore, Laplace's equation can be rewritten as

The solution V of this second-order differential equation must satisfy the following first-orderdifferential equation:

This differential equation can be rewritten as

The general solution of this first-order differential equation is

where b is a constant. If V = 0 at infinity then b must be equal to zero, and consequently



Laplace's equation in cylindrical coordinates is

If V is only a function of r then

and

Therefore, Laplace's equation can be rewritten as

The solution V of this second-order differential equation must satisfy the following first-orderdifferential equation:

This differential equation can be rewritten as

The general solution of this first-order differential equation is

where b is a constant. The constants a and b are determined by the boundary conditions.

3.1.4. Uniqueness Theorems

Consider a volume (see Figure 3.2) within which the charge density is equal to zero. Suppose that thevalue of the electrostatic potential is specified at every point on the surface of this volume. The firstuniqueness theorem states that in this case the solution of Laplace's equation is uniquely defined.



Figure 3.2. First Uniqueness Theorem To proof the first uniqueness theorem we will consider what happens when there are two solutions V1and V2 of Laplace's equation in the volume shown in Figure 3.2. Since V1 and V2 are solutions ofLaplace's equation we know that

and

Since both V1 and V2 are solutions, they must have the same value on the boundary. Thus V1 = V2 onthe boundary of the volume. Now consider a third function V3, which is the difference between V1 andV2

The function V3 is also a solution of Laplace's equation. This can be demonstrated easily:

The value of the function V3 is equal to zero on the boundary of the volume since V1 = V2 there.However, property 2 of any solution of Laplace's equation states that it can have no local maxima orminima and that the extreme values of the solution must occur at the boundaries. Since V3 is asolution of Laplace's equation and its value is zero everywhere on the boundary of the volume, themaximum and minimum value of V3 must be equal to zero. Therefore, V3 must be equal to zeroeverywhere. This immediately implies that

everywhere. This proves that there can be no two different functions V1 and V2 that are solutions ofLaplace's equation and satisfy the same boundary conditions. Therefore, the solution of Laplace'sequation is uniquely determined if its value is a specified function on all boundaries of the region.This also indicates that it does not matter how you come by your solution: if (a) it is a solution ofLaplace's equation, and (b) it has the correct value on the boundaries, then it is the right and onlysolution.



Figure 3.3. System with conductors. The first uniqueness theorem can only be applied in those regions that are free of charge andsurrounded by a boundary with a known potential (not necessarily constant). In the laboratory theboundaries are usually conductors connected to batteries to keep them at a fixed potential. In manyother electrostatic problems we do not know the potential at the boundaries of the system. Instead wemight know the total charge on the various conductors that make up the system (note: knowing thetotal charge on a conductor does not imply a knowledge of the charge distribution ρ since it isinfluenced by the presence of the other conductors). In addition to the conductors that make up thesystem, there might be a charge distribution ρ filling the regions between the conductors (see Figure3.3). For this type of system the first uniqueness theorem does not apply. The second uniquenesstheorem states that the electric field is uniquely determined if the total charge on each conductor isgiven and the charge distribution in the regions between the conductors is known. The proof of the second uniqueness theorem is similar to the proof of the first uniqueness theorem.Suppose that there are two fields and that are solutions of Poisson's equation in the regionbetween the conductors. Thus

and

where ρ is the charge density at the point where the electric field is evaluated. The surface integrals of and , evaluated using a surface that is just outside one of the conductors with charge Qi, are equal

to . Thus

The difference between and , , satisfies the following equations:



Consider the surface integral of , integrated over all surfaces (the surface of all conductors and theouter surface). Since the potential on the surface of any conductor is constant, the electrostaticpotential associated with and must also be constant on the surface of each conductor. Therefore,

will also be constant on the surface of each conductor. The surface integral of over thesurface of conductor i can be written as

Since the surface integral of over the surface of conductor i is equal to zero, the surface integral of over all conductor surfaces will also be equal to zero. The surface integral of over the outer

surface will also be equal to zero since on this surface. Thus

The surface integral of can be rewritten using Green's identity as

where the volume integration is over all space between the conductors and the outer surface. Since is always positive, the volume integral of can only be equal to zero if everywhere. Thisimplies immediately that everywhere, and proves the second uniqueness theorem.

3.2. Method of Images

Consider a point charge q held as a distance d above an infinite grounded conducting plane (seeFigure 3.4). The electrostatic potential of this system must satisfy the following two boundaryconditions:

A direct calculation of the electrostatic potential can not be carried out since the charge distribution onthe grounded conductor is unknown. Note: the charge distribution on the surface of a groundedconductor does not need to be zero.



Figure 3.4. Method of images. Consider a second system, consisting of two point charges with charges +q and -q, located at z = d andz = -d, respectively (see Figure 3.5). The electrostatic potential generated by these two charges can becalculated directly at any point in space. At a point P = (x, y, 0) on the xy plane the electrostaticpotential is equal to

Figure 3.5. Charge and image charge.The potential of this system at infinity will approach zero since the potential generated by each chargewill decrease as 1/r with increasing distance r. Therefore, the electrostatic potential generated by thetwo charges shown in Figure 3.5 satisfies the same boundary conditions as the system shown inFigure 3.4. Since the charge distribution in the region z > 0 (bounded by the xy plane boundary andthe boundary at infinity) for the two systems is identical, the corollary of the first uniqueness theoremstates that the electrostatic potential in this region is uniquely defined. Therefore, if we find anyfunction that satisfies the boundary conditions and Poisson's equation, it will be the right answer.Consider a point (x, y, z) with z > 0. The electrostatic potential at this point can be calculated easily forthe charge distribution shown in Figure 3.5. It is equal to

Since this solution satisfies the boundary conditions, it must be the correct solution in the region z > 0for the system shown in Figure 3.4. This technique of using image charges to obtain the electrostaticpotential in some region of space is called the method of images. The electrostatic potential can be used to calculate the charge distribution on the grounded conductor.Since the electric field inside the conductor is equal to zero, the boundary condition for (seeChapter 2) shows that the electric field right outside the conductor is equal to



where σ is the surface charge density and is the unit vector normal to the surface of the conductor.Expressing the electric field in terms of the electrostatic potential V we can rewrite this equation as

Substituting the solution for V in this equation we find

Only in the last step of this calculation have we substituted z = 0. The induced charge distribution isnegative and the charge density is greatest at (x = 0, y = 0, z = 0). The total charge on the conductorcan be calculated by surface integrating of σ:

where . Substituting the expression for σ in the integral we obtain

As a result of the induced surface charge on the conductor, the point charge q will be attractedtowards the conductor. Since the electrostatic potential generated by the charge image-charge systemis the same as the charge-conductor system in the region where z > 0, the associated electric field (andconsequently the force on point charge q) will also be the same. The force exerted on point charge qcan be obtained immediately by calculating the force exerted on the point charge by the image charge.This force is equal to

There is however one important difference between the image-charge system and the real system.This difference is the total electrostatic energy of the system. The electric field in the image-chargesystem is present everywhere, and the magnitude of the electric field at (x, y, z) will be the same as themagnitude of the electric field at (x, y, -z). On the other hand, in the real system the electric field willonly be non zero in the region with z > 0. Since the electrostatic energy of a system is proportional tothe volume integral of the electrostatic energy of the real system will be 1/2 of the electrostaticenergy of the image-charge system (only 1/2 of the total volume has a non-zero electric field in thereal system). The electrostatic energy of the image-charge system is equal to

The electrostatic energy of the real system is therefore equal to

The electrostatic energy of the real system can also be obtained by calculating the work required to bedone to assemble the system. In order to move the charge q to its final position we will have to exert a



force opposite to the force exerted on it by the grounded conductor. The work done to move thecharge from infinity along the z axis to z = d is equal to

which is identical to the result obtained using the electrostatic potential energy of the image-chargesystem.

Example: Example 3.2 + Problem 3.7 A point charge q is situated a distance s from the center of a grounded conducting sphere of radius R(see Figure 3.6).a) Find the potential everywhere.b) Find the induced surface charge on the sphere, as function of q. Integrate this to get the totalinduced charge.c) Calculate the electrostatic energy of this configuration.

Figure 3.6. Example 3.2 + Problem 3.7.a) Consider a system consisting of two charges q and q', located on the z axis at z = s and z = z',respectively. If the potential produced by this system is identical everywhere to the potential producedby the system shown in Figure 3.6 then the position of point charge q' must be chosen such that thepotential on the surface of a sphere of radius R, centered at the origin, is equal to zero (in this case theboundary conditions for the potential generated by both systems are identical). We will start with determining the correct position of point charge q'. The electrostatic potential at P(see Figure 3.7) is equal to


Figure 3.7. Image-charge system.The electrostatic potential at Q is equal to




Combining the two expression for q' we obtain

or


The position of the image charge is equal to

The value of the image charge is equal to

Now consider an arbitrary point P' on the circle. The distance between P' and charge q is d and thedistance between P' and charge q' is equal to d'. Using the cosine rule (see Figure 3.7) we can expressd and d' in terms of R, s, and θ:

The electrostatic potential at P' is equal to

Thus we conclude that the configuration of charge and image charge produces an electrostaticpotential that is zero at any point on a sphere with radius R and centered at the origin. Therefore, thischarge configuration produces an electrostatic potential that satisfies exactly the same boundaryconditions as the potential produced by the charge-sphere system. In the region outside the sphere, theelectrostatic potential is therefore equal to the electrostatic potential produced by the charge andimage charge. Consider an arbitrary point . The distance between this point and charge q is d andthe distance between this point and charge q' is equal to d'. These distances can be expressed in termsof r, s, and θ using the cosine rule:



The electrostatic potential at will therefore be equal to

b) The surface charge density σ on the sphere can be obtained from the boundary conditions of

where we have used the fact that the electric field inside the sphere is zero. This equation can berewritten as

Substituting the general expression for V into this equation we obtain

The total charge on the sphere can be obtained by integrating σ over the surface of the sphere. Theresult is

c) To obtain the electrostatic energy of the system we can determine the work it takes to assemble thesystem by calculating the path integral of the force that we need to exert in charge q in order to moveit from infinity to its final position (z = s). Charge q will feel an attractive force exerted by the inducedcharge on the sphere. The strength of this force is equal to the force on charge q exerted by the imagecharge q'. This force is equal to



The force that we must exert on q to move it from infinity to its current position is opposite to . Thetotal work required to move the charge is therefore equal to

Example: Problem 3.10 Two semi-infinite grounded conducting planes meet at right angles. In the region between them, thereis a point charge q, situated as shown in Figure 3.8. Set up the image configuration, and calculate thepotential in this region. What charges do you need, and where should they be located? What is theforce on q? How much work did it take to bring q in from infinity?

Consider the system of four charges shown in Figure 3.9. The electrostatic potential generated by thischarge distribution is zero at every point on the yz plane and at every point on the xz plane. Therefore,the electrostatic potential generated by this image charge distribution satisfies the same boundaryconditions as the electrostatic potential of the original system. The potential generated by the imagecharge distribution in the region where x > 0 and y > 0 will be identical to the potential of the originalsystem. The potential at a point P = (x, y, z) is equal to


Figure 3.9. Image charges for problem 3.10.The force exerted on q can be obtained by calculating the force exerted on q by the image charges.The total force is equal to the vector sum of the forces exerted by each of the three image charges. Theforce exerted by the image charge located at (-a, b, 0) is directed along the negative x axis and is equal



to

The force exerted by the image charge located at (a, -b, 0) is directed along the negative y axis and isequal to

The force exerted by the image charge located at (-a, -b, 0) is directed along the vector connecting (-a,-b, 0) and (a, b, 0) and is equal to

The total force on charge q is the vector sum of , and :

The electrostatic potential energy of the system can, in principle, be obtained by calculating the pathintegral of between infinity and (a, b, 0). However, this is not trivial since the force is a rathercomplex function of a and b. An easier technique is to calculate the electrostatic potential energy ofthe system with charge and image charges. The potential energy of this system is equal to

However, in the real system the electric field is only non-zero in the region where x > 0 and y > 0.Therefore, the total electrostatic potential energy of the real system is only 1/4 of the total electrostaticpotential energy of the image charge system. Thus

3.3. Separation of Variables

3.3.1. Separation of variables: Cartesian coordinates

A powerful technique very frequently used to solve partial differential equations is separation ofvariables. In this section we will demonstrate the power of this technique by discussing severalexamples.

Example: Example 3.3 (Griffiths) Two infinite, grounded, metal plates lie parallel to the xz plane, one at y = 0, the other at y = π (seeFigure 3.10). The left end, at x = 0, is closed off with an infinite strip insulated from the two platesand maintained at a specified potential . Find the potential inside this "slot".



