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Chapter 4
The Electric Potential
4.1 The Important Stuff
4.1.1 Electrical Potential Energy
A charge q moving in a constant electric field E experiences a force F = qE from that field. Also, as we know from our study of work and energy, the work done on the charge by the field as it moves from point r 1 to r 2 is
W =
∫ (^) r 2
r 1
F · ds
where we mean that we are summing up all the tiny elements of work dW = F · ds along the length of the path. When F is the electrostatic force, the work done is
W =
∫ (^) r 2
r 1
qE · ds = q
∫ (^) r 2
r 1
E · ds (4.1)
In Fig. 4.1, a charge is shown being moved from r 1 to r 2 along two different paths, with ds and E shown for a bit of each of the paths. Now it turns out that from the mathematical form of the electrostatic force, the work done by the force does not depend on the path taken to get from r 1 to r 2. As a result we say
E d s d s E
r 1
r 2
q
q
Figure 4.1: Charge is moved from r 1 to r 2 along two separate paths. Work done by the electric force involves the summing up E · ds along the path.
54 CHAPTER 4. THE ELECTRIC POTENTIAL
that the electric force is conservative and it allows us to calculate an electric potential energy, which as usual we will denote by U. As before, only the changes in the potential have any real meaning, and the change in potential energy is the negative of the work done by the electric force:
∆U = −W = −q
∫ (^) r 2
r 1
E · ds (4.2)
We usually want to discuss the potential energy of a charge at a particular point, that is, we would like a function U(r), but for this we need to make a definition for the potential energy at a particular point. Usually we will make the choice that the potential energy is zero when the charge is infinitely far away: U∞ = 0.
4.1.2 Electric Potential
Recall how we developed the concept of the electric field E: The force on a charge q 0 is always proportional to q 0 , so by dividing the charge out of F we get something which can conveniently give the force on any charge. Likewise, if we divide out the charge q from Eq. 4.2 we get a function which we can use to get the change in potential energy for any charge (simply by multiplying by the charge). This new function is called the electric potential, V :
∆V =
∆U
q
where ∆U is the change in potential energy of a charge q. Then Eq. 4.2 gives us the difference in electrical potential between points r 1 and r 2 :
∆V = −
∫ (^) r 2
r 1
E · ds (4.3)
The electric potential is a scalar. Recalling that it was defined by dividing potential energy by charge we see that its units are (^) CJ (joules per coulomb). The electric potential is of such great importance that we call this combination of units a volt^1. Thus:
1 volt = 1 V = 1 (^) CJ (4.4)
Of course, it is then true that a joule is equal to a coulomb-volt=C · V. In general, multiplying a charge times a potential difference gives an energy. It often happens that we are multiplying an elementary charge (e) (or some multiple thereof) and a potential difference in volts. It is then convenient to use the unit of energy given by the product of e and a volt; this unit is called the electron-volt:
1 eV = (e) · (1 V) = 1. 60 × 10 −^19 C · (1 V) = 1. 60 × 10 −^19 J (4.5)
Equation 4.3 can only give us the differences in the value of the electric potential between two points r 1 and r 2. To arrive at a function V (r) defined at all points we need to specify
(^1) Named in honor of the... uh... French physicist Jim Volt (1813–1743) who did some electrical experi-
ments in... um... Bologna. That’s it, Bologna.
56 CHAPTER 4. THE ELECTRIC POTENTIAL
Using Eq. 4.10, one can show that the potential due to an electric dipole with magnitude p at the origin (pointing upward along the z axis) is
V (r) =
4 π 0
p cos θ r^2
Here, r and θ have the usual meaning in spherical coordinates.
4.1.6 Potential Due to a Continuous Charge Distribution
To get the electrical potential due to a continuous distribution of charge (with V = 0 at infinity assumed), add up the contributions to the potential; the potential due to a charge dq at distance r is dV = (^4) π^10 dq r so that we must do the integral
V =
4 π 0
∫ (^) dq
r
4 π 0
∫
V
ρ(r)dτ r
In the last expression we are using the charge density ρ(r) of the distribution to get the element of charge dq for the volume element dτ.
