chapter 24 - kfupm 24.pdfchapter 24 electric potential 1 1. electric potential energy when an...
TRANSCRIPT
Chapter 24Electric Potential
1
1. Electric Potential Energy
When an electrostatic force acts between two or more charged particles within asystem of particle, we can assign an electric potential energy π to the system. Ifthe system changes its configuration from state π to state π, the electrostatic forcedoes work π on the particles. The resulting change βπ in potential energy of thesystem is
βπ = ππ β ππ = βπ.
The work done by the electrostatic force is path independent, since theelectrostatic force is conservative.
We take the reference configuration of a system of charged particles to be that inwhich the particles are all infinitely separated from each other. The energy of thisreference configuration is set to be zero.
2
1. Electric Potential Energy
Suppose that several charged particles come together from initially infiniteseparations (state π) to form a system of neighboring particles (state π). Let theinitial potential energy ππ be zero and let πβ represents the work done by theelectrostatic force between the particles during the process. The final potentialenergy π of the system is
π = βπβ.
3
1. Electric Potential Energy
4
(a) Negative work.
(b) Increases.π = ββπ
1. Electric Potential Energy
Example 1: Electrons are continually being knocked outof air molecules in the atmosphere by cosmic-rayparticles coming in from space. Once released, eachelectron experiences an electrostatic force πΉ due tothe electric field πΈ that is produced in the atmosphereby charged particles already on Earth. Near Earthβssurface the electric field has the magnitude πΈ = 150 N/C and is directed downward. What is the change βπin the electric potential energy of a released electronwhen the electrostatic force causes it to movevertically upward through a distance π = 520 m?
5
1. Electric Potential Energy
The work π done on the electron by the electrostaticforce is
π = πΉ β π = πΉπ cos 0 = πΈππ
= 150 N/C 1.60 Γ 10β19 C 520 m
= 1.2 Γ 10β14 J.
βπ = βπ = β1.2 Γ 10β14 J.
6
2. Electric Potential
The potential energy of a charged particle in an electric field depends on the chargemagnitude. The potential energy per unit charge, however, has a unique value atany point in an electric field.
For example, a proton that has potential energy of 2.40 Γ 10β17 J in an electricfield has potential per unit charge of
2.40 Γ 10β17 J
1.60 Γ 10β19 C= 150 J/C.
Suppose we next replace the proton with an alpha particle (charge +2π). The alphaparticle would have electric potential energy of 4.80 Γ 10β17 J. However, thepotential energy per unit charge would be the same 150 J/C.
7
2. Electric Potential
The potential energy per unit charge π/π is a characteristic only of the electricfield. The potential energy per unit charge at a point in an electric field is called theelectric potential π (or simply potential) at that point. Thus,
π =π
π.
Note that π is a scalar, not a vector.
The electric potential difference βπ between any two points π and π in an electricfield is equal to the difference in potential energy per unit charge between the twopoints:
βπ = ππ β ππ =πππβπππ=βπ
π.
8
2. Electric Potential
Substitutingβπ for βπ we find that
βπ = ππ β ππ = βπ
π.
The potential difference between two points is the negative of the work done bythe electrostatic force on a unit charge as it moved from one point to the other.
If we set ππ = 0 at infinity as our reference potential energy, π must be zero theretoo. The electric potential at any point in an electric field is thus
π =π
π= βπβπ.
The SI unit for potential π is joule per coulomb or volt (V):
1 volt = 1 joul per coulomb.9
2. Electric Potential
In terms of this unit, we can write the electric field in a more convenient unit:
1N
C= 1N
C
1 V
1 J/C
1 J
1 N β m= 1V
m.
We can also define an energy unit that is convenient for energy scale of the atomicphysics; the electron-volt (eV). One electron-volt is the energy equal to the workrequired to move a single elementary charge π through a potential difference ofone volt (πβπ). Thus
1 eV = π 1 V = 1.60 Γ 10β19 C 1 J/C = 1.60 Γ 10β19 J.
10
2. Electric Potential
Work Done by an Applied Force
Suppose we move a particle of charge π from point π to point π in an electric fieldby applying a force to it. By the work-kinetic energy theorem,
βπΎ = πapp +π,
whereπapp is the work done by the applied force andπ is the work done by theelectric field.
