Chapter 24 24.1 A particular 12V car battery can send a total charge ...

Chapter 24
24.1 A particular 12V car battery can send a total charge of 84 Ah through a circuit from one
terminal to the other. (a) How many coulombs of charge does this represent? (b) If this entire
charge undergoes a potential difference of 12V, how much energy is involved.
q = 84A ⋅ h ⋅ 3600s = 3.024 ×10 5 C
1h
ΔU = qΔV = 3.024 ×10 5 C ⋅12V = 3.63 ×10 6 J
24.2 The electric potential difference between the ground and a cloud n a particular thunderstorm is
1.2 ×10 9 V . In the unit electron volds, what is the magnitude of the change in the electric potential
energy of an electron that moves between the ground and the cloud.
U = qΔV
= (1e) ⋅(1.2 ×10 9 V )
= 1.2 ×10 9 eV
24.5 Two larfge, parallel, conducting plates are 12 cm apart and have chargs of equal magnitude
and opposide sign on their facign surfaces. An electrostatic force of 3.9 ×10 −15 N acts on an
electron palaced anywhere between the two plates (Neglect fringing.) (a) Find the electric field at
the position of the electron. (b) What is the potential difference between the plates.
Since the force on an electron is known everywhere, we can find the field
F = q E
E = F q = 3.9 × 10−15 N
1.6 × 10 −19 C = 2.4375 ×10 4 N / C
Since the field is due to parallel plates, we can now calculate the potential difference between the
plates from the field.
V = E ⋅ d
= 24,375 N C ⋅ 0.12m
= 2925V
24.6 The graph of the x component of the electric field as a function of x in a region of space is
shown in Figure 24.30. The y and x 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=2.0m (b) what is the
greatest positive value of the electric potential for point on the x axis for which 0 ≤ x ≤ 6.0 m and
(c)for what value of x is the electric potential zero.
The change in the electric potential can be computed by integrating the electric field.
That is

ΔV = −
x f
∫
x i
E ⋅ d r
In this problem, the only non-zero field is in the x direction, so we can simplify
ΔV = −
x f
∫
x i
E x
d x
This integral just represents the area under the curve. We can find the potential at x = 2m
V(2) −V (0) = −Area undercurve between x = 0 and x = 2
V(2) −10V = −( 1 2 ⋅ 2m ⋅ −20 N C ) = +20V
V(2) = 30V
Notice that as positive areas will make the potential decrease. So to get the largest final potential,
we need to find the largest possible negative area. This leads us to part b. The largest negative area
is between x=0 and x=3.
V(x) −V (0) = −Area undercurve between x = 0 and x
V(3) −10V = −( 1 2 ⋅ 3m ⋅ −20 N C ) = +30V
V(2) = 40V
To find where the potential reaches zero, we must find and area such that the final potential is zero.
V(x) −V (0) = −Area undercurve between x = 0 and x
0 −10V = −Area undercurve between x = 0 and x
( )
10 = (Area from 0 to 3) + (Area from 3 to 4) + (Area from 4 to x f
d = x f
− 4
10 = ( 1 2 ⋅ 3m ⋅ −20 N C ) + (1 2 ⋅1m⋅ 20 N ) + (d ⋅ 20)
C
10 = −30 +10 + (d ⋅ 20)
30 = (d ⋅ 20)
d = 3 2 m
x f
= 4 + 3 2 = 5.5m