Figure 3.10. Example 3.3 (Griffiths). The electrostatic potential in the slot must satisfy the three-dimensional Laplace equation. However,since V does not have a z dependence, the three-dimensional Laplace equation reduces to the two-dimensional Laplace equation:

The boundary conditions for the solution of Laplace's equation are:

1. V(x, y = 0) = 0 (grounded bottom plate).

2. V(x, y = π) = 0 (grounded top plate).

3. V(x = 0, y) = V0(y) (plate at x = 0).

4. V → 0 when x → ∞.

These four boundary conditions specify the value of the potential on all boundaries surrounding theslot and are therefore sufficient to uniquely determine the solution of Laplace's equation inside theslot. Therefore, if we find one solution of Laplace's equation satisfying these boundary conditions thanit must be the correct one. Consider solutions of the following form:

If this is a solution of the two-dimensional Laplace equation than we must require that


The first term of the left-hand side of this equation depends only on x while the second term dependsonly on y. Therefore, if this equation must hold for all x and y in the slot we must require that

and



The differential equation for X can be rewritten as

If C1 is a negative number than this equation can be rewritten as

where k2 = -C1 . The most general solution of this equation is

However, this function is an oscillatory function and does not satisfy boundary condition # 4, whichrequires that V approaches zero when x approaches infinity. We therefore conclude that C1 can not bea negative number. If C1 is a positive number then the differential equation for X can be written as

The most general solution of this equation is

This solution will approach zero when x approaches infinity if A = 0. Thus

The solution for Y can be obtained by solving the following differential equation:

The most general solution of this equation is

Therefore, the general solution for the electrostatic potential V(x,y) is equal to

where we have absorbed the constant B into the constants C and D. The constants C and D must bechosen such that the remaining three boundary conditions (1, 2, and 3) are satisfied. The firstboundary condition requires that V(x, y = 0) = 0:

which requires that C = 0. The second boundary condition requires that V(x, y = π) = 0:



which requires that . This condition limits the possible values of k to positive integers:

Note: negative values of k are not allowed since exp(-kx) approaches zero at infinity only if k > 0. Tosatisfy boundary condition # 3 we must require that

This last expression suggests that the only time at which we can find a solution of Laplace's equationthat satisfies all four boundary conditions has the form is when happens to havethe form . However, since k can take on an infinite number of values, there will be an infinitenumber of solutions of Laplace's equation satisfying boundary conditions # 1, # 2 and # 4. The mostgeneral form of the solution of Laplace's equation will be a linear superposition of all possiblesolutions. Thus

Boundary condition # 3 can now be written as

Multiplying both sides by sin(ny) and integrating each side between y = 0 and y = π we obtain

The integral on the left-hand side of this equation is equal to zero for all values of k except k = n. Thus

The coefficients Dk can thus be calculated easily:

The coefficients Dk are called the Fourier coefficients of . The solution of Laplace's equation inthe slot is therefore equal to

where

Now consider the special case in which . In this case the coefficients Dk are equal to



The solution of Laplace's equation is thus equal to

Example: Problem 3.12 Find the potential in the infinite slot of Example 3.3 (Griffiths) if the boundary at x = 0 consists to twometal stripes: one, from y = 0 to y = π/2, is held at constant potential , and the other, from y = π/2 toy = π is at potential .

The boundary condition at x = 0 is

The Fourier coefficients of the function are equal to

The values for the first four C coefficients are

It is easy to see that Ck + 4 = Ck and therefore we conclude that

The Fourier coefficients Ck are thus equal to

The electrostatic potential is thus equal to



Example: Problem 3.13 For the infinite slot (Example 3.3 Griffiths) determine the charge density on the strip at x=0,assuming it is a conductor at constant potential .

The electrostatic potential in the slot is equal to

The charge density at the plate at x = 0 can be obtained using the boundary condition for the electricfield at a boundary:

where is directed along the positive x axis. Since this boundary condition can be rewrittenas

Differentiating V(x,y) with respect to x we obtain

At the x = 0 boundary we obtain

The charge density σ on the x = 0 strip is therefore equal to

Example: Double infinite slots The slot of example 3.3 in Griffiths and its mirror image at negative x are separated by an insulatingstrip at x = 0. If the charge density σ(y) on the dividing strip is given, determine the potential in theslot.

The boundary condition at x = 0 requires that

where is directed along the positive x axis. Here we have used the symmetry of the configurationwhich requires that the electric field in the region x < 0 is the mirror image of the field in the region x> 0. Since this boundary condition can be rewritten as

We will first determine the potential in the x > 0 region. Following the same procedure as in Example



3 we obtain for the electrostatic potential

where the constants Dk must be chosen such that the boundary condition at x = 0 is satisfied. Thisrequires that

Thus

The constants Dk can be determined by multiplying both sides of this equation with andintegrating both sides with respect to y between y = 0 and y = π. The result is

The constants Ck are thus equal to

The electrostatic potential is thus equal to

3.3.2. Separation of variables: spherical coordinates

Consider a spherical symmetric system. If we want to solve Laplace's equation it is natural to usespherical coordinates. Assuming that the system has azimuthal symmetry ( ) Laplace'sequation reads

Multiplying both sides by r2 we obtain

Consider the possibility that the general solution of this equation is the product of a function ,which depends only on the distance r, and a function , which depends only on the angle θ:

Substituting this "solution" into Laplace's equation we obtain



Dividing each term of this equation by we obtain

The first term in this expression depends only on the distance r while the second term depends only onthe angle θ. This equation can only be true for all r and θ if

and

Consider a solution for R of the following form:

where A and k are arbitrary constants. Substituting this expression in the differential equation for R(r)we obtain

Therefore, the constant k must satisfy the following relation:

This equation gives us the following expression for k

The general solution for is thus given by

where A and B are arbitrary constants. The angle dependent part of the solution of Laplace's equation must satisfy the following equation

The solutions of this equation are known as the Legendre polynomial . The Legendrepolynomials have the following properties:



1. if m is even:

2. if m is odd:

3. for all m

4. or

Combining the solutions for and we obtain the most general solution of Laplace's equation ina spherical symmetric system with azimuthal symmetry:

Example: Problem 3.18 The potential at the surface of a sphere is given by

where k is some constant. Find the potential inside and outside the sphere, as well as the surfacecharge density on the sphere. (Assume that there is no charge inside or outside of the sphere.)

The most general solution of Laplace's equation in spherical coordinates is

First consider the region inside the sphere (r < R). In this region since otherwise wouldblow up at r = 0. Thus

The potential at r = R is therefore equal to

Using trigonometric relations we can rewrite as

Substituting this expression in the equation for we obtain

This equation immediately shows that unless . If then



The electrostatic potential inside the sphere is therefore equal to

Now consider the region outsider the sphere (r > R). In this region since otherwise wouldblow up at infinity. The solution of Laplace's equation in this region is therefore equal to

The potential at r = R is therefore equal to

The equation immediately shows that except when . If then

The electrostatic potential outside the sphere is thus equal to

The charge density on the sphere can be obtained using the boundary conditions for the electric fieldat a boundary:

Since this boundary condition can be rewritten as

The first term on the left-hand side of this equation can be calculated using the electrostatic potentialjust obtained:

In the same manner we obtain



Therefore,

The charge density on the sphere is thus equal to

Example: Problem 3.19 Suppose the potential at the surface of a sphere is specified, and there is no charge inside oroutside the sphere. Show that the charge density on the sphere is given by

where

Most of the solution of this problem is very similar to the solution of Problem 3.18. First consider theelectrostatic potential inside the sphere. The electrostatic potential in this region is given by

and the boundary condition is

The coefficients can be determined by multiplying both sides of this equation by andintegrating with respect to θ between θ = 0 and θ = π:

Thus

In the region outside the sphere the electrostatic potential is given by

and the boundary condition is

The coefficients are given by



The charge density on the surface of the sphere is equal to

Differentiating with respect to r in the region r > R we obtain

Differentiating with respect to r in the region r < R we obtain

The charge density is therefore equal to

Substituting the expressions for and into this equation we obtain

where

Example: Problem 3.23 Solve Laplace's equation by separation of variables in cylindrical coordinates, assuming there is nodependence on z (cylindrical symmetry). Make sure that you find all solutions to the radial equation.Does your result accommodate the case of an infinite line charge?

For a system with cylindrical symmetry the electrostatic potential does not depend on z. Thisimmediately implies that . Under this assumption Laplace's equation reads

Consider as a possible solution of V:



Substituting this solution into Laplace's equation we obtain

Multiplying each term in this equation by r2 and dividing by we obtain

The first term in this equation depends only on r while the second term in this equation depends onlyon φ. This equation can therefore be only valid for every r and every φ if each term is equal to aconstant. Thus we require that

and

First consider the case in which . The differential equation for can be rewritten as

The most general solution of this differential solution is

However, in cylindrical coordinates we require that any solution for a given φ is equal to the solutionfor φ + 2π. Obviously this condition is not satisfied for this solution, and we conclude that .The differential equation for can be rewritten as

The most general solution of this differential solution is

The condition that requires that m is an integer. Now consider the radial function .We will first consider the case in which . Consider the following solution for :

Substituting this solution into the previous differential equation we obtain



Therefore, the constant k can take on the following two values:

The most general solution for under the assumption that is therefore

Now consider the solutions for when . In this case we require that

or


If then the solution of this differential equation is

If then the solution of this differential equation is

Combining the solutions obtained for with the solutions obtained for we conclude that themost general solution for is given by

Therefore, the most general solution of Laplace's equation for a system with cylindrical symmetry is

Example: Problem 3.25 A charge density

is glued over the surface of an infinite cylinder of radius R. Find the potential inside and outside the



cylinder.

The electrostatic potential can be obtained using the general solution of Laplace's equation for asystem with cylindrical symmetry obtained in Problem 3.24. In the region inside the cylinder thecoefficient must be equal to zero since otherwise would blow up at . For the same reason

. Thus

In the region outside the cylinder the coefficients must be equal to zero since otherwise would blow up at infinity. For the same reason . Thus

Since must approach 0 when r approaches infinity, we must also require that is equal to 0.The charge density on the surface of the cylinder is equal to

Differentiating in the region r > R and setting r = R we obtain

Differentiating in the region r < R and setting r = R we obtain

The charge density on the surface of the cylinder is therefore equal to

Since the charge density is proportional to we can conclude immediately that forall and that for all except . Therefore

This requires that

A second relation between and can be obtained using the condition that the electrostaticpotential is continuous at any boundary. This requires that



Thus

and

We now have two equations with two unknown, and , which can be solved with the followingresult:

and

The electrostatic potential inside the cylinder is thus equal to

The electrostatic potential outside the cylinder is thus equal to

Example: Problem 3.37 A conducting sphere of radius a, at potential , is surrounded by a thin concentric spherical shell ofradius b, over which someone has glued a surface charge

where is a constant.a) Find the electrostatic potential in each region: i) r > b ii) a < r < bb) Find the induced surface charge on the conductor.c) What is the total charge of the system? Check that your answer is consistent with the behavior of Vat large r.

a) The system has spherical symmetry and we can therefore use the most general solution of Laplace'sequation in spherical coordinates:

In the region inside the sphere since otherwise would blow up at r = 0. Therefore



The boundary condition for is that it is equal to at r = a:

This immediately shows that for all except :

The electrostatic potential inside the sphere is thus given by

which should not come as a surprise. In the region outside the shell since otherwise would blow up at infinity. Thus

In the region between the sphere and the shell the most general solution for is given by

The boundary condition for at r = a is

This equation can only be satisfied if

The requirement that the electrostatic potential is continuous at r = b requires that

or

This condition can be rewritten as



The other boundary condition for the electrostatic potential at r = b is that it must produce the chargedistribution given in the problem. This requires that

This condition is satisfied if

Substituting the relation between the various coefficients obtained by applying the continuitycondition we obtain

These equations show that

Using these values for we can show that

The boundary condition for V at r = a shows that

These values for immediately fix the values for :



The potential in the region outside the shell is therefore equal to

The potential in the region between the sphere and the shell is equal to

b) The charge density on the surface of the sphere can be found by calculating the slope of theelectrostatic potential at this surface:

c) The total charge on the sphere is equal to

The total charge on the shell is equal to zero. Therefore the total charge of the system is equal to

The electrostatic potential at large distances will therefore be approximately equal to

This is equal to limit of the exact electrostatic potential when .

3.4. Multipole Expansions

Consider a given charge distribution ρ. The potential at a point P (see Figure 3.11) is equal to

where d is the distance between P and a infinitesimal segment of the charge distribution. Figure 3.11shows that d can be written as a function of r, r' and θ:



Figure 3.11. Charge distribution ρ. This equation can be rewritten as

At large distances from the charge distribution and consequently . Using the followingexpansion for :

we can rewrite 1/d as

Using this expansion of 1/d we can rewrite the electrostatic potential at P as

This expression is valid for all r (not only ). However, if then the potential at P will bedominated by the first non-zero term in this expansion. This expansion is known as the multipoleexpansion. In the limit of only the first terms in the expansion need to be considered:

The first term in this expression, proportional to 1/r, is called the monopole term. The second term inthis expression, proportional to 1/r2, is called the dipole term. The third term in this expression,proportional to 1/r3, is called the quadrupole term.

3.4.1. The monopole term.

If the total charge of the system is non zero then the electrostatic potential at large distances isdominated by the monopole term:

where Q is the total charge of the charge distribution. The electric field associated with the monopole term can be obtained by calculating the gradient of

:



3.4.2. The dipole term.

If the total charge of the charge distribution is equal to zero (Q = 0) then the monopole term in themultipole expansion will be equal to zero. In this case the dipole term will dominate the electrostaticpotential at large distances

Since θ is the angle between and we can rewrite as

The electrostatic potential at P can therefore be rewritten as

In this expression is the dipole moment of the charge distribution which is defined as

The electric field associated with the dipole term can be obtained by calculating the gradient of :

Example Consider a system of two point charges shown in Figure 3.12. The total charge of this system is zero,and therefore the monopole term is equal to zero. The dipole moment of this system is equal to

where is the vector pointing from -q to +q.

The dipole moment of a charge distribution depends on the origin of the coordinate system chosen.Consider a coordinate system S and a charge distribution ρ. The dipole moment of this chargedistribution is equal to

A second coordinate system S' is displaced by with respect to S:



The dipole moment of the charge distribution in S' is equal to

This equation shows that if the total charge of the system is zero (Q = 0) then the dipole moment ofthe charge distribution is independent of the choice of the origin of the coordinate system.

