Chapter 24: Electric Potential
Example Questions & Problems
ΔV =
−W
q
=
ΔU
q
f G
G
V = − ∫ E ⋅ ds
o
Vpoint =
1 q
4 πε0 r
G
∂V
E=− G
∂s
Example 24.1
a. If a negative charge is initially a rest in an electric field, will it move toward a region of
higher potential or lower potential? What about a positive charge? How does the
potential energy of the charge change in each of these two instances?
G
b. If the electric field E is uniform in a region, what can you infer about the electric
G
potential V? If V is uniform in a region of space, what can you infer about E ?
c. The earth has a net electric charge that causes an electric field at points near its
surface equal 150 N/C and directed in toward the center of the earth. Why would it
still be possible to adopt that the earth’s surface as a standard reference point of
potential and to assign the potential V = 0 to it?
d. How can you ensure that the electric potential in a given region of space will have the
same value through that region?
Example 24.2
An electric dipole consists of two point charges, q1 = +12nC and q2 =
−12nC, placed 10 cm apart. Compute the potentials at points A, C, and C.
The locations of points A, B, and C will be given by the instructor in lecture.
Example 24.3 (P24.54)
Proton in a well. The figure shows electric potential V along
an x axis. The scale of the vertical axis is set by VS = 10.0 V.
A proton is to be released at
with initial kinetic
energy 4.00 eV. (a) If it is initially moving in the negative
direction of the axis, does it reach a turning point (if so, what
is the x coordinate of that point) or does it escape from the
plotted region (if so, what is its speed at x = 0)? (b) If it is initially moving in the positive
direction of the axis, does it reach a turning point (if so, what is the x coordinate of that
point) or does it escape from the plotted region (if so, what is its speed at x = 6.0 cm)?
What are the (c) magnitude F and (d) direction (positive or negative direction of the x
axis) of the electric force on the proton if the proton moves just to the left of x = 3.0 cm?
What are (e) F and (f) the direction if the proton moves just to the right of x = 5.0 cm?
1
Example 24.4 (P24.58)
Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 =
1.00 m, uniform charges q1 = +2.00 μC and q2 = +1.00 μC, negligible thicknesses. What
is the magnitude of the electric field E at radial distance (a) r = 4.00m, (b) r = 0.70m, and
(c) r = 0.20m? With V = 0 at infinity, what is V at (d) r = 4.00m, (e) r = 1.00m, (f) r =
0.70m, (g) r = 0.50m, (h) r = 0.20m and (i) r = 0? (j) Sketch E(r) and V(r).
Example 24.5 (P24.115)
Three long parallel lines of charge, with the linear charge densities
shown, extend perpendicular to the page in both directions. Sketch some
of the (a) equipotential surfaces and (b) electric field lines. Explain your
reasoning.
2
Example A (24.8)
A graph of the x component of the electric field as a function of x in a
region of space is shown in the figure. The scale of the vertical axis is
set by EXS = 20.0 N/C. The y and z components of the electric field are
zero in this region. If the electric potential at the origin is 10 V, (a) what
is the electric potential at x = 2.0 m, (b) what is the greatest positive
value of the electric potential for points on the x axis for which 0 ≤ x ≤ 6.0 cm, and (c) for
what value of x is the electric potential zero?
Solution
a. By Eq. 24-18, the change in potential is the negative of the “area” under the curve.
Thus, using the area-of-a-triangle formula, we have
x =2 G
G 1
V − 10 = −
E ⋅ ds = ( 2 )( 20 ) ⎯⎯→ V = 30 V
0
2
G G
b. For any region within 0 < x < 3 m, − E ⋅ ds is positive, but for any region for which
∫
∫
x > 3 m it is negative. Therefore, V = Vmax occurs at x = 3 m.
x =3 G
G 1
Vmax − 10 = −
E ⋅ ds = ( 3 )( 20 ) ⎯⎯→ Vmax = 40 V
0
2
c. In view of our result in part (b), we see that now (to find V = 0) we are looking for
some X > 3 m such that the “area” from x = 3 m to x = X is 40 V. Using the formula
for a triangle (3 < x < 4) and a rectangle (4 < x < X), we require
1
(1)( 20 ) + ( X − 4 )( 20 ) = 40 ⎯⎯→ X = 5.5 m
2
∫
Example B (24.51)
Two tiny metal spheres A and B of mass mA = 5.00 g and mB = 10.0 g have equal
positive charge 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?