24.10 As a space shuttle moves through the dilute ionized gas of the Earth’s ionosphere, the
shuttle’s potential is typically changed by −1.0V during one revolution. Assuming the shuttle is a
sphere of radius 10m, estimate the amount of charge that it collects.
V =
q
4π ε 0
a
q = 4π ε 0
aV
= −1.11×10 −9 C
24.12 Consider a point charge q =1.0µC , point A at a distance d 1
= 2.0m from q and point B at a
distance d 2
=1.0m . (a) If these points are diametrically opposite each other as in 24-31a, what is
the electric potential difference V A
−V B
? (b) What is the electric potential difference if points A
and B are located as in 25-31b.
The potential at points A and B only depend on how far those points are from q, not the direction.
V A
−V B
=
The result is the same for both (a) and (b).
=
q
4πε 0
r A
−
q
4πε 0
r B
q
4πε 0
( 1 r A
− 1 r B
)
= 1×10−6 C
4πε 0
( 1
2m − 1
1m )
−4.5 ×10 3 V
24.15 In Fig 24-33, what is the net potential at point P due to the four point charges if V=0 at
infinity.
V = +q
4πε 0
d + +q
4πε 0
d +
−q
4πε 0
d +
= +q
4πε 0
d + −q
4πε 0
2d = +q
4πε 0
⋅ 2d
= 5.62 ×10 −4 V
−q
4πε 0
2d
24.19 The ammonia molecule NH3 has a permanent electric dipole moment equal to 1.47D
Calculate the electric potential due to an ammonia molecule at a point 52nm away along the axis of
the dipole
V =
pcosθ
4πε 0
r = 1.47 × 3.34 ×10−30 cos0°
=1.63 ×10 −5 V
2
4πε 0
(52 ×10 −9 ) 2

24.21 (a) Figure 25-36a shows a positively charged plastic rod of length L and uniform charge
density λ. Setting V=0 at infinity and considering Fig 25-13 and Eq 25-35, find the electric
potential at point P without written calculation. (b) Find the potential due to the second rod.
This problem is the problem worked out in the book with different limits... We can work it again.
Write an expression for the electric potential due to a small charge dq at the point p?
dV =
dq
4πε 0
r =
dq
4πε 0
x 2 + d 2
Write an expression for the dq and the r in terms of x and the distance y.
The potential at the point indicated?
dq = λdx
V =
L / 2
λdx
∫ = λ
4πε 0
x 2 + d 2 4πε 0
−L / 2
L /2
∫
− L /2
= λ ⎡
⋅ ln L / 2 + (L / 2)2 + d 2 ⎤
⎢
⎥
4πε 0 ⎣ ⎢ −L / 2 + (L / 2) 2 + d 2
⎦ ⎥
dx
x 2 + d 2
The potential for the second drawing is zero, since the equal and opposite charge contributions
cancel exactly.
24.23 A plastic rod has been formed into a circle of radius R. It has a positive charge +Q
uniformly distributed along one quarter of its circumference and a negative charge of -6Q
uniformly distributed along the rest of the circumference. With V=0 at infinity, what is the electric
potential at the center of the circle and (b) at point P, which is on the central axis of the circle at a
distance z form the center?
We will begin with (b) since (a) is a special case of b. We proceed in the usual way by defining r
and dq for bits of charge on the rod. We then integrate and simplify.

dV =
dq
4πε 0
r =
dq = λdl = λRdθ
dq
4πε 0
R 2 + z 2
+Q
λ +
=
2πR / 4
λ −
=
−6Q
6πR / 4
π / 2
λ
V =
+
Rdθ
∫ +
0 4πε 0
R 2 + z 2
=
=
=
=
=
2π
∫
π / 2
λ −
Rdθ
4πε 0
R 2 + z 2
λ +
R ⎡ π
4πε 0
R 2 + z 2 ⎣
⎢
2 −0 ⎤
⎦
⎥ + λ −
R ⎡
2π − π ⎤
4πε 0
R 2 + z 2 ⎣
⎢
2 ⎦
⎥
λ +
R ⎡ π⎤
4πε 0
R 2 + z 2 ⎣
⎢
2⎦
⎥ + λ −
R ⎡ 3π ⎤
4πε 0
R 2 + z 2 ⎣
⎢
2 ⎦
⎥
+Q
2πR / 4 ⋅ R ⎡ π⎤
4πε 0
R 2 + z 2 ⎣
⎢
2⎦
⎥ +
−6Q
6πR / 4 ⋅R ⎡ 3π
4πε 0
R 2 + z 2 ⎣
⎢
2
+Q
4πε 0
R 2 + z + −6Q
2 4πε 0
R 2 + z 2
−5Q
4πε 0
R 2 + z 2
⎤
⎦
⎥
We set z=0 to get the solution to part (a).
V =
−5Q
4πε 0
R
24.28 Figure 24-44 shows a thin plastic rod of length L = 12.0 cm and a uniform positive charge
of Q = 56.1 f C lying on the x-axis. With V=0 at infinity, find the electric potential at point P1 on
the axis at a distance d = 2.50cm from one end of the rod.
This problem is the problem worked out in the book with different limits... We can work it again.
Write an expression for the electric potential due to a small charge dq at the point p?
dV =
dq
4πε 0
r = dq
4πε 0
(x + d)
Write an expression for the dq and the r in terms of x and the distance y.