If βπΎ = 0, we obtain that
πapp = βπ.
Using that βπ = βπ, we obtain that
πapp = βπ = πβπ.
11
2. Electric Potential
12
(a) Positive work.
(b) Higher potential.
πapp = βπ = πβπ
3. Equipotential Surfaces
An equipotential surface is composed of adjacent points that have the sameelectric potential. No net workπ is done on a charged particle by an electric fieldwhen the particleβs initial point π and final point π lie on the same equipotentialsurface.
13
3. Equipotential Surfaces
14
π1 = 100 Vπ2 = 80 Vπ3 = 60 Vπ4 = 40 V
3. Equipotential Surfaces
15
3. Equipotential Surfaces
Equipotential surfaces are perpendicular to the electric field πΈ.
If πΈ were not perpendicular to an equipotential surface, it would have acomponent along the surface. This component would then do work on a chargedparticle as it moved along the surface.
16
4. Calculating the Potential from the Field
We can calculate the potential differencebetween any two points π and π if we knowthe electric field vector πΈ along any pathconnecting those points.
Consider the situation shown in the figure.The differential work ππ done on theparticle by the electric field as it movesthrough a differential displacement π π is
ππ = πΉ β π π = π0πΈ β π π .
17
4. Calculating the Potential from the Field
The total work π done between points πand π is
π = π0 π
π
πΈ β π π .
Using that π = ββπ = βπ0βπ we obtainthat
ππ β ππ = β π
π
πΈ β π π .
18
4. Calculating the Potential from the Field
If we set ππ = 0 and rewrite ππ as π we get
π = β π
π
πΈ β π π .
This equation gives us the potential π atany point π in an electric field relative tozero potential at point π. If we let point π beat infinity, this equation gives us thepotential π at any point π relative to zeropotential at infinity.
19
4. Calculating the Potential from the Field
If we set ππ = 0 and rewrite ππ as π we get
π = β π
π
πΈ β π π .
This equation gives us the potential π atany point π in an electric field relative tozero potential at point π. If we let point π beat infinity, this equation gives us thepotential π at any point π relative to zeropotential at infinity.
20
4. Calculating the Potential from the Field
Uniform Field: Letβs find the potentialdifference between points π and π in thecase of a uniform electric field. We obtain
ππ β ππ = β π
π
πΈ β π π = β π
π
πΈππ cos 0
= βπΈ π
π
ππ = βπΈβπ₯.
The is the change in voltage βπ between twoequipotential lines in a uniform field ofmagnitude πΈ, separated by distance βπ₯
βπ = βπΈβπ₯.21
4. Calculating the Potential from the Field
The electric field vector points from higherpotential toward lower potential. This is truefor any electric field.
22
4. Calculating the Potential from the Field
23
(a) Rightward. (b) Positive: 1,2,3,5. Negative: 4. (c) 3 then 1,2,5 tie, then 4.
πapp = πβπ = βπβπ
4. Calculating the Potential from the Field
Example 2:
(a) The figure shows two points π and π in a uniform electricfield πΈ. The points lie on the same electric field line (notshown) and are separated by a distance π. Find the potentialdifference ππ β ππ by moving a positive test charge π0 from π toπ along the path shown, which is parallel to the field direction.
πΈ β π π = πΈππ cos 0 = πΈππ .
ππ β ππ = β π
π
πΈ β π π = βπΈ π
π
ππ = βπΈπ.
The potential always decreases along a path that extends inthe direction of the electric field lines.
24
4. Calculating the Potential from the Field
Example 2:
(b) Now find the potential difference ππ β ππ bymoving the positive test charge π0 from π to π alongthe path πππ shown in the figure.
ππ β ππ = ππ β ππ .
Along path ππ
πΈ β π π = πΈππ cos 45Β° .
ππ β ππ = β π
π
πΈ β π π = βπΈ cos 45Β° π
π
ππ
= βπΈ cos 45Β° 2π = βπΈπ.