Figure 3.12. Electric dipole moment.

Example: Problem 3.40 A thin insulating rod, running from z = -a to z = +a, carries the following line charges:a)

b)

c)

In each case, find the leading term in the multipole expansion of the potential.

a) The total charge on the rod is equal to

Since , the monopole term will dominate the electrostatic potential at large distances. Thus

b) The total charge on the rod is equal to zero. Therefore, the electrostatic potential at large distanceswill be dominated by the dipole term (if non-zero). The dipole moment of the rod is equal to

Since the dipole moment of the rod is not equal to zero, the dipole term will dominate the electrostaticpotential at large distances. Therefore

c) For this charge distribution the total charge is equal to zero and the dipole moment is equal to zero.The electrostatic potential of this charge distribution is dominated by the quadrupole term.



The electrostatic potential at large distance from the rod will be equal to

Example: Problem 3.27 Four particles (one of charge q, one of charge 3q, and two of charge -2q) are placed as shown inFigure 3.12, each a distance d from the origin. Find a simple approximate formula for the electrostaticpotential, valid at a point P far from the origin.

The total charge of the system is equal to zero and therefore the monopole term in the multipoleexpansion is equal to zero. The dipole moment of this charge distribution is equal to

The Cartesian coordinates of P are

The scalar product between and is therefore

The electrostatic potential at P is therefore equal to


Example: Problem 3.38 A charge Q is distributed uniformly along the z axis from z = -a to z =a. Show that the electricpotential at a point is given by

for r > a.

The charge density along this segment of the z axis is equal to



Therefore, the nth moment of the charge distribution is equal to

This equation immediately shows that

The electrostatic potential at P is therefore equal to



Chapter 4. Electrostatic Fields in Matter4.1. Polarization4.2. The Field of a Polarized Object4.3. The Electric Displacement4.4. Linear Dielectrics4.5. Energy in dielectric systems4.6. Forces on dielectrics

Chapter 4. Electrostatic Fields in Matter

4.1. Polarization

A neutral atom, placed in an external electric field, will experience no net force. However, eventhough the atom as a whole is neutral, the positive charge is concentrated in the nucleus (radius = 10-14 m) while the negative charge forms an electron cloud (radius = 10-10 m) surrounding the nucleus(see Figure 4.1). The nucleus of the atom will experience a force pointing in the same direction as theexternal electric field (to the right in Figure 4.1) and of magnitude qEext. The negatively chargedelectron cloud will experience a force of the same magnitude, but pointed in a direction opposite tothe direction of the electric field. As a result of the external force, the nucleus will move in thedirection of the electric field until the external force on it is canceled by the force exerted on thenucleus by the electron cloud.

Figure 4.1. Atom in external electric field. Consider an electron cloud with a constant volume charge density ρ and a radius a. If the total chargeof the electron cloud is -q then the corresponding charge density ρ is equal to

The electric field inside the uniformly charged cloud is equal to

where r is the distance from the center of the cloud. Suppose that as a result of the external electricfield the nucleus moves by a distance d with respect to the center of the electron cloud. The electricforce exerted on the nucleus by the electron cloud is equal to

The equilibrium position of the nucleus is that position where the external force is canceled by the



force exerted on it by the electron cloud:

This expression can be rewritten as

The equilibrium distance d is thus equal to

The induced dipole moment of the atom is defined as

Therefore, the magnitude of the induced dipole moment is proportional to the magnitude of theexternal electric field, and its direction is equal to the direction of the external electric field. Theconstant of proportionality is called the atomic polarizability α and is defined as

Although this model of the atom is extremely crude, it produces results that are in reasonableagreement with direct measurements of the atomic polarizability.

Example: Problem 4.2 According to quantum mechanics, the electron cloud for a hydrogen atom in its ground state has acharge density equal to

where q is the charge of the electron and a is the Bohr radius. Find the atomic polarizability of such anatom.

As a result of an external electric field the nucleus of the atom will be displaced by a distance d withrespect to the center of the electron cloud. The force exerted on the nucleus by the electron cloud isequal to

where is the electric field generated by the electron cloud. The electric field generated by theelectron cloud can be calculated using Gauss's law:

The displacement of the nucleus will be very small compared to the size of the electron cloud (d«a).Therefore, we can expand exp(-2d/a) in terms of d/a:



The nucleus will be in an equilibrium position when the electric force exerted on it by the externalfield is equal to the electric force exerted on it by the electron cloud. This occurs when the electricfield at the position of the nucleus, generated by he electron cloud, is equal in magnitude to theexternally applied electric field, but pointing in the opposite direction. The dipole moment of thedipole can therefore be expressed in terms of the external field:

The electric polarizability of the material is therefore equal to

which is close to the result obtained using the classical model of the atom.

Figure 4.2. Torque on dipole in an electric field. Besides polarizing the atoms of a material, the external electric field can align its molecules. Somemolecules, like water, have a permanent dipole moment. Normally, the dipole moments of the watermolecules will be directed randomly, and the average dipole moment is zero. When the water isexposed to an external electric field, a torque is exerted on the water molecule, and it will try to alignits dipole moment with the external electric field. This is schematically illustrated in Figure 4.2.Figure 4.2 shows a dipole placed in an electric field, directed along the x axis. The net force onthe dipole is zero since the net charge is equal to zero. The torque on the dipole with respect to itscenter is equal to

As a result of this torque, the dipole will try to align itself with the electric field. When the dipolemoment is pointing in the same direction as the electric field the torque on the dipole will be equal tozero.

Example: Problem 4.6 A dipole with dipole moment is situated a distance d above an infinite grounded conducting plane(see Figure 4.3). The dipole makes and angle θ with the perpendicular to the plane. Find the torque on . If the dipole is free to rotate, in what direction will it come to rest?



Figure 4.3. Problem 4.6

Figure 4.4. Method of images (Problem 4.6). This problem can be solved using the method of images (see Figure 4.4a). Note that the method ofimages, when applied to a dipole, does not produce an exact mirror image of the dipole. After definingthe image dipole, we chose a new coordinate system such that the image dipole is located at theorigin, and pointing upwards (along the positive z axis, see Figure 4.4b). The electric field at theposition of the real dipole due to the image dipole is equal to

The torque on the real dipole is equal to

The torque on the dipole is positive when 0 < θ < π/2 and as a consequence the dipole will rotatecounter clockwise towards the stable orientation of θ = 0. The torque on the dipole is negative whenπ/2 < θ < π and as a consequence the dipole will rotate clockwise towards the stable orientation of θ =π.

Example: Problem 4.7 Show that the energy of a dipole in an electric field is given by



Figure 4.5. Problem 4.7. Consider the dipole located at the origin of a coordinate system. The z axis of the coordinate systemcoincides with the direction of the electric field and the angle between the dipole and the z axis isequal to θ (see Figure 4.5). The energy of the system can be determined by calculating the work to bedone to move the dipole from infinity to its present location. Assume the dipole is initially orientedparallel to the x axis and is first moved from infinity along the x axis to r = 0. The force exerted on thedipole by the electric field is directed perpendicular to the displacement and therefore the work doneby this force is equal to zero. The dipole is then rotated to its final position (from π/2 to θ). The torqueexerted by the electric field on the dipole is equal to

In order to rotate the dipole I must supply a torque opposite to :

Therefore, the work done by me is equal to

The potential energy of the dipole is therefore equal to

and reaches a minimum when is parallel to (the dipole is aligned with the electric field).

4.2. The Field of a Polarized Object

Consider a piece of polarized material with a dipole moment per unit volume equal to . Theelectrostatic potential generated by this material is equal to

where . Using the following relation (one of the product rules of the vector operator)

we can rewrite the expression for the electric potential as



where

and

Here the unit vector is perpendicular to the integration surface (and pointing outwards). Theequation for the electrostatic potential shows that the potential (and therefore also the electric field)generated by a polarized object is equal to the potential generated by an object with surface chargedensity and volume charge density .

Example: Problem 4.10 A sphere of radius R carries a polarization

where k is a constant and is the vector from the center.a) Calculate the bound charges and .b) Find the field inside and outside the sphere.

a) The unit vector on the surface of the sphere is equal to the radial unit vector. The bound surfacecharge is equal to

The bound volume charge is equal to

b) First consider the region outside the sphere. The electric field in this region due to the surfacecharge is equal to

The electric field in this region due to the volume charge is equal to

Therefore, the total electric field outside the sphere is equal to zero. Now consider the region inside the sphere. The electric field in this region due to the surface charge is



equal to zero. The electric field due to the volume charge is equal to

The bound charges introduced in this Section are not just mathematical artifacts, but are real charges,bound to the individual dipoles of the material. Consider for example the three dipoles shown inFigure 4.6a. When they are aligned (lengthwise) the center charges cancel, and the system looks like asingle dipole with dipole moment 3dq (see Figure 4.6b).

Figure 4.6. Aligned dipoles. In a uniformly polarized material of thickness s and polarization all dipoles are perfectly aligned(see Figure 4.7). The net result of the alignment of the individual dipoles is a positive surface chargeon one side of the material and negative surface charge on the opposite side. Consider a cylinder withsurface area A whose axis is aligned with the direction of polarization of the polarized material. Thetotal dipole moment of this cylinder is equal to

Figure. 4.7. Uniform polarization.Since the only charge of the system resides on the end caps of the cylinder (volume charges cancel ina uniformly polarized material: see Figure 4.6), the net charge there must be equal to

The charge density on the surface is therefore equal to

If the surface of the material is not perpendicular to the direction of polarization then surface chargedensity will be less than P (surface charge distributed over a larger area) and equal to

where is the unit vector perpendicular to the surface of the material, pointing outwards. For thematerial shown in Figure 4.7 this equation immediately shows that a positive surface charge resides



on the right surface ( parallel to ) and a negative surface charge resides on the left surface ( antiparallel to ). Since these charges reside on the surface and are bound to the dipoles they are calledthe bound surface charge or . If the material is uniformly polarized then the volume charge density is equal to zero (see Figure 4.6).However, if the polarization is not uniform then there will be a net volume charge inside the material.Consider a system of three aligned dipoles (see Figure 4.8). If the polarization is not uniform then thestrength of the individual dipoles will vary. Assuming that the physical size (length) of the dipolesshown in Figure 4.8 is the same, then the varying dipole strength is a result of variations in the chargeon the ends of the dipoles. Since the net charge on the polarized material must be equal to zero, thesum of the volume charges and surface charges must be equal to zero. Thus

This equation can be rewritten by substituting the expression for the surface charge density andapplying the fundamental theorem of divergences:

Since this relation holds for any volume we can conclude that

Figure 4.8. Non-uniform polarization.

Example: Problem 4.31 A dielectric cube of side s, centered at the origin, carries a "frozen-in" polarization , where k is aconstant. Find all the bound charges, and check that they add up to zero.

The bound volume charge density is equal to

Since the bound volume charge density is constant, the total bound volume charge in the cube is equalto product of the charge density and the volume:

The surface charge density is equal to

The scalar product between and can be evaluate easily (see Figure 4.9) and is equal to



Figure 4.9. Problem 4.31.Therefore the surface charge density is equal to

The surface charge density is constant across the surface of the cube and consequently the totalsurface charge on the cube is equal to the product of the surface charge density and the total surfacearea of the cube:

The total bound charge on the cube is equal to

4.3. The Electric Displacement

The electric field generated by a polarized material is equal to the electric field produced by its boundcharges. If free charges are also present then the total electric field produced by this system is equal tothe vector sum of the electric fields produced by the bound charges and by the free charges. Gauss'slaw can also be used for this type of systems to calculate the electric field as long as we include bothfree and bound charges:

where is the polarization of the material. This expression can be rewritten as

The expression in parenthesis is called the electric displacement which is defined as

In terms of , Gauss's law can be rewritten as



and

These two versions of Gauss's law are particularly useful since they make reference only to freecharges, which are the charges we can control. Although it seems that the displacement has properties similar to the electric field there are somevery significant differences. For example, the curl of is equal to

and is in general not equal to zero. Since the curl of is not necessarily equal to zero, there is ingeneral no potential that generates . The Helmholtz theorem tell us that if we know the curl and the divergence of a vector function thenthis is sufficient information to uniquely define the vector function . Therefore, the electric field isuniquely defined by Gauss's law since we know that he curl of is zero, everywhere. Thedisplacement current on the other hand is not uniquely determined by the free charge distribution,but requires additional information (like for example ).