4.1.7 Potential Energy of a System of Charges
The potential energy of a pair of point charges (i.e. the work W needed to bring point charges q 1 and q 2 from infinite separation to a separation r) is
U = W =
4 π 0
q 1 q 2 r
For a larger set of charges the potential energy is given by the sum
U = U 12 + U 23 + U 13 +... =
4 π 0
∑ pairs ij
qiqj rij
Here rij is the distance between charges qi and qj. Each pair is only counted once in the sum.
4.2 Worked Examples
4.2.1 Electric Potential
- The electric potential difference between the ground and a cloud in a partic- ular thunderstorm is 1. 2 × 109 V. What is the magnitude of the change in energy (in multiples of the electron-volt) of an electron that moves between the ground and the cloud?
4.2. WORKED EXAMPLES 57
The magnitude of the change in potential as the electron moves between ground and cloud (we don’t care which way) is |∆V | = 1. 2 × 109 V. Multiplying by the magnitude of the electron’s charge gives the magnitude of the change in potential energy. Note that lumping “e” and “V” together gives the eV (electron-volt), a unit of energy:
|∆U| = |q∆V | = e(1. 2 × 109 V) = 1. 2 × 109 eV = 1.2 GeV
- An infinite nonconducting sheet has a surface charge density σ = 0. 10 μC/m^2 on one side. How far apart are equipotential surfaces whose potentials differ by 50 V?
In Chapter 3, we encountered the formula for the electric field due a nonconducting sheet of charge. From Eq. 3.5, we had: Ez = σ/(2 0 ), where σ is the charge density of the sheet, which lies in the xy plane. So the plane of charge in this problem gives rise to an E field:
Ez =
σ 2 0
=
(0. 10 × 10 −6 C m 2 2(8. 85 × 10 −^12 C 2 N·m^2 )
= 5. 64 × 10 3 N C
Here the E field is uniform and also Ex = Ey = 0. Now, from Eq. 4.7 we have
∂V ∂z
= −Ez = − 5. 64 × 10 3 N C.
and when the rate of change of some quantity (in this case, with respect to the z coordinate) is constant we can write the relation in terms of finite changes, that is, with “∆”s:
∆V ∆z
= −Ez = − 5. 64 × 10 3 N C
and from this result we can find the change in z corresponding to any change in V. If we are interested in ∆V = 50 V, then
∆z = −
∆V
Ez
(50 V)
(5. 64 × 10 3 N C )
= − 8. 8 × 10 −^3 m = − 8 .8 mm
i.e. to get a change in potential of +50 V we need a change in z coordinate of − 8 .8 mm. Since the potential only depends on the distance from the plane, the equipotential surfaces are planes. The distance between planes whose potential differs by 50 V is 8.8 mm.
- Two large, parallel conducting plates are 12 cm apart and have charges of equal magnitude and opposite sign on their facing surfaces. An electrostatic force of 3. 9 × 10 −^15 N acts on an electron placed anywhere between the two plates.
4.2. WORKED EXAMPLES 59
r= r r= 4
x
Figure 4.2: Path of integration for Example 5. Integration goes from r′^ = ∞ to r′^ = r.
(b) Using the result of part (a), the difference between values of V (r) on the sphere’s surface and at its center is
V (R) − V (0) = −
qR^2 8 π 0 R^3
q 8 π 0 R
(c) For q positive, the answer to part (b) is a negative number, so the center of the sphere must be at a higher potential.
- A charge q is distributed uniformly throughout a spherical volume of radius R. (a) Setting V = 0 at infinity, show that the potential at a distance r from the center, where r < R, is given by
V =
q(3R^2 − r^2 ) 8 π 0 R^3
(b) Why does this result differ from that of the previous example? (c) What is the potential difference between a point of the surface and the sphere’s center? (d) Why doesn’t this result differ from that of the previous example?