25
5. Potential Due to a Point Charge
We here want to derive an expression for the electricpotential π for the space around a charged particle, relativeto zero potential at infinity.
To do that, we move a positive test charge π0 from point π toinfinity.
We choose the simplest path connecting the two pointsβaradial straight line. Evaluating the dot product gives
πΈ β π π = πΈππ cos 0 = πΈ ππ =π
4ππ0
ππ
π2.
26
5. Potential Due to a Point Charge
The line integral becomes
ππ β ππ = β π
π π
4ππ0
ππ
π2= β
π
4ππ0 π
βππ
π2
= βπ
4ππ0β1
π π
β
= βπ
4ππ0
1
π .
Setting ππ = 0 (atβ), ππ = π and switching π to π we get
π =1
4ππ0
π
π.
Note that π can be positive or negative and that π has thesame sign as π.
27
5. Potential Due to a Point Charge
A positively charged particle produces a positive electricpotential. A negatively charged particle produces a negativeelectric potential.
Also, note that the expression for π holds outside or on thesurface of a spherically symmetric charge distribution.
28
6. Potential Due to a Group of Point Charges
The net potential due to π point charges at a given point is
π = π=1
π
ππ =1
4ππ0 π=1
π ππππ.
Here ππ is the charge (positive or negative) of the πth particle, and ππ is the radialdistance of the given point from the πth charge.
An important advantage of potential over electric field is that we sum scalarquantities instead of vector quantities.
29
6. Potential Due to a Group of Point Charges
All the same.
30
6. Potential Due to a Group of Point Charges
Example 3: What is the electric potential at point P,located at the center of the square of point chargesshown in the figure? The distance π = 1.3 m, and thecharges are
π1 = 12 nC, π2 = 31 nC,
π3 = β24 nC, π4 = 17 nC.
31
6. Potential Due to a Group of Point Charges
Example 3: What is the electric potential at point P,located at the center of the square of point chargesshown in the figure? The distance π = 1.3 m, and thecharges are
π1 = 12 nC, π2 = 31 nC,
π3 = β24 nC, π4 = 17 nC.
π = π=1
4
ππ =1
4ππ0
π1π+π2π+π3π+π4π
=1
4ππ0
π1 + π2 + π3 + π4π
32
6. Potential Due to a Group of Point Charges
π1 + π2 + π3 + π4 = 12 + 31 β 24 + 17 Γ 10β9 C
= 36 Γ 10β9 C.
π =2
2π.
Substituting gives
π = 8.99 Γ109N β m2
C23.6 Γ 10β8 C
1.3/ 2m
= 350 V.
33
6. Potential Due to a Group of Point Charges
Example 4: (a) In the figure electrons (of charge βπ) areequally spaced and fixed around a circle of radius π .Relative to π = 0 at infinity, what are the electricpotential and electric field at the center πΆ of the circledue to these electrons?
π = π=1
12
ππ = β1
4ππ0
12π
π
By the symmetry of the problem, πΈ = 0.
34
6. Potential Due to a Group of Point Charges
Example 4: (b) If the electrons are moved along the circleuntil they are nonuniformly spaced over a 120Β° arc (seethe figure), what then is the potential at πΆ? How does theelectric field at πΆ change (if at all)?
The potential is not changed because the distancesbetween πΆ and each charge are not changed.
The electric field is no longer zero, however, because thearrangement is no longer symmetric. A net field is nowdirected toward the charge distribution.
35
7. Potential Due to an Electric Dipole
The potential due to an electric dipole is the net potentialdue to the positive and negative charges.
At point π, the net potential is
π = π=1
2
ππ = π + + π β =1
4ππ0
π
π +βπ
π β
=π
4ππ0
π β β π +π + π β
.
36
7. Potential Due to an Electric Dipole
For π β« π, we can make the following approximation (seethe figure):
π β β π + β π cos π ,
andπ β π + β π
2.
Substituting in the expression for π gives
π =π
4ππ0
π cos π
π2=1
4ππ0
π cos π
π2.
37
7. Potential Due to an Electric Dipole
Point π, π then π.