Example: Problem 4.16 Suppose the field inside a large piece of dielectric is , so that the electric displacement is equal to

.a) Now, a small spherical cavity is hollowed out of the material. Find the field at the center of thecavity in terms of and . Also find the displacement at the center of the cavity in terms of and .b) Do the same for a long needle-shaped cavity running parallel to .c) Do the same for a thin wafer-shaped cavity perpendicular to .

a) Consider a large piece of dielectric material with polarization and a small sphere withpolarization superimposed on it. The field generated by this system is equal to the field generatedby the dielectric material with a small spherical cavity hollowed out (principle of superposition). Theelectric field inside a sphere with polarization is uniform and equal to

(see Example 2 of Griffiths). The field at the center of the cavity is therefore equal to

The corresponding electric displacement at the center of the cavity is equal to

b) Consider a large piece of dielectric material with polarization and a small long needle-shapedpiece with polarization superimposed on it. The field generated by this system is equal to the fieldgenerated by the dielectric material with a small long needle-shaped cavity hollowed out (principle ofsuperposition). The electric field of a polarized needle of length s is equal to that of two point charges(+q and -q) located a distance s apart. The charge on top of the needle will be negative, while thecharge on the bottom of the needle will be positive. The charge density on the end caps of the needle



is equal to P. Therefore,

where A is the surface area of the end caps of the needle. The electric field generated by the needle atits center is equal to

In the needle limit and therefore . Thus at the center of the needle cavity

The electric displacement at this point is equal to

c) Consider a large piece of dielectric material with polarization and a thin wafer-shaped piece ofdielectric material with polarization superimposed on it. The field generated by this system isequal to the field generated by the dielectric material with a thin wafer-shaped cavity hollowed out(principle of superposition). The electric field inside the wafer will be that of two parallel plates withcharge densities equal to -σ on the top and +σ on the bottom. For a thin wafer-shaped cavity theelectric field between the plates will be equal to the field of a parallel-plate capacitor with infinitelylarge plates. Thus

The net electric field in the center of the cavity is therefore equal to

The electric displacement at the center of the cavity is equal to

4.4. Linear Dielectrics

Most dielectric materials become polarized when they are placed in an external electric field. In manymaterials the polarization is proportional to the electric field:

where is the total electric field (external + internal). The constant of proportionality, , is calledthe electric susceptibility. Materials in which the induced polarization is proportional to the electricfield are called linear dielectrics. The electric displacement in a linear dielectric is also proportional to the total electric field:



where ε is called the permittivity of the material which is equal to

The constant is called the dielectric constant K of the material. Consider a volume V entirely filled with linear dielectric material with dielectric constant K. Thepolarization of this material is equal to

and is therefore proportional to everywhere. Therefore

and consequently

The electric displacement therefore satisfies the following two conditions:

and

The electric field generated by the free charges when the dielectric is not present satisfies thefollowing two equations:

and

Comparing the two sets of differential equations for and we conclude that

The displacement can also be expressed in terms of the total field inside the dielectric:

These two equations show that

The presence of the dielectric material therefore reduces the electric field by a factor K.

Example: Problem 4.18 The space between the plates of a parallel-plate capacitor (see Figure 4.10) is filled with two slabs oflinear dielectric material. Each slab has thickness s, so that the total distance between the plates is 2s.



Slab 1 has a dielectric constant of 2, and slab 2 has a dielectric constant of 1.5. The free chargedensity on the top plate is σ and on the bottom plate is -σ.a) Find the electric displacement in each slab.b) Find the electric field in each slab.c) Find the polarization in each slab.d) Find the potential difference between the plates.e) Find the location and amount of all bound charge.f) Now that you know all charges (free and bound), recalculate the field in each slab, and comparewith your answers to b).

Figure 4.10. Problem 4.18.a) The electric displacement in slab 1 can be calculated using "Gauss's law". Consider a cylinderwith cross sectional area A and axis parallel to the z axis, being used as a Gaussian surface. The top ofhe cylinder is located inside the top metal plate (where the electric displacement is zero) and thebottom of the cylinder is located inside the dielectric of slab 1. The electric displacement is directedparallel to the z axis and pointed downwards. Therefore, the displacement flux through this surface isequal to

The free charge enclosed by this surface is equal to

Combining these two equations we obtain

In vector notation

In the same way we obtained for slab 2

b) The electric field in slab 1 is equal to

The electric field in slab 2 is equal to



c) Once and are known, the polarization can be calculated:

Therefore, the polarization of slab 1 is equal to

The polarization of slab 2 is equal to

d) The potential difference between the top plate and the bottom plate is equal to

e) There are no bound volume charges (constant polarization). The bound surface charge density onthe surface of a dielectric with polarization is equal to . For slab 1 the polarization is equal to

The surface charge density on the top of slab 1 is equal to

The surface charge density on the bottom of slab 1 is equal to

For slab 2 the polarization is equal to

The surface charge density on the top of slab 2 is equal to

The surface charge density on the bottom of slab 2 is equal to

f) The total charge above slab 1 is equal to σ - σ/2 = σ/2. This charge will produce an electric field inslab 1 equal to



The total charge below slab 1 is equal to σ/2 - σ/3 + σ/3 - σ = - σ/2. This charge will produce anelectric field in slab 1 equal to

The total electric field in slab 1 is the vector sum of these two fields and is equal to

The total charge above slab 2 is equal to σ - σ/2 + σ/2 - σ/3 = 2σ/3. This charge will produce anelectric field in slab 2 equal to

The total charge below slab 1 is equal to σ/3 - σ = - 2σ/3. This charge will produce an electric field inslab 1 equal to

The total electric field in slab 1 is the vector sum of these two fields and is equal to

These answers are in agreement with the results obtained in part b).

Example: Problem 4.20 A sphere of linear dielectric material has embedded in it a uniform free charge density ρ. Find thepotential at the center of the sphere, if its radius is R and its dielectric constant is K.

The system has spherical symmetry and therefore the electric displacement is easy to calculatesince and . The calculation of is very similar to the calculation of usingGauss's law:

The corresponding electric field is equal to

Here we have used the fact that K = 1 in the region outside the sphere (r > R). The potential at thecenter of the sphere can be calculated using this electric field:



The examples of calculations involving polarized material that have been discussed so far are eitherartificial, in the sense that the polarization is specified at the start, or highly symmetric, so that theelectric displacement can be obtained directly from the free charge. In the next couple of examples wewill encounter systems where these special conditions do not apply.

Example: Example 7 (Griffiths) and Problem 4.23 A sphere of linear dielectric material (dielectric constant K) is placed in an originally uniform electricfield (note: we will assume that this electric field is directed along the positive z axis). a) Find the new field inside the sphere.b) Solve for the field inside the sphere by the method of separation of variables. Note that: (1) V iscontinuous at R; (2) the discontinuity in the normal derivative of V at the surface is equal to ;(3) because the dielectric is linear .

a) Suppose the electric field inside the sphere is equal to . Since the material is a linear dielectric thepolarization is proportional to the total electric field:

However, a uniformly polarized sphere with polarization produces an internal electric field equal to

The electric field produced by the polarization of the sphere will therefore reduce the electric fieldinside the sphere by

This change in the electric field will change the polarization of the sphere by

This change in the polarization of the sphere will again change the electric field inside the sphere.This change of the electric field strength is equal to

This iterative process will continue indefinitely, and the final electric field will be equal to



The final polarization of the sphere is therefore equal to

b) Since the dielectric will be uniformly polarized, all the bound charge will reside on the surface ofthe sphere:

Therefore, the charge density is zero everywhere except on the surface of the sphere. The electrostaticpotential of this system must therefore satisfy Laplace's equation (see Chapter 3). The most generalsolution of Laplace's equation for this system is

Note that the potential does not approach zero when r approaches infinity since the electric field atinfinity is equal to . The electrostatic potential has to be continuous at r = R. Thus

This relation requires that

These two equations can be rewritten as

The normal derivative of V at the surface of the sphere must satisfy the following boundary condition:

Note that since the sphere is neutral, there is no free charge present. Therefore, the total surface chargeon the sphere is equal to the bound surface charge. Substituting the general solution for V in thisequation we obtain

This equation can be rewritten by using the expressions for Bn in terms of An with the followingresult:



The bound charge is determined by the electric field, and therefore by the gradient of the potential:

Combining the last two equations we obtain

This equation shows that for n = 1:

and for n ≠ 1:

These two equations can be rewritten as

and for n ≠ 1

The electrostatic potential since the sphere is thus equal to

We conclude that the electrostatic potential inside the sphere only depends on the z coordinate. Theelectric field inside the sphere can be obtained from the gradient of the electrostatic potential:

which is identical to the result we obtained in a).

Example: Problem 4.35 Prove the following uniqueness theorem: A region S contains a specified free charge distribution and various pieces of linear dielectric material, with the susceptibility of each one given. If thepotential is specified on the boundary of S (and V = 0 at infinity) then the potential throughout S isuniquely defined.

Suppose that there are two different solutions and . The corresponding electric fields are and , respectively. The corresponding electric displacements are and . Consider



a third function . Since and must have the same value on the border, there. Nowconsider the volume integral (over volume S) of :

since on the surface of volume S. The left-hand side of this equation can be rewritten as

But the divergence of is equal to zero since

Therefore

The integral on the right-hand side of this equation can be rewritten in terms of using the followingrelations:

and

Therefore,

Since this equation can only be satisfied if . This requires that

and

everywhere. We therefore conclude that there are no two different electrostatic potentials that satisfythe same boundary conditions. The electrostatic potential is therefore uniquely defined if its value isspecified on the surface of the volume S.

Example: Problem 4.36 A conducting sphere at potential is half embedded in linear dielectric material of susceptibility ,which occupies the region (see Figure 4.11). Claim: the potential everywhere is exactly the sameas it would have been in the absence of the dielectric! Check this claim as follows:a) Write down the formula for the suggested potential , in terms of , R, and r. Use it to determinethe field, the polarization, the bound charge, and the free charge distribution on the sphere.b) Show that the total charge configuration would indeed produce the potential .



c) Appeal to the uniqueness theorem in Problem 4.35 to complete the argument.

Figure 4.11. Problem 4.36.a) In the absence of the dielectric, the electrostatic potential of this system is constant inside thesphere and is given by

in the region outside the sphere. The electric field in the region outside the sphere is equal to thegradient of V and is therefore given by

If this solution satisfies the boundary conditions on the surface of the sphere when the dielectric ispresent then it is the only solution (uniqueness theorem of problem 4.35). The boundary conditions forthe electrostatic potential are:

1. V is continuous on the surface of the sphere. This boundary condition is satisfied by the proposedsolution.

2. The difference in the normal derivative of V on the surface of the sphere is equal to

For the proposed solution this requires that

which shows that the total charge is uniformly distributed across the surface of the sphere.

The polarization of the dielectric material in the region z < 0 (and r > R) can be obtained from theelectric field:

In the region z > 0 the polarization is equal to zero since no dielectric material is present there. Thebound surface charge on the surface of the dielectric is equal to

where is the surface vector (perpendicular to the surface and pointing out of the dielectric). For thespherical surface and thus



There is no bound charge on the flat surface (z = 0) of the dielectric since is there. The boundvolume charge is zero everywhere since

The free charge on the surface of the sphere can be determined from the electric displacement . Theelectric displacement can be obtained from the electric field. In the region above the dielectric (z >0) and outside the sphere (r > R) the electric displacement is equal to

In the region z < 0 and outside the sphere (r > R) the electric displacement is equal to

The free charge on the bottom hemisphere and part of the z = 0 plane (see Figure 4.12a) is equal to

There is no contribution to the surface charge from the z = 0 plane since there. The freecharge density on the bottom hemisphere is therefore equal to

In the same manner we can calculate the free charge density on the top hemisphere:

The total charge density (bound charge + free charge) on the surface is therefore equal to

Therefore, the total charge on the surface of the sphere is distributed uniformly, and has a valueconsistent with the boundary condition for the normal derivative of V. Since the proposed solutionsatisfies the boundary conditions for V it will be the only correct solution.

Figure 4.12. Determination of free charge in Problem 4.36.



4.5. Energy in dielectric systems

Consider a capacitor with capacitance C and charged up to a potential V. The total energy stored in thecapacitor is equal to the work done during the charging process:

If the capacitor is filled with a linear dielectric (dielectric constant K) than the total capacitance willincrease by a factor K:

and consequently the energy stored in the capacitor (when held at a constant potential) is increased bya factor K. A general expression for the energy of a capacitor with dielectric materials present can befound by studying the charging process in detail. Consider a free charge held at a potential V.During the charging process the free charge is increased by . The work done on the extra freecharge is equal to

Since the divergence of the electric displacement is equal to the free charge density , thedivergence of is equal to . Therefore,

Using the following relation

we can rewrite the expression for ∆W as

The first term on the right-hand side of this equation can be rewritten as

since the product of potential and electric displacement approach zero faster than 1/r2 when rapproached infinity. Therefore,

Assuming that the materials present in the system are linear dielectrics then

This relation can be used to rewrite :



The expression for can thus be rewritten as

The total work done during the charging process is therefore equal to

Note: this equation can be used to calculate the energy for a system that contains linear dielectrics. Ifsome materials in the system are non-linear dielectrics than the derivation given above is not correct (

for non-linear dielectrics).

Example: Problem 4.26 A spherical conductor, of radius a, carries a charge Q. It is surrounded by linear dielectric material ofsusceptibility χe, out to a radius b. Find the energy of this configuration.

Since the system has spherical symmetry the electric displacement is completely determined by thefree charge. It is equal to

Since we are dealing with linear dielectrics, the electric field is equal to . Taking intoaccount that the susceptibility of vacuum is zero and the susceptibility of a conductor is infinite weobtain for :

The scalar product is equal to since and are parallel, everywhere. The energy of thesystem is equal to



4.6. Forces on dielectrics

A dielectric slab placed partly between the plates of a parallel-plate capacitor will be pulled inside thecapacitor. This force is a result of the fringing fields around the edges of the parallel-plate capacitor(see Figure 4.13). Note: the field outside the capacitor can not be zero since otherwise the lineintegral of the electric field around a closed loop, partly inside the capacitor and partly outside thecapacitor, would not be equal to zero.

Figure 4.13. Fringing fields.Inside the capacitor the electric field is uniform. The electric force exerted by the field on the positivebound charge of the dielectric is directed upwards and is canceled by the electric force on the negativebound charge (see Figure 4.14). Outside the capacitor the electric field is not uniform and the electricforce acting on the positive bound charge will not be canceled by the electric force acting on thenegative bound charge. For the system shown in Figure 4.14 the vertical components of the two forces(outside the capacitor) will cancel, but the horizontal components are pointing in the same directionand therefore do not cancel. The result is a net force acting on the slab, directed towards the center ofthe capacitor.