(a) We find the function V (r) just as we did the last example, but this time the reference point (the place where V = 0) is at r = ∞. So we will evaluate:
V (r) = −
∫ (^) r
rref
E · ds = −
∫ (^) r
∞
Er (r′) dr′^. (4.15)
The integration path is shown in Fig. 4.2. We note that the integration (from r′^ = ∞ to r′^ = r with r < R) is over values of r both outside and inside the sphere. Just as before, the E field for points inside the sphere is
Er, in(r) =
qr 4 π 0 R^3
but now we will also need the value of the E field outside the sphere. By Gauss‘(s) law the external E field is that same as that due to a point charge q at distance r, so:
Er, out(r) =
q 4 π 0 r^2
Because Er(r) has two different forms for the interior and exterior of the sphere, we will have to split up the integral in Eq. 4.15 into two parts. When we go from ∞ to R we need
60 CHAPTER 4. THE ELECTRIC POTENTIAL
to use Eq. 4.17 for Er(r′). When we go from R to r we need to use Eq. 4.16 for Er(r′). So from Eq. 4.15 we now have
V (r) = −
∫ (^) R
∞
Er, out(r′) dr′^ −
∫ (^) r
R
Er, in(r′) dr′
∫ (^) R
∞
( q 4 π 0 r′^2
) dr′^ −
∫ (^) r
R
( qr′ 4 π 0 R^3
) dr′
q 4 π 0
{∫ R ∞
dr′ r′^2
∫ (^) r
R
r′ R^3
dr′
}
Now do the individual integrals and we’re done:
V (r) = −
q 4 π 0
−^
r′
∣∣ ∣∣ ∣
R
∞
r′^2 2 R^3
∣∣ ∣∣ ∣
r
R
q 4 π 0
{ −
R
(r^2 − R^2 ) 2 R^3
}
q 4 π 0
( 2 R^2 2 R^3
(R^2 − r^2 ) 2 R^3
)
q(3R^2 − r^2 ) 8 π 0 R^3
(b) The difference between this result and that of the previous example is due to the different choice of reference point. There is no problem here since it is only the differences in electrical potential that have any meaning in physics.
(c) using the result of part (a), we calculate:
V (R) − V (0) =
q(2R^2 ) 8 π 0 R^3
q(3R^2 ) 8 π 0 R^3
= −
qR^2 8 π 0 R^3
q 8 π 0 R
This is the same as the corresponding result in the previous example.
(d) Differences in the electrical potential will not depend on the choice of the reference point, the answer should be the same as in the previous example... if V (r) is calculated correctly!
- What are (a) the charge and (b) the charge density on the surface of a con- ducting sphere of radius 0 .15 m whose potential is 200 V (with V = 0 at infinity)?
(a) We are given the radius R of the conducting sphere; we are asked to find its charge Q. From our work with Gauss’(s) law we know that the electric field outside the sphere is the same as that of a point charge Q at the sphere’s center. Then if we were to use Eq. 4.
62 CHAPTER 4. THE ELECTRIC POTENTIAL
V=400 V
r = 4
Q R
Figure 4.4: Conducting charged sphere, has potential of 400 V (with V = 0 at r = ∞), from Example 8.
so that with Er = 0, V must be constant throughout the interior of the spherical conductor. Since the value of V on the sphere itself is +400 V, V then must also equal +400 V at the center. So V = +400 V at the center of the sphere. (There was no calculating to do on this problem!)
- What is the excess charge on a conducting sphere of radius R = 0.15 m if the potential of the sphere is 1500 V and V = 0 at infinity?
The problem is diagrammed in Fig. 4.4. If the sphere has net charge Q then from Gauss’ law the radial component of the electric field for points outside the sphere is
Er = k
Q
r^2
Using Eq. 4.6 with r = ∞ as the reference point, the potential at distance R from the sphere’s center is:
V (R) = −
∫ (^) r
∞
Erdr = −
∫ (^) r
∞
kQ r^2
dr
kQ r
∣∣ ∣∣ ∣
R
∞
kQ R
kQ R
(Note that the integration takes place over values of r outside the sphere so that the expres- sion for Er is the correct one. Er is zero inside the sphere.) We are given that V (R) = 400 V, so from kQ/R = 400 V we solve for Q and get:
Q =
R(400 V)
k
(0.15 m)(400 V) (8. 99 × 10 9 N·m
2 C^2 )
= 2. 5 × 10 −^8 C
- The electric potential at points in an xy plane is given by
V = (2. (^0) mV 2 )x^2 − (3. (^0) mV 2 )y^2.
4.2. WORKED EXAMPLES 63
What are the magnitude and direction of the electric field at the point (3.0 m, 2 .0 m)?
Equations 4.7 show how to get the components of the E field if we have the electric potential V as a function of x and y. Taking partial derivatives, we find:
Ex = −
∂V
∂x
= −(4. (^0) mV 2 )x and Ey = −
∂V
∂y
= +(6. (^0) mV 2 )y.
