gauss’ law - carleton university · 2008-09-08 · gauss’ law is true for any surface enclosing...

Gauss’ Law Field Lines Electric Flux

Recall that we defined the electric field to be the force per unit charge at a particular point:

For a source point charge Q and a test charge q at point P

Q

q

at P

P

If Q is positive, then the field is directed radially away from the charge.

Note: direction of arrows Note: spacing of lines Note: straight lines

+

Note: direction of arrows Note: spacing of lines Note: straight lines

If Q is negative, then the field is directed radially towards the charge.

Negative Q implies anti-parallel to

-

Field lines were introduced by Michael Faraday to help visualize the direction and magnitude of the electric field. The direction of the field at any point is given by the direction of the field line, while the magnitude of the field is given qualitatively by the density of field lines. In the above diagrams, the simplest examples are given where the field is spherically symmetric. The direction of the field is apparent in the figures. At a point charge, field lines converge so that their density is large - the density scales in proportion to the inverse of the distance squared, as does the field. As is apparent in the diagrams, field lines start on positive charges and end on negative charges. This is all convention, but it nonetheless useful to remember.

+ +

This figure portrays several useful concepts. For example, near the point charges (that is, at a distance that is small compared to their separation), the field becomes spherically symmetric. This makes sense - near a charge, the field from that one charge certainly should dominate the net electric field since it is so large. Along a line (more accurately, a plane) bisecting the line joining the charges, we see that the field is directed along the -x direction as shown.

- +

In this case, we see the zero-field region precisely between the two charges, and we also see a fairly rapid convergence on a spherically symmetric distribution of field lines.

+ +

Example (Question) The figure shows the electric field lines for a system of

two point charges.

(a) What are the relative magnitudes of the charges? (b) What are the signs of the charges? (c) In what regions of space is the electric field strong? weak?

Example (Solution) The figure shows the electric field lines for a system of two point charges.


a) There are 32 lines coming from the charge on the left, while there are 8 converging on that on the right. Thus, the one on the left is 4 times larger than the one on the right.



• The one on the left is positive; field lines leave it. The one on the right is negative; field lines end on it.



• The field is strong near both charges. It is strongest on a line connecting the charges. Few field lines are drawn there, but this is for clarity. The field is weakest to the right of the right hand charge.

END

How would you measure 'the density of electric field lines' in a vicinity of space? First think only of a discrete set of electric field lines. One o b v i o u s a n s w e r t o t h e question is that you would count the number of lines passing through an imaginary geometr ica l (not rea l ! ) surface.

The first two are obvious, and the following diagram will indicate the last:

Uniform electric field - the lines have constant density and are all parallel

Surface is inserted in three different orientations and intersects the electric field lines.

The maximum number o f f i e l d l i n e s i s intercepted when the surface is perpendicular to the field, while no field lines will pass through it when the surface is parallel to the field.

N u m b e r o f f i e l d l i n e s passing through surface is large

N u m b e r o f f i e l d l i n e s passing through surface is zero

Number of field l i n e s p a s s i n g through surface is between zero and the maximum. Depends on the orientation angle θ.

Electric flux a function of surface orientation, … relative to the electric field

In general, the number of field lines passing through an area A is directly proportional to cos(θ), where θ is the angle between the electric field vector and the normal to the surface vector.

Use this symbol to represent electric flux through a surface

The electric flux through the blue surface is the same in these two figures. Electric flux is proportional to the product of electric field and surface area A.

Increasing electric field

Increasing surface area

The number of field lines passing through a geometrical surface of given area depends on three things: the field strength, the area, and the orientation of the surface.

[E]-Electric field; Newton/Coulomb {N/C}

[A]-Surface area; meter2 {m2}

[θ]-Angle; degrees or radians {o, rad }

[ΦE]-Electric flux; Newton meter2/Coulomb {Nm2/C}

Example (Question) A uniform electric field of 2.1 kN/C passes through a rectangular area 22 cm by 28 cm. The field makes an angle of 30o with the normal to the area. Determine the electric flux through the rectangle.

Area A

Example (Solution) A uniform electric field of 2.1 kN/C passes through a rectangular area 22 cm by 28 cm. The field makes an angle of 30o with the normal to the area. Determine the electric flux through the rectangle.