38
π =1
4ππ0
π cos π
π2
7. Potential Due to an Electric Dipole
39
Induced Dipole Moment
7. Potential Due to an Electric Dipole
40
Induced Dipole Moment
8. Calculating the Field from the Potential
Let us see graphically how we can calculate the field from the potential. If we knowthe potential π at all points near a charge distribution, we can draw a family ofequipotential surfaces. The electric field lines, sketched perpendicular to thosesurfaces, reveal the variation of πΈ.
We want now to write the mathematical equivalence of this graphical procedure.
41
8. Calculating the Field from the Potential
The figure shows cross sections of a family of closelyspaced equipotential surfaces.
The potential difference between each pair ofadjacent surfaces is ππ. The electric field πΈ at anypoint π is perpendicular to the equipotential surfacethrough π.
Suppose that a positive test charge π0 movesthrough a displacement π π from an equipotentialsurface to the adjacent surface.
The work ππ done by the electric field on the testcharge is
ππ = βπ0ππ.
42
8. Calculating the Field from the Potential
The work ππ can also be written as
ππ = πΉ β π π = π0πΈ β π π = π0πΈππ cos π .
Equating the two expression for ππ yields
βπ0ππ = π0πΈππ cos π ,
or
πΈ cos π = βππ
ππ .
πΈ cos π is the component of πΈ in the direction ofπ π . Thus,
πΈπ = βππ
ππ .
43
8. Calculating the Field from the Potential
The component of πΈ in any direction is the negativeof the rate at which the electric potential changeswith distance in that direction.
If we take ππ to be along the π₯, π¦ and π§ axes, weobtain the π₯, π¦ and π§ components of πΈ at any point:
πΈπ₯ = βππ
ππ₯; πΈπ¦ = β
ππ
ππ¦; πΈz = β
ππ
ππ§.
If we know the function π π₯, π¦, π§ , we can find thecomponents of πΈ at any point.
44
8. Calculating the Field from the Potential
When the electric field πΈ is uniform, we write that
πΈ = ββπ
βπ .
where π is perpendicular to the equipotential surfaces. The component of theelectric field is zero in any direction parallel to the equipotential surfaces becausethere is no change in potential along an equipotential surface.
45
8. Calculating the Field from the Potential
πΈ = ββπ
βπ
(a) 2, then 3 & 1.
(b) 3.
(c) Acceleratesleftward.
46
8. Calculating the Field from the Potential
Example 5: Starting with the expression for the potential due to an electric dipole,
π =1
4ππ0
π cos π
π2.
derive an expression for the radial component of electric field due to the electricdipole.
πΈπ = βππ
ππ= βπ cos π
4ππ0
π
ππ
1
π2
= βπ cos π
4ππ0β2
π3=1
2ππ0
π cos π
π3.
47
8. Calculating the Field from the Potential
Example 6: The electric potential at any point on the central axis of a uniformlycharged disk is given by
π =π
2π0π§2 + π 2 β π§ .
Starting with this expression, derive an expression for the electric field at any pointon the axis of the disk.
πΈπ§ = βππ
ππ§=π
2π0
π
ππ§π§ β π§2 + π 2
=π
2π01 β
π§
π§2 + π 2.
48
9. Electric Potential Energy of a System of ChargesThe electric potential energy of a system of fixed point charges is equal to the workthat must be done by an external agent to assemble the system, bringing eachcharge in from an infinite distance.
We assume that the charges are stationary at both the initial and final states.
50
9. Electric Potential Energy of a System of ChargesLet us find the electric potential energy of the systemshown in the figure. We can build the system by bringingπ2 from infinity and put it at distance π from π1. The workwe do to bring π2 near π1 is π2π, where π is the potentialthat has been set up by π1. Thus, the electric potential ofthe pair of point charges shown in the figure is
π = πapp = π2π =1
4ππ0
π1π2π.
π has the same sign as that of π1π2.
The total potential energy of a system of several particlesis the sum of the potential energies for every pair ofparticles in the system.
51
9. Electric Potential Energy of a System of ChargesExample 7: The figure shows three point charges held infixed positions by forces that are not shown. What is theelectric potential energy π of this system of charges?Assume that π = 12 cm and that
π1 = +π, π2 = β4π, π3 = +2π,
and π = 150 nC.