Plugging in the given values of x = 3.0 m and y = 2.0 m we get:
Ex = − 12 V m and Ey = − 12 V m
So the magnitude of the E field at the given is
E =
√ (12.0)^2 + (12.0)2 V m = 17 (^) mV
and its direction is given by
θ = tan−^1
( Ey Ex
) = tan−^1 (1.0) = 135◦
where for θ we have made the proper choice so that it lies in the second quadrant.
4.2.2 Potential Energy of a System of Charges
- (a) What is the electric potential energy of two electrons separated by 2 .00 nm? (b) If the separation increases, does the potential energy increase or decrease?
Since the charge of an electron is −e, using Eq. 4.13 we find:
U =
4 π 0
(−e)(−e) r
=
4 π(8. 85 × 10 −^12 C
2 N·m^2 )
(1. 60 × 10 −^19 C)^2
(2. 00 × 10 −^9 m) = 1. 15 × 10 −^19 J
As the charges are both positive, the potential energy is a positive number and is inversely proportional to r. So the potential energy decreases as r increases.
- Derive an expression for the work required to set up the four-charge config- uration of Fig. 4.5, assuming the charges are initially infinitely far apart.
The work required to set up these charges is the same as the potential energy of a set of point charges, given in Eq. 4.14. (That is, sum the potential energies k q riijqj over all pairs of
4.2. WORKED EXAMPLES 65
A
B
q 1
q 2
Figure 4.7: Charge configuration for Example 12.
q^2 4 π 0 a
( −2 +
)
So now add up all the W ’s to get the total work done:
WTotal = W 1 + W 2 + W 3 + W 4
=
q^2 4 π 0 a
( − 1 − 1 +
)
q^2 4 π 0 a
( −4 +
)
This is a nice analytic answer; if we combine all the numerical factors (including the 4π) we get:
WTotal =
(− 0 .21)q^2 0 a
This is the same result as we’d get by using Eq. 4.14.
- In the rectangle of Fig. 4.7, the sides have lengths 5 .0 cm and 15 cm, q 1 = − 5. 0 μC and q 2 = +2. 0 μC. With V = 0 at infinity, what are the electric potentials (a) at corner A and (b) corner B? (c) How much work is required to move a third charge q 3 = +3. 0 μC from B to A along a diagonal of the rectangle? (d) Does this work increase or decrease the electric energy of the three–charge system? Is more, less or the same work required if q 3 is moved along paths that are (e) inside the rectangle but not on the diagonal and (f) outside the rectangle?
(a) To find the electric potential due to a group of point charges, use Eq. 4.10. Since point A is 15 cm away from the − 5. 0 μC charge and 5.0 cm away from the +2. 0 μC charge, we get:
V =
4 π 0
[ q 1 r 1
q 2 r 2
]
= (8. 99 × 10 9 N·m
2 C^2 )
[ (− 5. 0 × 10 −^6 C) (15 × 10 −^2 m)
(+2. 0 × 10 −^6 C)
(5. 0 × 10 −^2 m)
] = 6. 0 × 104 V
66 CHAPTER 4. THE ELECTRIC POTENTIAL
(b) Perform the same calculation as in part (a). The charges q 1 and q 2 are at different distances from point B so we get a different answer:
V = (8. 99 × 10 9 N·m
2 C^2 )
[ (− 5. 0 × 10 −^6 C) (5. 0 × 10 −^2 m)
(+2. 0 × 10 −^6 C)
(15 × 10 −^2 m)
] = − 7. 8 × 105 V
(c) Using the results of part (a) and (b), calculate the change in potential V as we move from point B to point A:
∆V = VA − VB = 6. 0 × 104 V − (− 7. 8 × 105 V) = 8. 4 × 105 V
The change in potential energy for a +3. 0 μC charge to move from B to A is
∆U = q∆V = (3. 0 × 10 −^6 C)(8. 4 × 105 V) = 2.5 J
(d) Since a positive amount of work is done by the outside agency in moving the charge from B to A, the electric energy of the system has increased. We can see that this must be the case because the +3. 0 μC charge has been moved closer to another positive charge and farther away from a negative charge.