22 cm 28 cm

30o 2.1kN/C

0.22 m X0.28 m = 0.062 m2

END

It is useful also to represent the area A by a vector. The length of the vector is given by the area (a scalar quantity), while the orientation is perpendicular to the area. With this definition, the flux can be defined as:

recall

A slightly different notation for electric flux:

Unit vector perpendicular to surface of area A

q

Surface

Element of surface area dA

WHAT IF THE SURFACE IS CURVED AND/OR THE FIELD VARIES WITH POSITION?

Consider the surface as made up of small elements over which the

electric field is uniform.

Flux for Non-Uniform Fields / Flux for Non-Uniform Surface

You might have noticed that all these equations really only work for uniform electric fields.

We can use them here provided we make them pertain to differential (small) area elements, and over a differential area

the field is uniform.

We then need to integrate (sum) to get the total flux through an extended surface in a non-uniform field.

As before, dA is a vector oriented perpendicular to the area, and the area itself is differential (i.e., it's infinitesimally small and it's shape doesn't usually matter). The total electric flux can be evaluated by integrating this differential flux over the surface.


The differential electric flux passing through a differential area is given by:


q

Surface

Divide surface into small elements dA

Flux for one segment

THE INTEGRAL IS TAKEN OVER THE ENTIRE SURFACE.

Total flux through surface


THE BASIC DEFINITION OF ELECTRIC FLUX

Example (Question)

A positive test charge of magnitude 3.0µC is surrounded by a sphere with radius 0.20 m centered on the charge. Find the electric flux through the sphere due to this charge.

+

At any point on the sphere the magnitude of the electric field is:

Point charge electric field at distance r from charge

E = 6.75 X 105 N/C

+

From symmetry the field is perpendicular to the spherical surface at every point.

Same direction on the surface

+

The electric field has the same magnitude over the spherical surface. It is thus constant

and may be taken outside the integral

The integral that remains is just the surface area of the sphere.

+

The electric flux through the spherical surface is:

E = 6.75 X 105 N/C

END

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Example (Question)

A positive test charge of magnitude 3.0µC is surrounded by a sphere with radius 0.20 m centered on the charge. Find the electric flux through the sphere due to this charge.

Consider the same example again, but this time we save the calculations for the end.

+

At any point on the sphere the magnitude of the electric field is:

Point charge electric field at distance r from charge

+

The electric field has the same magnitude over the spherical surface. It is thus constant

and may be taken outside the integral

The integral that remains is just the surface area of the sphere.

+

The electric flux through the spherical surface is:

€

ΦE =qεo

+

END

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Gauss' law is a generalization of the results discussed above for the single charge and spherical surface.

It relates the electric flux passing through any surface enclosing a charge distribution to the net charge enclosed.

GAUSS’ LAW

It relates the electric flux passing through any surface enclosing a

charge distribution to the net charge enclosed.

The flux through the two spheres is the same since they enclose the

same charge.

GAUSS’ LAW

It relates the electric flux passing through any surface

enclosing a charge distribution to the net charge

enclosed.

The flux through the two surfaces is the same since they

enclose the same charge.

GAUSS’ LAW

Each field line that enters the surface also exits the surface.

The flux through the surface is zero since no charge is

enclosed.

GAUSS’ LAW

+

Flux through spherical surface is zero.

YOU CAN REASON

Each field line that exits the surface eventually re-enters the

surface.

The net charge enclosed is zero.

GAUSS’ LAW

- +

IN GAUSS’ LAW, THE CLOSED SURFACE IS REFERRED TO AS A GAUSSIAN SURFACE +

USING GAUSS’ LAW

Gauss’ law is true for any surface enclosing any charge distribution. When the charge distribution has sufficient symmetry we can chose a surface --- Gaussian Surface --- over which the evaluation of the flux integral becomes simple. Gauss’ law allows us to calculate the field far more easily than we could using Coulomb’s law and superposition.

Step 1: Study the symmetry to see if you can construct a Gaussian surface on which the field magnitude and its direction relative to the surface are constant.

Using Gauss’ Law

Through step 1 the dot product in the integral can be replaced by:

Then Ecos(θ) is constant and can be removed from inside the integral:

Step 2: Evaluate the flux. This should be easy because of your choice of the Gaussian surface.