The potential at the point indicated?
dq = λdx
λ = 56.1 f C
0.12m = 4.675 ×10−13 C / m
L
λ dx
V = ∫ = λ
0
4πε 0
(x + d) 4πε 0
= λ ⎡
⋅ ln L + d ⎤
4πε 0
⎣
⎢
d ⎦
⎥
= 7.39 ×10 −3 V
L / 2
∫
−L / 2
dx
(x + d)
24.31 The electric potential at points in an xy plane is given by
V = (2.0V / m 2 ) x 2 − (3.0V / m 2 ) y 2
In unit-vector notation, what is the electric field at the point (3.0 m, 2.0m).
E x
= − ∂V
∂x = −(4.0V / m2 ) x
E y
= − ∂V
∂y = (6.0V / m2 ) y
E = −(4.0V / m 2 ) x ˆ i + (6.0V / m 2 ) y ˆ j
= −(12V / m) i ˆ + (12.0V / m) ˆ j
24.37 How much work is required to set up the arrangement of Fig. 24-46 if q = 2.30 ×10 −12 C ,
a = 64.0cm and the particles are initially infinitely fare apart and at rest.
The work required to assemble the charges is the energy stored in the charge array. We can
calculate this by summing over all the energies of all the pairs of charges.
U =
= 2 ⋅
(+q)( +q)
4π ε 0
( 2a) + (−q)(−q)
4 π ε 0
( 2a) + ( +q)(−q)
4π ε 0
a
+ (+q)(−q)
4π ε 0
a
+ (+q)(−q)
4 π ε 0
a + (+q)(−q)
4π ε 0
a
q 2
4π ε 0
( 2a) − 4 ⋅ q 2
= −1.92 ×10 −13 J
W = −U =1.92 ×10 −13 J
4π ε 0
a

24.45 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.
This is an energy conservation problem. The energy at infinity is all kinetic. The energy of the
stopped electron is all potential We can used conservation of energy to find the initial velocity at
infinity.
a
a
E i
= 1 2 mv2
E f
=
E f
= E i
1
2 mv2 =
v =
q 2
4π ε 0
a + q 2
4π ε 0
a
q 2
4π ε 0
a + q 2
4π ε 0
a
q 2
mπ ε 0
a
= 318.3m / s
24.54 An empty hollow metal sphere has a potential of +400V with respect to ground What is the
electric potential at the center.
Since there is no electric field inside the sphere, the potential must remain constant throughout the
sphere. So the potential at the center =potential at the surface =400V
24.57 A metal sphere of radius 15 cm has a net charge of 3.0 ×10 −8 C . (a) What is the electric
field at the sphere’s surface? (b)( If V=0 at infinity , what is the electric potential at the sphere’s
surface? (c) At what distance from the sphere’s surface has the electric potential decreased by
500V?
We can treat this sphere as a point charge so long as we are outside. After finding the voltage on
the surface, we can subtract 500V and solve for r. Note that the answer in the back of the book
shows the change in r, not the r.
E =
V =
q
4π ε 0
a 2
=11983.4V / m
q
4π ε 0
a = 1797.5V
1297.5V =
q
4 π ε 0
r ⇒ r = 20.8cm