Figure 4.14. Forces on dielectric.A direct calculation of this force requires a knowledge of the fringing fields of the capacitor which areoften not well known and difficult to calculate. An alternative method that can be used is to determinethis force is to calculate the change in the energy of the system when the dielectric is displaced by adistance ds. The work to be done to pull the dielectric out by an infinitesimal distance ds is equal to

where is the force provided by us to pull the slab out of the capacitor. This force must just be equalin magnitude but directed in a direction opposite to the force exerted by the electric field on theslab. Thus

The work done by us to move the slab must be equal to the change in the energy of the capacitor(conservation of energy). Consider the situation shown in Figure 4.15 where the slab of dielectric isinserted to a depth s in the capacitor. The capacitance of this system is equal to



Figure 4.15. Calculation of . If the total charge on the top plate is Q then the energy stored in the capacitor is equal to

The force on the dielectric can now be calculated and is equal to

Example: Problem 4.28 Two long coaxial cylindrical metal tubes (inner radius a, outer radius b) stand vertically in a tank ofdielectric oil (susceptibility χe, mass density ρ). The inner one is maintained at potential V, and theouter one is grounded. To what height h does the oil rise in the space between the tubes?

The height of the oil is such that the electric force on the oil balances the gravitational force. Thecapacitance of an empty cylindrical capacitor of height H is equal to

If the oil rises to a height h then the capacitance of the capacitor is equal to

The electric force on the dielectric (the oil) is equal to

and is directed upwards The gravitational force acting on the oil is equal to

and is directed downwards. In the equilibrium position . Thus

or



For a linear dielectric, the polarization is proportional to the total macroscopic field :

The polarization of the dielectric is equal to the vector sum of the polarization of the individualatoms or molecules:

where N is the number of atoms or molecules per unit volume. The polarization of an individual atomor molecule is proportional to the microscopic field at the position of the atom or molecule due toeverything except the particular atom or molecule under consideration:

The dipole moment of the atom or molecule will generate an electric field at its center equal to

where R is the radius of the atom or molecule. The total macroscopic field seen by the atom ormolecule is there for equal to

where N is the number of atoms per unit volume. The total polarization of the dielectric is thus equalto

Therefore

This equation can be rewritten in terms of the dielectric constants K as

or



This equation shows that a measurement of the macroscopic parameter K can be used to obtaininformation about the microscopic parameter α. This equation is known as the Clausius-Mossottiformula or the Lorentz-Lorenz equation.



Chapter 5. Magnetostatics5.1. The Magnetic Field5.2. The Biot-Savart Law5.3. The Divergence and Curl of B.5.4. The Vector Potential5.5. The Three Fundamental Quantities of Magnetostatics5.6. The Boundary Conditions of B5.7. The Multipole Expansion of the Magnetic Field

Chapter 5. Magnetostatics

5.1. The Magnetic Field

Consider two parallel straight wires in which current is flowing. The wires are neutral and thereforethere is no net electric force between the wires. Nevertheless, if the current in both wires is flowing inthe same direction, the wires are found to attract each other. If the current in one of the wires isreversed, the wires are found to repel each other. The force responsible for the attraction and repulsionis called the magnetic force. The magnetic force acting on a moving charge q is defined in terms ofthe magnetic field:

The vector product is required since observations show that the force acting on a moving charge isperpendicular to the direction of the moving charge. In a region where there is an electric field and amagnetic field the total force on the moving force is equal to

This equation is called the Lorentz force law and provides us with the total electromagnetic forceacting on q. An important difference between the electric field and the magnetic field is that theelectric field does work on a charged particle (it produces acceleration or deceleration) while themagnetic field does not do any work on the moving charge. This is a direct consequence of theLorentz force law:

We conclude that the magnetic force can alter the direction in which a particle moves, but can notchange its velocity.

Example: Problem 5.1 A particle of charge q enters the region of uniform magnetic field (pointing into the page). The fielddeflects the particle a distance d above the original line of flight, as shown in Figure 5.1. Is the chargepositive or negative? In terms of a, d, B, and q, find the momentum of the particle.

In order to produce the observed deflection, the force on q at the entrance of the field region must bedirected upwards (see Figure 5.1). Since direction of motion of the particle and the direction of themagnetic field are known, the Lorentz force law can be used to determine the direction of themagnetic force acting on a positive charge and on a negative charge. The vector product between and points upwards in Figure 5.1 (use the right-hand rule). This shows that the charge of the particleis positive.



Figure 1. Problem 5.1.The magnitude of the force acting on the moving charge is equal to

As a result of the magnetic force, the charged particle will follow a spherical trajectory. The radius ofthe trajectory is determined by the requirement that the magnetic force provides the centripetal force:

In this equation r is the radius of the circle that describes the circular part of the trajectory of charge q.The equation can be used to calculate r:

where p is the momentum of the particle. Figure 5.2 shows the following relation between r, d and a:

This equation can be used to express r in terms of d and a:

The momentum of the charge q is therefore equal to

Figure 2. Problem 5.2. The electric current in a wire is due to the motion of the electrons in the wire. The direction of currentis defined to be the direction in which the positive charges move. Therefore, in a conductor the currentis directed opposite to the direction of the electrons. The magnitude of the current is defined as the



total charge per unit time passing a given point of the wire (I = dq/dt). If the current flows in a regionwith a non-zero magnetic field then each electron will experience a magnetic force. Consider a tinysegment of the wire of length dl. Assume that the electron density is -λ C/m and that each electron ismoving with a velocity v. The magnetic force exerted by the magnetic field on a single electron isequal to

A segment of the wire of length dl contains λ dl/e electrons. Therefore the magnetic force acting inthis segment is equal to

Here we have used the definition of the current I in terms of dq and dt:

In this derivation we have defined the direction of to be equal to the direction of the current (andtherefore opposite to the direction of the velocity of the electrons). The total force on the wire istherefore equal to

Here I have assumed that the current is constant throughout the wire. If the current is flowing over asurface, it is usually described by a surface current density , which is the current per unit length-perpendicular-to-flow. The force on a surface current is equal to

If the current flows through a volume, is it is usually described in terms of a volume current density . The magnetic force on a volume current is equal to

The surface integral of the current density across the surface of a volume V is equal to the totalcharge leaving the volume per unit time (charge conservation):

Using the divergence theorem we can rewrite this expression as

Since this must hold for any volume V we must require that

This equation is known as the continuity equation.



5.2. The Biot-Savart Law

In this Section we will discuss the magnetic field produced by a steady current. A steady current is aflow of charge that has been going on forever, and will be going on forever. These currents producemagnetic fields that are constant in time. The magnetic field produced by a steady line current is givenby the Biot-Savart Law:

where is an element of the wire, is the vector connecting the element of the wire and P, and isthe permeability constant which is equal to

The unit of the magnetic field is the Tesla (T). For surface and volume currents the Biot-Savart lawcan be rewritten as

and

Example: Problem 5.9 Find the magnetic field at point P for each of the steady current configurations shown in Figure 5.3.

a) The total magnetic field at P is the vector sum of the magnetic fields produced by the four segmentsof the current loop. Along the two straight sections of the loop, and are parallel or opposite, andthus . Therefore, the magnetic field produced by these two straight segments is equal to zero.Along the two circular segments and are perpendicular. Using the right-hand rule it is easy toshow that

and

where is pointing out of the paper. The total magnetic field at P is therefore equal to



Figure 5.3. Problem 5.9. b) The magnetic field at P produced by the circular segment of the current loop is equal to

where is pointing out of the paper. The magnetic field produced at P by each of the two linearsegments will also be directed along the negative z axis. The magnitude of the magnetic fieldproduced by each linear segment is just half of the field produced by an infinitely long straight wire(see Example 5 in Griffiths):

The total field at P is therefore equal to

Example: Problem 5.12 Suppose you have two infinite straight-line charges λ, a distance d apart, moving along at a constant v(see Figure 5.4). How fast would v have to be in order for the magnetic attraction to balance theelectrical repulsion?

Figure 5.4. Problem 5.12. When a line charge moves it looks like a current of magnitude I = λv. The two parallel currents attracteach other, and the attractive force per unit length is

and is attractive. The electric generated by one of the wires can be found using Gauss' law and is equalto

The electric force per unit length acting on the other wire is equal to



and is repulsive (like charges). The electric and magnetic forces are balanced when

or

This requires that

This requires that the speed v is equal to the speed of light, and this can therefore never be achieved.Therefore, at all velocities the electric force will dominate.

5.3. The Divergence and Curl of B.

Using the Biot-Savart law for a volume current we can calculate the divergence and curl of :

and

This last equation is called Ampere's law in differential form. This equation can be rewritten, usingStokes' law, as

This equation is called Ampere's law in integral form. The direction of evaluation of the lineintegral and the direction of the surface element vector must be consistent with the right-hand rule. Ampere's law is always true, but is only a useful tool to evaluate the magnetic field if the symmetry ofthe system enables you to pull outside the line integral. The configurations that can be handled byAmpere's law are: 1. Infinite straight lines 2. Infinite planes 3. Infinite solenoids 4. Toroids

Example: Problem 5.14 A thick slab extending from z = -a to z = a carries a uniform volume current . Find the magneticfield both inside and outside the slab.



Figure 5.5. Problem 5.14 Because of the symmetry of the problem the magnetic field will be directed parallel to the y axis. Themagnetic field in the region above the xy plane (z > 0) will be the mirror image of the field in theregion below the xy plane (z < 0). The magnetic field in the xy plane (z = 0) will be equal to zero.Consider the Amperian loop shown in Figure 5.5. The current is flowing out of the paper, and wechoice the direction of to be parallel to the direction of . Therefore,

The direction of evaluation of the line integral of must be consistent with our choice of the directionof (right-hand rule). This requires that the line integral of must be evaluated in a counter-clockwise direction. The line integral of is equal to

Applying Ampere's law we obtain for :

Thus

5.4. The Vector Potential

The magnetic field generated by a static current distribution is uniquely defined by the so-calledMaxwell equations for magnetostatics:



Similarly, the electric field generated by a static charge distribution is uniquely defined by the so-called Maxwell equations for electrostatics:

The fact that the divergence of is equal to zero suggests that there are no point charges for .Magnetic field lines therefore do not begin or end anywhere (in contrast to electric field lines that starton positive point charges and end on negative point charges). Since a magnetic field is created bymoving charges, a magnetic field can never be present without an electric field being present. Incontrast, only an electric field will exist if the charges do not move. Maxwell's equations for magnetostatics show that if the current density is known, both the divergenceand the curl of the magnetic field are known. The Helmholtz theorem indicates that in that case thereis a vector potential such that

However, the vector potential is not uniquely defined. We can add to it the gradient of any scalarfunction f without changing its curl:

The divergence of is equal to

It turns out that we can always find a scalar function f such that the vector potential is divergence-less. The main reason for imposing the requirement that is that it simplifies many equationsinvolving the vector potential. For example, Ampere's law rewritten in terms of is

or

This equation is similar to Poisson's equation for a charge distribution ρ:

Therefore, the vector potential can be calculated from the current in a manner similar to how weobtained V from ρ. Thus



Note: these solutions require that the currents go to zero at infinity (similar to the requirement that ρgoes to zero at infinity).

Example: Problem 5.22 Find the magnetic vector potential of a finite segment of straight wire carrying a current I. Check thatyour answer is consistent with eq. (5.35) of Griffiths.

The current at infinity is zero in this problem, and therefore we can use the expression for in termsof the line integral of the current I. Consider the wire located along the z axis between z1 and z2 (seeFigure 5.6) and use cylindrical coordinates. The vector potential at a point P is independent of φ(cylindrical symmetry) and equal to

Here we have assumed that the origin of the coordinate system is chosen such that P has z = 0. Themagnetic field at P can be obtained from the vector potential and is equal to

where θ1 and θ2 are defined in Figure 5.6. This result is identical to the result of Example 5 inGriffiths.


Example: Problem 5.24 If is uniform, show that , where is the vector from the origin to the point in question.That is check that and .



The curl of is equal to

Since is uniform it is independent of r, θ, and φ and therefore the second and third term on the right-hand side of this equation are zero. The first term, expressed in Cartesian coordinates, is equal to

The fourth term, expressed in Cartesian coordinates, is equal to

Therefore, the curl of is equal to

The divergence of is equal to

Example: Problem 5.26 Find the vector potential above and below the plane surface current of Example 5.8 in Griffiths.

In Example 5.8 of Griffiths a uniform surface current is flowing in the xy plane, directed parallel tothe x axis:

However, since the surface current extends to infinity, we can not use the surface integral of tocalculate and an alternative method must be used to obtain . Since Example 8 showed that isuniform above the plane of the surface current and is uniform below the plane of the surfacecurrent, we can use the result of Problem 5.27 to calculate :

In the region above the xy plane (z > 0) the magnetic field is equal to

Therefore,



In the region below the xy plane (z < 0) the magnetic field is equal to

Therefore,

We can verify that our solution for is correct by calculating the curl of (which must be equal tothe magnetic field). For z > 0:

The vector potential is however not uniquely defined. For example, and are also possible solutions that generate the same magnetic field. These solutions also satisfy therequirement that .

5.5. The Three Fundamental Quantities of Magnetostatics

Our discussion of the magnetic fields produced by steady currents has shown that there are threefundamental quantities of magnetostatics: 1. The current density 2. The magnetic field 3. The vector potential These three quantities are related and if one of them is known, the other two can be calculated. Thefollowing table summarizes the relations between , , and :

5.6. The Boundary Conditions of B

In Chapter 2 we studied the boundary conditions of the electric field and concluded that the electricfield suffers a discontinuity at a surface charge. Similarly, the magnetic field suffers a discontinuity at



a surface current.