Using Gauss’ Law

Step 3: Evaluate the enclosed charge. This is not the same as the total charge if the Gaussian surface lies within the charge distribution.

Using Gauss’ Law

The Gaussian surface is an imaginary surface that you place in the charge distribution. Choosing the Gauusian surface shape and location depends on the symmetry of the charge distribution AND on what exactly you want to calculate.

Step 4: Equate the flux

and solve for E. The direction of the electric field vector should be evident for the symmetry and polarity of the charges.

Using Gauss’ Law

In step 1 you reasoned the direction of E. Now go back and use that information.

ELECTRIC FIELD FROM A SPHERICAL CHARGE DISTRIBUTION.

Charged sphere ρV

In this example we will follow the steps in applying Gauss’ law in order to obtain an expression for the electric field at a point external to the uniform spherical charge distribution.

Finding the electric field using Coulomb’s law and superposition is lengthy and involves messy math. Using Gauss’ law makes the task easy, but, we must assume the field is radial.

R

Example (Question)


Charged sphere ρV

P

Step 1: Study the symmetry to see if you can construct a Gaussian surface on which the field magnitude and its direction relative to the surface are constant. Imaginary Gaussian surface (sphere of radius r)

Assume a radial electric field

normal to r

Example (Solution)


Step 2: Evaluate the flux. This should be easy because of your choice of the Gaussian surface. The integral is just the area of the Gaussian surface.

P dA

Example (Solution)



P

r

R

Charged sphere ρV Imaginary Gaussian surface

All of the spherical charge distribution is contained inside the spherical Gaussian surface.

Total charge enclosed

Example (Solution)


P

r

R



Example (Solution)


Combining results and solving for the electric field

P

Direction of E from symmetry of charge distribution

Example (Solution)



P

R r

Integral over charge distribution

Integral over Gaussian surface

SOLVING FOR THE ELECTRIC FIELD E

Example (Solution)


Writing electric field expression in terms of total charge q inside Gaussian surface

P

R r

Integral over charge distribution



END

Example (Solution)


Charged sphere ρV

In this example we will follow the steps in applying Gauss’ law in order to obtain an expression for the electric field at a point internal to the uniform spherical charge distribution.

Finding the electric field using Coulomb’s law and superposition is lengthy and involves messy math. Using Gauss’ law makes the task easy, but, we must assume the field is radial.

R

Example (Question)


Charged sphere ρV

Step 1: Study the symmetry to see if you can construct a Gaussian surface on which the field magnitude and its direction relative to the surface are constant. Imaginary Gaussian surface (sphere of radius r)

Assume a radial electric field

normal to

Example (Solution)

P

R

r


Step 2: Evaluate the flux. This should be easy because of your choice of the Gaussian surface. The integral is just the area of the Gaussian surface.

dA

Example (Solution)



Charged sphere ρV Imaginary Gaussian surface

Only the charge inside the Gaussian surface is enclosed by the Gaussian surface. All charge external to the Gaussian surface is not included.

Total charge enclosed

Example (Solution)

P

R

r

Volume enclosed by Gaussian surface



Only the charge inside the Gaussian surface is enclosed by the Gaussian surface. All charge external to the Gaussian surface is not included.

Example (Solution)

P

R

r External charge does not contribute to flux

External charge

Gaussian surface

E field lines intersect surface twice contribution to flux cancel


P

r

R



Example (Solution)

V = volume enclosed by Gaussian surface



P

Direction of E from symmetry of charge distribution

Example (Solution)



I n t e g r a l o v e r c h a r g e distribution contained inside Gaussian surface



Example (Solution)

P R

r

END

P P

r R

Electric field expression at a point outside and inside the spherical charge distribution

OUTSIDE INSIDE

Direction of E is radial

R r

ELECTRIC FIELD FROM A INFINITE FLAT CHARGE DISTRIBUTION.

Charged surface ρS

In this example we will follow the steps in applying Gauss’ law in order to obtain an expression for the electric field at a point above a uniformly charged infinite flat surface.

Finding the electric field using Coulomb’s law and superposition is lengthy and involves messy math. Using Gauss’ law makes the task easy, but, we must assume the field is uniform and normal to the flat surface.

Example (Question)

x

y z

Charged surface ρS

In this example we will follow the steps in applying Gauss’ law in order to obtain an expression for the electric field at a point above a uniformly charged infinite flat surface.