(e) The force which a point charge (or set of point charges) exerts on a another charge is a conservative force. So the work which it does (or likewise the work required of some outside force) as the charge moves from one point to another is independent of the path taken. Therefore we would require the same amount of work if the path taken was some other path inside the rectangle.
(f) Since the work done is independent of the path taken, we require the same amount of work even if the path from A to B goes outside the rectangle.
- Two tiny metal spheres A and B of mass mA = 5.00 g and mB = 10.0 g have equal positive charges q = 5. 00 μC. The spheres are connected by a massless nonconducting string of length d = 1.00 m, which is much greater than the radii of the spheres. (a) What is the electric potential energy of the system? (b) Suppose you cut the string. At that instant, what is the acceleration of each sphere? (c) A long time after you cut the string, what is the speed of each sphere?
(a) The initial configuration of the charges in shown in Fig. 4.8(a). The electrostatic potential energy of this system (i.e. the work needed to bring the charges together from far away is
U =
4 π 0
q 1 q 2 r
= (8. 99 × 10 9 N·m 2 C^2 )
(5. 00 × 10 −^6 C)^2
(1.00 m)
= 0.225 J
We are justified in using formulae for point charges because the problem states that the sizes of the spheres are small compared to the length of the string (1.00 m).
68 CHAPTER 4. THE ELECTRIC POTENTIAL
Momentum conservation gives us the other equation that we need. If mass B has x– velocity vB then mass A has x–velocity −vA (it moves in the other direction. The system begins and ends with a total momentum of zero so then:
−mAvA + mB vB = 0 =⇒ vB =
mA mB
vA
Substitute this result into 4.18 and get:
1 2 mAv
2 A +^
1 2 mB
( m^2 A m^2 B
) v A^2 = 0.225 J
Factor out v A^2 on the left side and plug in some numbers:
1 2
( mA +
m^2 A mB
) v^2 A = (^12)
( 5 .00 g +
(5.00 g)^2 (10.0 g)
) v^2 A = (3. 75 × 10 −^3 kg)v A^2 = 0.225 J
So then we get the final speed of A:
v A^2 =
0 .225 J
- 75 × 10 −^3 kg
= 60. 0 m 2 s^2 =⇒^ vA^ = 7.^75
m s
and the speed of B:
vB =
mA mB
vA =
( 5 .00 g 10 .0 g
)
75 m s = 3. 87 m s
Two electrons are fixed 2 .0 cm apart. Another electron is shot from infinity and stops midway between the two. What is its initial speed?
The problem is diagrammed in Fig. 4.9(a) and (b). Since the electrostatic force is a conservative force, we know that energy is conserved between configurations (a) and (b). In picture (a) there is energy stored in the repulsion of the pair of electrons as well as the kinetic energy of the third electron. (Initially the third electron is too far away to “feel” the first two electrons.) In picture (b) there is no kinetic energy but the electrical potential energy has increased due to the repulsion between the third electron and the first two. If we can calculate the change in potential energy ∆U then by using energy conservation, ∆U + ∆K = 0 we can find the initial speed of the electron. The potential energy of a set of point charges (with V = 0 at ∞) is given in Eq. 4.14. When the third electron comes from infinity and stops at the midpoint, the increase in potential energy the contribution given by the third electron as it “sees” its new neighbors. With r = 1.0 cm, this increase is
∆U =
4 π 0
(−e)(−e) r
4 π 0
(−e)(−e) r
e^2 2 π 0 r
4.2. WORKED EXAMPLES 69
(a)
2.0 cm -e
-e
-e
v 0
1.0 cm
-e
-e
-e
(b)
v=
Figure 4.9: (a) Electron flies in from ∞ with speed v 0. (b) Electron comes to rest midway between the other two electrons.
The change in kinetic energy is ∆K = − 12 mv^20. Then energy conservation gives:
∆K = −∆U =⇒ − 12 mv 02 = −
e^2 2 π 0 r
Solve for v 0 :
v^20 =
e^2 π 0 mr
=
(1. 60 × 10 −^19 C)^2
π(8. 85 × 10 −^12 C
2 N·m^2 )(9.^11 ×^10
− (^31) kg)(1. 0 × 10 − (^2) m) = 1.^01 ×^10
5 m^2 s^2
which gives v 0 = 3. 18 × 10 2 m s