Figure 5.7. Boundary conditions for . Consider the surface current (see Figure 5.7). The surface integral of over a wafer thin pillbox isequal to

where A is the area of the top and bottom of the pill box. The surface integral of can be rewrittenusing the divergence theorem:

since for any magnetic field . Therefore, the perpendicular component of the magnetic fieldis continuous at a surface current:

The line integral of around the loop shown in Figure 5.8 (in the limit ε → 0) is equal to

According to Ampere's law the line integral of around this loop is equal to

Figure 5.8. Boundary conditions for .Therefore, the boundary condition for the component of , parallel to the surface and perpendicular tothe current, is equal to

The boundary conditions for can be combined into one equation:

where is a unit vector perpendicular to the surface and the surface current and pointing "upward".The vector potential is continuous at a surface current, but its normal derivative is not:



5.7. The Multipole Expansion of the Magnetic Field

To calculate the vector potential of a localized current distribution at large distances we can use themultipole expansion. Consider a current loop with current I. The vector potential of this current loopcan be written as

At large distance only the first couple of terms of the multipole expansion need to be considered:

The first term is called the monopole term and is equal to zero (since the line integral of is equal tozero for any closed loop). The second term, called the dipole term, is usually the dominant term. Thevector potential generated by the dipole terms is equal to


where is called the magnetic dipole moment of the current loop. It is defined as

If the current loop is a plane loop (current located on the surface of a plane) then is the areaof the triangle shown in Figure 5.9. Therefore,

where a is the area enclosed by the current loop. In this case, the dipole moment of the current loop isequal to

where the direction of must be consistent with the direction of the current in the loop (right-handrule).



Figure 5.9. Calculation of . Assuming that the magnetic dipole is located at the origin of our coordinate system and that is

pointing along the positive z axis, we obtain for :

The corresponding magnetic field is equal to

The shape of the field generated by a magnetic dipole is identical to the shape of the field generatedby an electric dipole.

Example: Problem 5.33 Show that the magnetic field of a dipole can be written in the following coordinate free form:

Figure 5.10. Problem 5.33. Consider the configuration shown in Figure 5.10. The scalar product between and is equal to

The scalar product between and is equal to

Therefore,

Example: Problem 5.34 A circular loop of wire, with radius R, lies in the xy plane, centered at the origin, and carries a currentI running counterclockwise as viewed from the positive z axis.a) What is its magnetic dipole moment?



b) What is its (approximate) magnetic field at points far from the origin?c) Show that, for points on the z axis, your answer is consistent with the exact field as calculated inExample 6 of Griffiths.

a) Since the current loop is a plane loop, its dipole moment is easy to calculate. It is equal to

b) The magnetic field at large distances is approximately equal to

c) For points on the positive z axis θ = 0°. Therefore, for z>0

Fore points on the negative z axis θ = 180°. Therefore, for z<0

The exact solution for on the positive z axis is

For z » R the field is approximately equal to

which is consistent with the dipole field of the current loop.

Example: Problem 5.35 A phonograph record of radius R, carrying a uniform surface charge σ, is rotating at constant angularvelocity ω. Find its magnetic dipole moment.

The rotational period of the disk is equal to

Consider the disk to consist of a large number of thin rings. Consider a single ring of inner radius rand with dr. The charge on such a ring is equal to

Since the charge is rotating, the moving charge corresponds to a current dI:

The dipole moment of this ring is therefore equal to



The total dipole moment of the disk is equal to



Chapter 6. Magnetostatic Fields in Matter6.1. Magnetization

6.1.1. Paramagnetism6.1.2. Diamagnetism

6.2. The Field of a Magnetized ObjectExample: Problem 6.7

6.3. The Auxiliary Field HExample: Problem 6.12Example: Problem 6.14

6.4. Linear MediaExample: Example 6.3Example: Problem 6.18

6.5. Nonlinear Media

Chapter 6. Magnetostatic Fields in Matter

6.1. Magnetization

Any macroscopic object consists of many atoms or molecules, each having electric charges in motion.With each electron in an atom or molecule we can associate a tiny magnetic dipole moment (due to itsspin). Ordinarily, the individual dipoles cancel each other because of the random orientation of theirdirection. However, when a magnetic field is applied, a net alignment of these magnetic dipolesoccurs, and the material becomes magnetized. The state of magnetic polarization of a material isdescribed by the parameter M which is called the magnetization of the material and is defined as

M = magnetic dipole moment per unit volume

In some material the magnetization is parallel to . These materials are called paramagnetic. In othermaterials the magnetization is opposite to . These materials are called diamagnetic. A third group ofmaterials, also called Ferro magnetic materials, retain a substantial magnetization indefinitely after theexternal field has been removed.

Figure 6.1. Force on a rectangular current loop.

6.1.1. Paramagnetism

Consider a rectangular current loop, with sides s1 and s2, located in a uniform magnetic field, pointingalong the z axis. The magnetic dipole moment of the current loop makes an angle θ with the z axis(see Figure 6.1a). The magnetic forces on the left and right sides of the current loop have the samemagnitude but point in opposite directions (see Figure 6.1b). The net force acting on the left and right



side of the current loop is therefore equal to zero. The force on the top and bottom part of the currentloop (see Figure 6.1a) also have the same magnitude and point in opposite directions. However sincethese forces are not collinear, the corresponding torque is not equal to zero. The torque generated bymagnetic forces acting on the top and the bottom of the current loop is equal to

The magnitude of the force F is equal to

Therefore, the torque on the current loop is equal to

where is the magnetic dipole moment of the current loop. As a result of the torque on the currentloop, it will rotate until its dipole moment is aligned with that of the external magnetic field. In atoms we can associate a dipole moment with each electron (spin). An external magnetic field willline up the dipole moment of the individual electrons (where not excluded by the Pauli principle). Theinduced magnetization is therefore parallel to the direction of the external magnetic field. It is thismechanism that is responsible for paramagnetism. In a uniform magnetic field the net force on any current loop is equal to zero:

since the line integral of is equal to zero around any closed loop. If the magnetic field is non-uniform then, in general, there will be a net force on the current loop.Consider an infinitesimal small current square of side ε, located in the yz plane and with a currentflowing in a counter-clockwise direction (see Figure 6.2). The force acting on the current loop is thevector sum of the forces acting on each side:



Figure 6.2. Infinitesimal square current loop.In this derivation we have used a first-order Taylor expansion of :

and

Assuming that the current loop is so small that the derivatives of are constant over the boundaries ofthe loop we can evaluate the integrals and obtain for the total force:

where m is the magnetic dipole moment of the current loop. In this derivation we have used the factthat the divergence of is equal to zero for any magnetic field and this requires that

The magnetic dipole moment m of the current loop is equal to

Therefore, the equation for the force acting on the current loop can be rewritten in terms of :

Any current loop can be build up of infinitesimal current loops and therefore

for any current loop.

Example: Problem 6.1.a) Calculate the torque exerted on the square loop shown in Figure 6.3 due to the circular loop(assume r is much larger than a or s).b) If the square loop is free to rotate, what will its equilibrium orientation be?

a) The dipole moment of the current loop is equal to

where we have defined the z axis to be the direction of the dipole. The magnetic field at the positionof the square loop, assuming that r»a, will be a dipole field with θ = 90°:

The dipole moment of the square loop is equal to



Figure 6.3. Problem 6.1.Here we have assumed that the x axis coincides with the line connecting the center of the currentcircle and the center of the current square. The torque on the square loop is equal to

b) Suppose the dipole moment of the square loop is equal to

The torque on this dipole is equal to

In the equilibrium position, the torque on the current loop must be equal to zero. This thereforerequires that

Thus, in the equilibrium position the dipole will have its dipole moment directed along the z axis. Theenergy of a magnetic dipole in a magnetic field is equal to

The system will minimize its energy if the dipole moment and the magnetic field are parallel. Sincethe magnetic field at the position of the square loop is pointing down, the equilibrium position of thecurrent loop will be with its magnetic dipole moment pointing down (along the negative z axis).

6.1.2. Diamagnetism

Consider a very classical picture of a Hydrogen atom consisting of an electron revolving in a circularorbit of radius r around a nucleus (see Figure 6.4). Suppose that the velocity of the electron is equal tov. Since the velocity of the electron is very high, the revolving electron looks like a steady current ofmagnitude



Figure 6.4. Electron in orbit. The direction of the current is in a direction opposite to that of the electron. The dipole moment of this

current is equal to

If the atom is placed in a magnetic field, it will be subject to a torque. However, it is very difficult totilt the entire orbit. Instead the electron will try to reduce its torque by changing its velocity. With nomagnetic field present, the velocity of the electron can be obtained by requiring that the centripetalforce is sustained by just the electric force:

In a magnetic field, the centripetal force will be sustained by both the electric and the magnetic field:

Here we have assumed that the magnetic field is pointing along the positive z axis (in a directionopposite to the direction of the magnetic dipole moment). We have also assumed that the size of theorbit (r) does not change when the magnetic field is applied. Combining the last two equations weobtain

or

Assuming that the change in the velocity is small we can use the following approximations:

and

Therefore,



This equation shows that the presence of the magnetic field will increase the speed of the electron. Anincrease in the velocity of the electron will increase the magnitude of the dipole moment of therevolving electron. The change in is opposite to the direction of . If the electron would have beenorbiting the other way, it would have been slowed down by the magnetic field. Again the change inthe dipole moment is opposite to the direction of . In the presence of an external magnetic field the dipole moment of each orbit will be slightlymodified, and all these changes are anti-parallel to the external magnetic field. This is the magnetismthat is responsible for diamagnetism. Diamagnetism is present in all materials, but is in general muchweaker than paramagnetism. It can therefore only be observed in those materials whereparamagnetism is not present.

6.2. The Field of a Magnetized Object

Consider a magnetized material with magnetization . The associated vector potential is equal to

Following the same procedure used in Chapter 4 to calculate the electrostatic potential of a polarizedmaterial, we obtain for :

where is the bound volume current and is the bound surface current. If the materialhas a uniform magnetization then the bound volume current is zero. The field produced by amagnetized object is equal to the field produced by the bound currents.

Figure 6.5. Bound surface current. Consider a uniformly magnetized thin slab of material of thickness t. The material can be cut up intotiny current loops (see Figure 6.5). If each current loop has an area a then the dipole moment due to asurface current I is equal to

The volume of the current loop is at and therefore its dipole moment must be equal to

This requires that the surface current of the current loop is equal to



Since the magnetization is uniform, the current in each of the current loops will be constant andflowing in the same direction. Therefore, all volume currents cancel, and the only current remainingwill be a surface current, flowing on the surface of the material. The current flowing on the surface ofthe material will be equal to the current in each of the current loops. Therefore, the current density onthe surface is equal to

In vector notation:

This equation is also consistent with the fact that there is no current flowing on the top and bottomsurfaces (where ).


An infinitely long circular cylinder carries a uniform magnetization parallel to its axis. Find themagnetic field (due to ) inside and outside the cylinder.

Consider a coordinate system S in which the z axis coincides with the axis of the cylinder. Themagnetization of the material is equal to

Since the material is uniformly magnetized, its bound volume current is equal to zero. The boundsurface current is equal to

This current distribution is identical to the current distribution in an infinitely long solenoid. Themagnetic field outside an infinitely long solenoid is equal to zero (see Example 9, Chapter 5 ofGriffiths), and therefore also the field outside the magnetized cylinder will be equal to zero. Themagnetic field inside an infinitely long solenoid can be calculated easily using Ampere's law (seeExample 9, Chapter 5 of Griffiths). It is equal to

6.3. The Auxiliary Field H

The magnetic field in a system containing magnetized materials and free currents can be obtained bycalculating the field produced by the total current where

This approach is very similar to the approach taken in electrostatics where the total electric fieldproduced by a system containing dielectric materials is equal to the electric field produced by a chargedistribution σ where



To calculate the magnetic field produced by a system containing magnetized materials we have to usethe following form of Ampere's law:


The quantity in parenthesis is called the H-field

plays a role in magnetostatics analogous to in electrostatics. Ampere's law in terms of reads

and

Ampere's law for tells us that the curl of is equal to the free current density. However, aknowledge of the free current density is not sufficient to determine . The Helmholtz theorem showsthat besides knowing the curl of a vector function, we also need to know the divergence of that vectorfunction before it is uniquely defined. Although the divergence of is zero for any magnetic field(and therefore Ampere's law for defines uniquely) the divergence of is not necessarily zero:

Therefore, only for those systems where can we use Ampere's law for directly to calculate . The divergence of will be zero only for systems with cylindrical, plane, solenoidal, or toroidal

symmetry. The field is a quantity that is used in the laboratory more often that the field. This is a result ofthe dependence of on only the free currents (which are easy to control). The field depends bothon the free and on the bound currents, and thus requires a detailed knowledge of the magneticproperties of the materials used. In electrostatics, the electric field can be obtained immediately fromthe potential difference (which is easy to control). The electric displacement depends only on thefree charge distribution, but in most cases a direct measurement of the free charge distribution is verydifficult to carry out. Therefore, in electrostatics the electric field is in most cases a more usefulparameter then the electric displacement .