Example (Question)

View of the flat surface with electric field lines drawn. Field is uniform and normal to the flat surface.

ELECTRIC FIELD FROM A INFINITE FLAT CHARGE DISTRIBUTION.

+ + + + + +

+ + + + + +

+ + + + + +


Example (Solution) ELECTRIC FIELD FROM A INFINITE FLAT CHARGE DISTRIBUTION.

One possibility is to use a cylindrical surface with the top and bottom surfaces parallel to the charged plane. The E field will be normal to the top and bottom surfaces.

The E field is parallel to the side of the cylinder and as such no field lines (flux) pass through it.

This type of surface is often called a Gaussian pill box.

+ + + + + +

+ + + + + +

+ + + + + +

A1 A2 A3



Surface S contains three parts (Top A3, Bottom A1 and Side A2)

Top Bottom Side

+ + + + + +

+ + + + + +

+ + + + + +

A1 A2 A3



Top flux

parallel Uniform over A3

Outward pointing

+ + + + + +

+ + + + + +

+ + + + + +

A1 A2 A3



Bottom flux

parallel Uniform over A1

Outward pointing

+ + + + + +

+ + + + + +

+ + + + + +

A1 A2 A3



Side flux

perpendicular Contained in side surface

Outward pointing

+ + + + + +

+ + + + + +

+ + + + + +

A1 A2 A3



Top Bottom Side

Total flux through Gaussian surface is equal to flux through top an bottom surfaces.

+ + + + + +

+ + + + + +

+ + + + + + A1

A2 A3



T h e t o t a l c h a r g e enclosed consists of only the charge on the disk contained within the Gaussian Surface.

Charged surface ρS

Area A=A1=A3

Disk of charged contained inside Gaussian surface

+ + + + + +

+ + + + + +

+ + + + + +

A1 A2 A3




+ + + + + +

+ + + + + +

+ + + + + +

A1 A2 A3


Solve for electric field

Direction of E determined from symmetry

END

+ + + + + +

+ + + + + +

+ + + + + +

A1 A2 A3

ELECTRIC FIELD FROM A LONG STRAIGHT LINE OF CHARGE.

In one of the examples of lecture 2 we calculated the magnitude and direction of the electric field from a long straight line of charge using Coulomb’s law and superposition. Now we will show how Gauss’ law can be used to obtain the same result, in a few simple steps.

Example (Question)

Uniform charge density

ELECTRIC FIELD FROM A LONG STRAIGHT LINE OF CHARGE.


Imaginary Gaussian surface

Chose a cylinder of length and radius R E normal and constant to surface A2 E parallel and to surfaces A1 and A3

Example (Solution)

ELECTRIC FIELD FROM A LONG STRAIGHT LINE OF CHARGE. Step 2: Evaluate the flux. This should be easy because of your choice of the Gaussian surface.

Surface S contains three parts (Front A1, Back A3 and

Side A2)

Front Back Side

Example (Solution)


Flux front

perpendicular Contained in front surface

Outward pointing

Example (Solution)


Flux back

perpendicular Contained in back surface

Outward pointing

Example (Solution)


Flux side

parallel Normal to side surface (constant on surface)

Outward pointing

Example (Solution)

ELECTRIC FIELD FROM A LONG STRAIGHT LINE OF CHARGE. Step 3: Evaluate the enclosed charge. This is not the same as the total charge if the Gaussian surface lies within the charge distribution.

Example (Solution)


The total enclosed by the Gaussian surface is only the segment of the line within the bounds of the front and back surfaces.

ELECTRIC FIELD FROM A LONG STRAIGHT LINE OF CHARGE. Example (Solution)




ELECTRIC FIELD FROM A LONG STRAIGHT LINE OF CHARGE. Step 4: Equate the flux and solve for E.

Example (Solution)

ELECTRIC FIELD FROM A LONG STRAIGHT LINE OF CHARGE. Step 4: Equate the flux and solve for E. The direction of the electric field vector should be evident for the symmetry and polarity of the charges.

Example (Solution)

€

E =ρ

εoA2

Surface area of a cylinder

Note R radial distance from charged line

END

gauss’ law - carleton university · 2008-09-08 · gauss’ law is true for any surface enclosing...

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