Solution
a. The potential energy is
(
)(
)
2
8.99 × 109 N ⋅ m2 C2 5.0 × 10−6 C
q2
=
= 0.225 J = U
U=
4πε0 d
1.00 m
relative to the potential energy at infinite separation.
b. Each sphere repels the other with a force that has magnitude
(
)(
)
2
8.99 × 109 N ⋅ m2 C2 5.0 × 10−6 C
q2
F=
=
= 0.225 N.
2
4πε0 d2
(1.00 m )
According to Newton’s second law the acceleration of each sphere is the force
divided by the mass of the sphere. Let mA and mB be the masses of the spheres. The
acceleration of sphere A and B is
3
aA =
F
0.225 N
=
= 45.0 m s2 = a A
m A 5.0 × 10−3 kg
and
aB =
F
= 22.5 m s2 = aB
mB
c. Energy and momentum is conserved.
Conservation of energy
The initial potential energy is U0 = 0.225 J, as calculated in part (a). The initial kinetic
energy is zero since the spheres start from rest. The final potential energy is zero
since the spheres are then far apart. The final kinetic energy is 21 m A v 2A + 21 mB v B2 ,
where vA and vB are the final velocities. Thus,
E0 = E f ⎯⎯→ U0 + K 0 = Uf + K f ⎯⎯→ U0 =
1
1
mA v 2A + mB v B2 .
2
2
Conservation of momentum
solving for vB
P0 = Pf ⎯⎯→ 0 = mA v A + mB v B ⎯⎯⎯⎯⎯→
vB = −(mA / mB )v A
Substituting vB from the momentum equation into the energy equation, and collecting
terms, we obtain
U0 = 21 (m A / mB )(m A + mB )v 2A .
Therefore,
vA =
2UmB
m A (m A + mB )
=
2(0.225 J)(10 × 10 −3 kg)
(5.0 × 10 −3 kg)(5.0 × 10 −3 kg + 10 × 10 −3 kg)
= 7.75 m/s = v A
and
vB = −
mA
mB
⎛ 5.0 × 10−3 kg ⎞
⎟ (7.75 m/s) = −3.87 m/s = v B
−3
⎝ 10 × 10 kg ⎠
vA = − ⎜
Example C (24.65)
Two metal spheres, each of radius 3.0 cm, have a center-to-center separation of 2.0 m.
Sphere 1 has charge +1.0 × 10−8 C; sphere 2 has charge −3.0 × 10−8 C. Assume that the
separation is large enough for us to assume that the charge on each sphere is uniformly
distributed (the spheres do not affect each other). With V = 0 at infinity, calculate (a) the
potential at the point halfway between the centers and the potential on the surface of (b)
sphere 1 and (c) sphere 2. SSM
Solution
a. The electric potential is the sum of the contributions of the individual spheres. Let q1
be the charge on one, q2 be the charge on the other, and d be their separation. The
point halfway between them is the same distance d/2 (= 1.0 m) from the center of
each sphere, so the potential at the halfway point is
8.99 × 109 N ⋅ m2 C2 1.0 × 10 −8 C − 3.0 × 10 −8 C
q1 + q2
V=
=
4πε 0 d 2
1.0 m
(
)(
)
= −1.8 × 102 V = V
b. The distance from the center of one sphere to the surface of the other is d – R,
where R is the radius of either sphere. The potential of either one of the spheres is
due to the charge on that sphere and the charge on the other sphere. The potential
at the surface of sphere 1 is
4
V1 =
−8
1 ⎡ q1
q2 ⎤
3.0 × 10 −8 C ⎤
9
2
2 ⎡ 1.0 × 10 C
+
=
8.99
×
10
N
⋅
m
C
−
⎢ 0.030 m
⎥
4πε 0 ⎢⎣ R d − R ⎥⎦
2.0 m − 0.030 m ⎦
⎣
(
)
= 2.9 × 103 V = V1
c. The potential at the surface of sphere 2 is
q2 ⎤
1 ⎡ q1
1.0 × 10 −8 C
3.0 × 10 −8 C ⎤
9
2
2 ⎡
=
+
=
×
⋅
−
V2
8.99 10 N m C ⎢
⎥
4 πε 0 ⎢⎣ d − R R ⎥⎦
0.030 m ⎦
⎣ 2.0 m − 0.030 m
(
)
= −8.9 × 103 V = V2
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