An infinitely long cylinder, of radius R, carries a "frozen-in" magnetization, parallel to the axis,



where k is a constant and r is the distance from the axis (there is no free current anywhere). Find themagnetic field inside and outside the cylinder by two different methods:a) Locate all the bound currents, and calculate the field they produce.b) Use Ampere's law to find , and then get .

a) The magnetization of the material is directed along the z axis and is equal to

The bound volume current is equal to

The bound surface current is equal to

Figure 6.6. Problem 6.12.The bound currents produce a solenoidal field. The field outside the cylinder will be equal to zero andthe field inside the cylinder will be directed along the z axis. Its magnitude can be obtained usingAmpere's law. Consider the Amperian loop shown in Figure 6.6. The line integral of along theAmperian loop is equal to

The current intercepted by the Amperian loop is equal to

Ampere's law can now be used to calculate the magnetic field:



b) The divergence of is equal to zero. Therefore, Ampere's law uniquely defines . The field ispointing in the z direction. Using Ampere's law, in terms of the field, we immediately conclude thatfor the Amperian loop shown in Figure 6.6

since there is no free current This can only be true if . This implies that

Therefore, the magnetic field is equal to

In the region outside the cylinder the magnetization is equal to zero and therefore the magnetic field isequal to

In the region inside the cylinder the magnetization is equal to

and therefore the magnetic field is equal to

which is identical to the result obtained in part a).


Suppose the field inside a large piece of material is , and the corresponding field is equal to

a) A small spherical cavity is hollowed out of the material. Find the field at the center of the cavityin terms of and . Also find the field at the center of the cavity in terms of and .b) Do the same for a long needle-shaped cavity running parallel to .c) Do the same for a thin wafer-shaped cavity perpendicular to .Assume the cavities are small enough so that , , and are essentially constant.

a) The field in the spherical cavity is the superposition of the field and the field produced by asphere with magnetization . The bound volume current in the sphere is equal to zero (uniformmagnetization). The bound surface current is equal to

Here we have assumed that the magnetization of the sphere is directed along the z axis. Now consider



a uniformly charged sphere, rotating with an angular velocity ω around the z axis. The system carriesa surface current equal to

Comparing these two equations for the surface current we conclude that

In Example 11 of Chapter 5 the magnetic field produced by a uniformly charged, rotating sphere wascalculated. The magnetic field inside the sphere was found to be uniform and equal to

But since we can rewrite this expression as

The magnetic field inside the spherical cavity is therefore equal to

The corresponding field is equal to

Here we have used the fact that since no materials are present there.

b) The magnetic field inside the needle-shaped cavity is equal to the vector sum of the field and thefield produced by a needle-shaped cylindrical piece of material with magnetization . The fieldinside a needle-shaped cylinder of magnetization is approximately equal to the field inside aninfinitely long solenoid. This field was calculated in Problem 6.7, and for a cylinder with a uniformmagnetization it is equal to

The magnetic field inside the needle-shaped cavity is thus equal to


c) The magnetic field in the center of a thin wafer-shaped cavity is equal to the vector sum of andthe magnetic field inside a waver-shaped material with magnetization . Since the thickness of thewafer approaches zero, the total surface current on the material approaches zero, and consequently themagnetic field inside the waver approaches zero. Therefore, the magnetic field inside the cavity willbe equal to




6.4. Linear Media

In paramagnetic and diamagnetic materials, the magnetization is maintained by the external magneticfield. The magnetization disappears when the field is removed. Most paramagnetic and diamagneticmaterials are linear; that is their magnetization is proportional to the field:

The constant of proportionality is called the magnetic susceptibility of the material. In vacuum themagnetic susceptibility is zero. In a linear medium, there is linear relation between the magnetic fieldand the field:

where is called the permeability of the material. The permeability of free space is equal to . The linear relation between and does not automatically imply that the divergence of is zero.The divergence of will only be equal to zero inside a linear material, but will be non-zero at theinterface between two materials of different permeability. Consider for example the interface betweena linear material and vacuum (see Figure 6.7). The surface integral of across the surface of theGaussian pillbox shown in Figure 6.7 is definitely not equal to zero. According to the divergencetheorem the surface integral of is equal to the volume integral of :

Figure 6.7. Interface of linear materials.Therefore, if the surface integral of is not equal to zero, the divergence of can not be zeroeverywhere.

Example: Example 6.3

An infinite solenoid (N turns per unit length, current I) is filled with linear material of susceptibilityχm. Find the magnetic field inside the solenoid.



Figure 6.8. Example 6.3. Because of the symmetry of the problem, the divergence of will be equal to zero, everywhere.Therefore, the field can be obtained directly from Ampere's law. Consider the Amperian loopshown in Figure 6.8. The line integral of around the loop is equal to

where the line integral is evaluate in the direction shown in Figure 6.8, and it is assumed that the field is directed along the z axis. The free current intercepted by the Amperian loop is equal to

Ampere's law for the field immediately shows that

The magnetic field inside the solenoid is equal to

The magnetization of the material is equal to

and is uniform. Therefore, there will be no bound volume currents in the material. The bound surfacecurrent is equal to

This last equation shows that the bound surface current flows in the same direction (paramagneticmaterials) or in an opposite direction (diamagnetic materials) as the free current.


A sphere of linear magnetic material is placed in an originally uniform magnetic field . Find the newfield inside the sphere.

This problem can be solved using a method similar to the method used in example 7 of Griffiths(Chapter 4). The external field will magnetize the sphere:



This magnetization will produce a uniform magnetic field inside the sphere (see Example 6.1 ofGriffiths, Chapter 6):

This additional magnetic field magnetizes the sphere by an additional amount:

This additional magnetization produces an additional magnetic field inside the sphere:

The total magnetic field inside the sphere is therefore equal to

When can check the consistency of this answer by calculating the magnetization of the sphere:

The magnetic field inside the sphere due to the magnetization is equal to

The total magnetic field inside the sphere is therefore equal to

which is consistent with our assumption.

6.5. Nonlinear Media

The best known nonlinear media are the ferromagnetic materials. Ferromagnetic materials do notrequire external fields to sustain their magnetization (therefore, the magnetization definitely dependsin a nonlinear way on the field). The magnetization in ferromagnetic materials involves the alignmentof the dipole moments associated with the spin of unpaired electrons. The difference betweenferromagnetic materials and paramagnetic materials is that in ferromagnetic materials the interactionbetween nearby dipoles makes them want to point in the same direction, even when the magnetic fieldis removed. However, the alignment occurs in relative small patches, called domains. When a



ferromagnetic material is not located in a magnetic field, the dipole moments of the various domainsare not aligned, and the material as a whole is not magnetized. When the ferromagnetic material is putinto a magnetic field, the boundaries of the domain parallel to the field will increase at the expense ofneighboring boundaries. If the field is strong enough, one domain takes over the entirely, and theferromagnetic material is said to be saturated (all unpaired electrons are aligned and therefore themagnetization reaches a maximum value). The magnetic susceptibility of ferromagnetic materials isaround 103 (roughly eighth orders of magnitude larger than the susceptibility of paramagneticmaterials). When the magnetic field is removed some magnetization remains (and we have created apermanent magnet). For any ferromagnetic material, the magnetization depends not only on theapplied magnetic field but also on the magnetization history. The alignment of dipoles in aferromagnet can be destroyed by random thermal motion. The destruction of the alignment occurs at aprecise temperature (called the Curie point). When a ferromagnetic material is heated above its Curietemperature it becomes paramagnetic.



Chapter 7. Electrodynamics7.1. Electromotive Force7.2. Faraday's Law7.3. Inductance7.4. The Maxwell Equations

Chapter 7. Electrodynamics

7.1. Electromotive Force

An electric current is flowing when the electric charges are in motion. In order to sustain an electriccurrent we have to apply a force on these charges. In most materials the current density isproportional to the force per unit charge:

The constant of proportionality σ is called the conductivity of the material. Instead of specifying theconductivity, it is more common to specify the resistivity ρ:

For conductors the resistivity is typically 10-8 Ω-m; for semiconductor it varies between 0.01 Ωm and1 Ω-m, and for insulators it varies between 105 Ω-m and 106 Ω-m. In most cases the force on thecharges is the electromagnetic force. In that case the current density is equal to:

If the velocity of the charges is small the second term can be ignored, and the equation for reducesto Ohm's Law:

Consider a wire of cross-sectional area A and length L. If a potential difference V is applied betweenthe ends of the wire, it will produce an electric field inside the wire of magnitude

The current density in the wire is therefore equal to

The total current flowing through the wire is therefore equal to

This equation shows that the current flowing from one electrode to the other electrode is proportionalto the potential difference between them. This is a rather surprising result since the charge carriers are



constantly accelerating. However, the proportionality between the current and the potential differencehas been found to be correct for most materials. This relation can be written as

The constant of proportionality R is called the resistance of the material. It is in general a function ofthe geometry of the system and the conductivity of the materials between the electrodes. The unit ofresistance is the ohm (Ω). The resistance of the wire is equal to

To create a current we have to do work. The work required to move a unit of charge across a potentialdifference V is equal to V. To establish a current I, we need to deliver a power P where

The unit of power is the Watt (1 W = 1 J/s). The work done by the electric force on the chargecarriers is converted into heat (Joule heating).

Example: Problem 7.1 Two concentric metal spherical shells, of radius a and b, respectively, are separated by weaklyconducting material of conductivity σ.a) If they are maintained at a potential difference V, what current flows from one to the other?b) What is the resistance between the shells?

a) Suppose a charge Q is placed on the inner shell. The electric field in the region between the shellswill be equal to

The corresponding potential difference between the spheres is equal to

Therefore, in order to maintain a potential difference V between the spheres, we must place a chargeQ equal to

on the center shell. The total current flowing between the two shells is equal to

b) The resistance between the shells can be obtained from Ohm's law:



Example: Problem 7.2a) Two metal objects are embedded in weakly conducting material of conductivity σ (see Figure 7.1).Show that the resistance between them is related to the capacitance of the arrangement by

b) Suppose you connected a battery between 1 and 2 and charged them up to a potential difference V0.If you then disconnect the battery, the charge will gradually leak off. Show that V(t) = V0 exp(- t/τ),and find the time constant τ in terms of ε0 and σ.

a) Suppose a charge Q is placed on the positively charged conductor. The current flowing from thepositively charged conductor is equal to

where the surface integral is taken over a surface that encloses the positively charged conductor (forexample, the dashed surface in Figure 7.1). The expression for I can be rewritten in terms of theelectric field as

Figure 7.1. Problem 7.2.Using Gauss's law to express the surface integral of in terms of the total enclosed charge we obtain

The charge on the conductor is related to the capacitance of the arrangement and the potentialdifference between the conductors:

The current I is therefore equal to

The resistance of the system can be calculated using Ohm's law:



b) The charge Q residing on the positively charged conductor is equal to


and has the following solution:

The potential difference V is equal to

The decay constant τ is equal to

In any electric circuit a current will only exist if a driving force is available. The most commonsources of the driving force are batteries and generators. When a circuit is hooked up to a powersource a current will start to flow. In a single-loop circuit the current will be the same everywhere.Consider the situation in which the currents are not the same (see Figure 7.2). If Iin > Iout thenpositive charge will accumulate in the middle. This accumulation of positive charge will generate anelectric field (see Figure 7.2) that slows down the incoming charges and speeds up the outgoingcharges. A reduction in the velocity of the incoming charges will reduce the incoming current. Anincrease in the velocity of the outgoing charges will increase the outgoing current. The current willchange until Iin = Iout. The total force f on the charge carriers (per unit charge) is equal to the sum of the source force, fs, andthe electric force:

Figure 7.2. Current flow.The work required to move one unit of charge once around the circuit is equal to



where ε is called the electromotive force or emf. The emf determines the current flowing through thecircuit. This can be most easily seen bye rewriting the force on the charge carriers in terms of thecurrent density

Here, a is the cross-sectional area of the wire (perpendicular to the direction of the current).

Example: Problem 7.5a) Show that electrostatic force alone cannot be used to drive current around a circuit.b) A rectangular loop of wire is situate so that one end is between the plates of a parallel-platecapacitor (see Figure 7.3), oriented parallel to the field E = σ/ε0. The other end is way outside, wherethe field is essentially zero. If the width of the loop is h and its total resistance is R, what currentflows? Explain.

Figure 7.3. Problem 7.5.a) If only electrostatic forces are present then the force per unit charge is equal to the electrostaticforce:

The associated emf is therefore equal to

for any electrostatic field.b) The only force on the charge carriers in the wire loop is the electric force. However, in part a) weconcluded that the emf associate with an electric force, generated by an electrostatic field, is equal tozero. Therefore, the emf in the wire loop is equal to zero, and consequently the current in the loop isalso equal to zero. Note: at first sight it might appear that there is a net emf, if we assume that theelectric field generated by the capacitor is that of an ideal capacitor (that is a homogeneous fieldinside and no field outside). Under that assumption, the emf is equal to

The contribution of the path integral from c to d is equal to zero since the electric field is zero there,and the contribution of the path integrals between b and c and between a and d is equal to zero sincethe electric field and the displacement are perpendicular there. Clearly the calculated emf is non-zero,and disagrees with the result of part a). The disagreement is a result of our incorrect assumption thatthe electric field outside the capacitor is equal to zero (there are fringing fields).

An important source of emf is the generator. In these devices the emf arises from the motion of aconducting wire through a magnetic field. Consider the system shown in Figure 7.4 (note: the



magnetic field is only present in the region left of the dashed line). Consider the free charges on theconductor. Since it is moving with a velocity v in a magnetic field it will experience a magnetic force.The force on a positive charge q located ion segment ab of the wire loop is equal to

Figure 7.4. The generator.The magnetic force per unit charge is therefore equal to

Since there are no other forces acting on the charges, the emf generated will be entirely due to thismagnetic force. The emf will be equal to

The magnetic flux intercepted by the wire loop is equal to

The rate of change of the magnetic flux is equal to

Comparing the rate of change of enclosed magnetic flux and the induced emf we can conclude that

This relation is called the flux rule for motional emf.