The electric potential energy of the system (energy requiredto assemble the system) is
π = π12 + π13 + π23
=1
4ππ0
π1π2π+1
4ππ0
π1π3π+1
4ππ0
π2π3π.
52
9. Electric Potential Energy of a System of Charges
π =1
4ππ0
π β4π
π+1
4ππ0
π 2π
π+1
4ππ0
β4π 2π
π
= β10π2
4ππ0π= 8.99 Γ 1012
N β m2
C210 150 Γ 10β9C
0.12 m
= β17 mJ.
53
9. Electric Potential Energy of a System of ChargesExample 8: An alpha particle (two protons, two neutrons)moves into a stationary gold atom (79 protons, 118neutrons), passing through the electron region thatsurrounds the gold nucleus like a shell and headeddirectly toward the nucleus. The alpha particle slows untilit momentarily stops when its center is at radial distance π= 9.23 fm from the nuclear center. Then it moves backalong its incoming path. (Because the gold nucleus ismuch more massive than the alpha particle, we canassume the gold nucleus does not move.) What was thekinetic energy πΎπ of the alpha particle when it was initiallyfar away (hence external to the gold atom)? Assume thatthe only force acting between the alpha particle and thegold nucleus is the (electrostatic) Coulomb force.54
9. Electric Potential Energy of a System of ChargesDuring the entire process, the mechanical energy of thealpha particle-gold atom system is conserved.
πΎπ + ππ = πΎπ + ππ .
ππ is zero because the atom is neutral. ππ is the electricpotential energy of the alpha particle-nucleus system. Theelectron shell produces zero electric field inside it. πΎπ iszero.
πΎπ = ππ =1
4ππ0
2π 79π
9.23 Γ 10β15 m= 3.94 Γ 10β9 J
= 24.6 MeV.
55
10. Potential of a Charged Isolated Conductor
In Ch. 23, we concluded that πΈ = 0 within an isolated conductor. We then usedGaussβ law to prove that all excess charge placed on a conductor lies entirely on itssurface. Here we use the first of these facts to prove an extension of the second:
βAn excess charge placed on an isolated conductor will distribute itself on thesurface of that conductor so that all points of the conductorβwhether on thesurface or insideβcome to the same potential. This is true even if the conductorhas an internal cavity and even if that cavity contains a net charge.β
To prove this fact we use that
ππ β ππ = β π
π
πΈ β π π .
Since πΈ = 0 for all point within a conductor, ππ = ππ for all possible pairs of points πand π in the conductor.56
10. Potential of a Charged Isolated Conductor
The figure shows the potential against radialdistance π from the center from an isolatedspherical conducting shell of radius π = 1.0 m andcharge π = 0.10 πC.
For points outside the shell up to its surface, π π= π π/π. Now assume that we push a test chargethough a hole in the shell. No extra work is requiredto do this because πΈ becomes zero once the testcharge gets inside the shell. Thus, the potential atall points inside the shell has the same value as thatof its surface.
57
10. Potential of a Charged Isolated Conductor
The figure shows the variation of the electric fieldwith radial distance for the same shell. Everywhereinside the shell πΈ = 0. Outside the shell πΈ = π π/π2. The electric field can be obtained from thepotential using πΈ = βππ/ππ.
58
10. Potential of a Charged Isolated Conductor
Spark Discharge from a Charged Conductor
On nonspherical conductors, a surface charge doesnot distribute itself uniformly over the surface ofthe conductor. At sharp points or sharp edges, thesurface charge densityβand thus the externalelectric field, which is proportional to itβmay reachvery high values. The air around such sharp pointsor edges may become ionized, producing thecorona discharge.
59
10. Potential of a Charged Isolated Conductor
Isolated Conductor in an External Electric Field
If an isolated conductor is placed in an externalelectric field, all points of the conductor still cometo a single potential regardless of whether theconductor has an excess charge or not. The freeconduction electrons distribute themselves on thesurface in such a way that the electric field theyproduce at interior points cancels the externalelectric field. Furthermore, the electron distributioncauses the net electric field at all points on thesurface to be perpendicular to the surface.
64