7.2. Faraday's Law

When a conducting wire moves in a constant magnetic field an emf is generated equal to

In this case, the magnetic force is responsible for the emf. However, the same emf is generated whenthe wire is stationary and the magnetic field is moving. In this case, the magnetic force does not play arole (since v = 0) and an electric field is responsible for the emf. This electric field is not anelectrostatic field (since electrostatic fields can not generate an emf; see Problem 7.5) but is induced



by the changing magnetic field. The line integral of this electric field is equal to

This equation can be rewritten by applying Stoke's theorem:

Since we have not made any assumption about the surface, this equation can only be true if

This relation is called Faraday's law in differential form. The direction of the currents generated bythe changing magnetic field can be obtained most easily using Lenz's law which states that

“ If a current flows, it will be in such a direction that the magnetic field it produces tends to counteractthe change in flux that induced the emf. “

Example: Problem 7.14 A long solenoid of radius a, carrying N turns per unit length, is looped by a wire of resistance R (seeFigure 7.5).a) If the current in the solenoid is increasing,

what current flows in the loop, and which way (left or right) does it pass through the resistor.b) If the current I in the solenoid is constant but the solenoid is pulled out of the loop and reinserted inthe opposite direction what total charge passes through the resistor?

Figure 7.5. Problem 7.14.a) Assume that the solenoid is an ideal solenoid; that is

If the current in the solenoid increases, the strength of the magnetic field also increases. The rate ofchange in the strength of the magnetic field is equal to

The magnetic flux intercepted by the wire loop is equal to



The corresponding rate of change of the magnetic flux is equal to

The induced emf can be obtained from the flux law:

The current induced in the wire loop is equal to

The solenoidal magnetic field points from left to right. An increase in the strength of the magneticfield will induce a current in the loop directed such that the magnetic field it produces point from rightto left (Lenz's law). Therefore, the current flows from left to right through the resistor.

b) The change in the magnetic flux enclosed by the wire loop is equal to

The current flowing through the resistor is equal to

This relation shows that

Substituting the expression for we obtain

7.3. Inductance

Consider two loops: loop 1 and loop 2 (see Figure 7.6). A current I1 flowing through loop 1 willproduce a magnetic field at the position of loop 2 equal to

The magnetic flux through loop 2 is equal to



Figure 7.6. Inductance. Here, M21 is called the mutual inductance of the two loops. It is a purely geometrical quantity that

depends on the sizes, shapes and relative positions of the two loops. It does not change if we switchthe role of loop 1 and loop 2: the flux through loop 2 when we run a current I around loop 1 is exactlythe same as the flux through loop 1 when we send the same current I around loop 2. Besides inducing an emf in a nearby loop, the changing current in loop 1 also induces an emf in loop1. The flux through loop 1 generated by the current in loop 1 is equal to

The constant of proportionality is called the self inductance. The unit of inductance is the Henrie(H).

Example: Problem 7.19 A square loop of wire, of side s, lies midway between two long wires, 3s apart and in the same place.(Actually, the long wires are sides of a large rectangular loop, but the short ends are so far away thatthey can be neglected.) A clockwise current I in the square loop is gradually increasing: dI/dt = k =constant. Find the emf induced in the big loop. Which way will the induced current flow?

Figure 7.7. Problem 7.19. The system is schematically shown in Figure 7.7. To calculate the emf induces in the large loop weneed to determine the mutual inductance M. It is hard to calculate M in terms of a current in the squareloop since the magnetic field generated by this loop is rather complicated (and therefore difficult tointegrate). However, exploiting the equality of the mutual inductances, we can also evaluate theinductance in terms of a current I in the large loop. The magnetic field generated by the top wire ofthe large loop is that of an infinitely long straight wire carrying a current I is equal to

The flux associated with this magnetic field and intercepted by the square loop is equal to

The magnetic field generated by the bottom wire at the position of the square loop will be pointing inthe same direction as the magnetic field generated by the top wire at the position of the square loop.Since the square loop is located at the same distance from the top wire as it is from the bottom wire,the flux intercepted by the square loop is equal to



Therefore, the mutual inductance of the two loops is equal to

The induced emf in the large loop can now be calculated easily:

The direction of the flux through the large loop is pointing into the page. This can be seen most easilyby considering the magnetic field lines. Inside the square loop, the field lines point into the page(right-hand rule). Since the field lines form closed loops, they must be pointing out of the pageanywhere outside the square loop. However, the large wire loop only covers a limited fraction ofspace, and therefore definitely will not intercept all field lines outside the square loop. Therefore,there will be more field lines pointing into the page then there are field lines pointing out of the page.Consequently, the net magnetic flux will be pointing into the page. When the current in the squareloop increases the flux intercepted by the large loop will increase. The induced emf will produce amagnetic field that counteracts this increase in flux, and therefore produces a flux pointing out of thepaper. The right-hand rule shows that the direction of the current induced in the large loop must beflowing in a counter-clockwise direction.

7.4. The Maxwell Equations

The electric and magnetic fields in electrostatics and magnetostatics are described by the followingfour equations:

In systems with non-steady currents not all of these equations are valid anymore. For example,

for every vector function. However, according to Ampere's law

which is only zero for steady currents (for which is a constant, independent of position). For non-steady currents

We thus conclude that Ampere's law does not hold for non-steady currents. The failure of Ampere'slaw can also be observed in a system in which a capacitor is being charged (see Figure 7.8). During



the charging process a current I is flowing through the wire, and consequently there will be a magneticfield present. The magnetic field generated by the charging current can be calculated using Ampere'slaw. When we are far away from the capacitor the generated magnetic field will be that of a linecurrent. Consider an Amperian loop of radius r, centered on the wire. The line integral of aroundthis loop is equal to

According to Ampere's law the line integral of around a closed loop is proportional to the currentintercepted by a surface spanned by this loop. For the system shown in Figure 7.8 the interceptedcurrent is ill defined. Consider first surface 1. The current intercepted by surface 1 is equal to I.Surface 2 is also spanned by the Amperian loop, but the current intercepted by this loop is zero. Wethus conclude that Ampere's law does not apply in systems where the current is not continuous.

Figure 7.8. Charging a capacitor. Maxwell modified Ampere's law in the following manner:

The term added by Maxwell is called the displacement current. It is defined as

Consider the region between the capacitor plates in Figure 7.8. The electric field in this region is equalto

where we have assumed that the field produced is that of an ideal capacitor with surface area A andthe z axis is in the direction of the current. The rate of change of the electric field is equal to

The surface integral of across surface 2 is therefore equal to

The surface integral of across surface 2 is equal to



The modification of Ampere's law by Maxwell insures that the surface integral of is independentof the surface chosen. In electrostatics and magnetostatics the electric and magnetic fields are constantin time, and therefore, the new form of Ampere's law reduces to the form of Ampere's law we havebeen using so far. In a region where there are no free charges or free currents Maxwell's equations become verysymmetric

The symmetry is broken when electric charges are present, unless besides electric charges there aremagnetic monopoles. If the magnetic charge density is equal to η and the magnetic current is equal to then Maxwell's equation become

To obtain Maxwell's equations that describe the electric and magnetic fields in matter we must takethe bound charges and bound currents into account:

In the non-static case, the polarization can be time dependent. Therefore, also the bound chargedensity is time dependent, and a net current can be associated with the change in the bound chargedensity. This current is called the polarization current and is equal to

Maxwell's equations in matter are therefore equal to

It is common to rewrite Maxwell's equations in terms of the parameters we can control (the freecharge density and the free current density). Gauss's law can be rewritten as

where is called the electric displacement. Ampere's law can be rewritten as



where is called the H field. The most general form of Maxwell's equations, in terms of the freecharges and free currents, is given by


http://teacher.pas.rochester.edu/PHY217/Exams/MidtermExam1/MidtermExam1.html 1/1

Midterm Exam # 1, Physics 217October 17, 2001, 8.30 am – 9.50 am

Problem 1 (35 points)Check Stoke's theorem for the vector function

using the triangular surface shown in Figure 1. Express all your answers in terms of a.

Figure 1. Problem 1.

Problem 2 (30 points)A spherical shell of radius R has a uniform surface charge density. The total charge on the shell is q.Find the electrostatic energy of this system in two different ways. Express all your answers in terms ofq and R.

Problem 3 (35 points)Consider the following three vector functions:

(1)

(2)

(3)

where α is a constant with the appropriate units.a) Which of these three vector functions can describe an electrostatic field?b) For the vector function of part a) that can describe an electrostatic field find the correspondingelectrostatic potential at point P(x, y, z), using the origin as your reference point.c) Find the charge distribution that produces the electrostatic potential obtained in part b).Express all your answers in terms of α, x, y, and z.

12/21/2015 Homework


Midterm Exam # 2, Physics 217November 28, 2001, 8.30 am – 9.50 am

Problem 1 (35 points)A certain coaxial cable consists of a copper wire, of radius a, surrounded by an infinitesimal thinconcentric copper tube of radius c (see Figure 1). The charge on the wire is λ C/m and the charge onthe tube is -λ C/m. The space between the wire and the tube is partially filled (from b to c) with alinear dielectric of susceptibility χe.a) What is the magnitude and direction of the electric displacement in the three regions a < r < b, b < r< c, and c < r?b) What is the magnitude and direction of the electric field in the three regions a < r < b, b < r < c, andc < r?c) What is the capacitance per unit length of this cable?


Problem 2 (35 points)Consider a circular current loop of radius R, lying in the xy plane, and carrying a current I in thedirection indicated (see Figure 2).a) Find the exact magnetic field (magnitude and direction) a distance z above the center of the currentloop.b) What is the magnetic dipole moment of the current loop?c) Verify that for z » R the exact magnetic field calculated in a) is consistent with the field of amagnetic dipole.


Problem 3 (35 points)A uniform line charge λ is placed on an infinite straight wire, a distance d above a groundedconducting plane. The wire runs parallel to the x axis and directly above it. The conducting plane isthe xy plane.

12/21/2015 Homework


a) Find the potential in the region above the grounded plane.b) Find the charge density σ induced on the conducting plane.Note: The potential generated by a uniform line charge λ on an infinite straight wire is equal to

where r is the distance from the line charge.

12/21/2015 Homework

http://teacher.pas.rochester.edu/PHY217/Exams/FinalExam/FinalExam.html 1/4

Final Exam, Physics 217Problem 1 (20 points)Problem 2 (20 points)Problem 3 (20 points)Problem 4 (20 points)Problem 5 (20 points)Problem 6 (20 points)

Final Exam, Physics 217

December 20, 2001, 12.30 pm – 3.30 pm1. Answer questions 1, 2, and 3 in exam book # 1. Answer questions 4, 5, and 6 in exam book # 2.

2. Each question is worth 20 points: Yes, your algebra is correct, and the total points you can score is120 (a 20 point bonus if you answer all 6 questions).

3. Each answer needs to be well motivated. You will not receive any credit for just the answer (even ifit is correct) if no motivation is provided.

4. Here are some useful relations:

The general solution of the following differential equation

is

where a is a constant.

5. The grades will be distributed via email on or before 12/24.

6. Have a good and save holiday, and best wishes for 2002.

7. All complaints/comments/questions about the exam and the course should be directed to theinstructor.

Problem 1 (20 points)

12/21/2015 Homework


Find the force on the charge +q, shown in Figure 1, which is located a distance of 3d above aninfinitely large grounded conducting plane, which is located in the xy plane.

Figure 1. Problem 1.Express your answer in terms of d and q. Make sure you specify both the magnitude and the directionof the force.


Consider an infinitely long straight wire of radius R, directed along the z axis, which carries a currentI.

a) What is the magnetic field at a distance r from the center of the wire (r > R)?

b) Find the vector potential at a distance r from the center of the wire (r > R).

c) Show that your solution in part b) is correct by verifying that .

d) Show that your solution in part b) is correct by verifying that .

Express all your answers in terms of R, I, and r.


The electrostatic potential of some charge configuration is given by the expression

where A and λ are constants.

a) Find the electric field .

12/21/2015 Homework


b) Find the charge density .

c) Find the total charge Q.

Express all your answers in terms of A and λ.


A point charge q is imbedded at the center of a sphere of linear dielectric material (with susceptibilityχe and radius R).

1. Find the electric field inside the sphere (r < R).

b) Find the polarization inside the sphere (r < R).

c) Find the bound volume and surface charge densities ρb and σb.

d) What is the total bound surface charge on the surface of the sphere?

e) Where is the opposing bound volume charge located?

Express all your answers in terms of q, χe, and R.


A certain transmission line in constructed from two thin metal “ribbons” of width w, a very smalldistance h << w apart (see Figure 2). A current I travels down one strip and back along the other. Ineach case, the current density across the ribbon is uniform.


a) Find the capacitance per unit length C.

b) Find the inductance per unit length L.

c) What is the product of C and L?

12/21/2015 Homework


d) If the strips are insulated from one another by a non-conducting material of permittivity ε andpermeability μ, what will the product of C and L be?

Express your answers in terms of h, w, ε, and μ.


An infinitely long cylinder, of radius R, carries a "frozen-in" magnetization, parallel to the axis,

where k is a constant and r is the distance from the axis (there is no free current anywhere). Find themagnetic field inside and outside the cylinder by two different methods:

a) Locate all the bound currents, and calculate the field they produce.

b) Use Ampere's law to find , and then get .

Express you answers in terms of k and R.

sem5 em electomagnet ph217 note2to7 exam1n2nfinal

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