Problem 1. A 4.5 × 10–9 C charge is located 3.2 m from a –2.8 × 10–9 C charge. Find the
electrostatic force exerted by one charge on the other.
Solution: Since the charges have opposite signs, the force is one of attraction .
Its magnitude is
2
4.5 × 10−9 C )( 2.8 × 10−9 C )
ke q1q2 ⎛
9 N ⋅m ⎞ (
=
8.
99
×
10
=
⎜⎝
r2
C 2 ⎟⎠
( 3.2 m ) 2
F=
1.1 × 10−8 N
Problem 2. Two neighboring cells are ionized (charged) by x-radiation. (a) If the charge on
each cell is equal to 1.60 × 10–14 C and the cells are 2.5 μm apart, what is the force exerted by
each cell? (b) If the distance between the cells is doubled, what happens to the force between
them?
Solution: (a) The force is
−14
k q2 ⎛
N ⋅ m 2 ⎞ ( 1.60 × 10 C )
=
F = e 2 = ⎜ 8.99 × 109
C 2 ⎟⎠ ( 2.5 × 10−6 m ) 2
r
⎝
2
3.7 × 10−7 N
( repulsion)
(b) With r two times larger, r2 is four times larger and the force is
reduced to one fourth ofits previous value .
Problem 3. An electron is accelerated by a constant electric field of magnitude 300 N/C. (a)
Find the acceleration of the electron. (b) Use the equations of motion with constant
acceleration to find the electron’s speed after 1.00 × 10–8 s, assuming it starts from rest.
Solution: (a) The magnitude of the force on the electron is F = q E = eE , and the
acceleration is
a=
−19
F eE ( 1.60 × 10 C ) ( 300 N C )
=
=
= 5.27 × 1013 m s2
9.11 × 10-31 kg
me me
(
)(
)
(b) v = vi + at= 0 + 5.27 × 1013 m s2 1.00 × 10−8 s = 5.27 × 105 m s
Problem 4. Two identical conducting spheres are placed with their centers 0.30 m apart. One
is given a charge of 12 x 10–9 C and the other a charge of –18 × 10–9 C. (a) Find the
electrostatic force exerted by one sphere on the other. (b) The spheres are connected by a
conducting wire. After equilibrium has occurred, find the electrostatic force between the two.
Solution: (a) The spherically symmetric charge distributions behave as if all charge
was located at the centers of the spheres. Therefore, the magnitude of the
attractive force is
2
12 × 10−9 C )( 18 × 10−9 C )
keq1 q2 ⎛
9 N ⋅m ⎞ (
F=
= ⎜ 8.99 × 10
=
r2
C 2 ⎟⎠
⎝
( 0.30 m ) 2
2.2 × 10−5 N
(b) When the spheres are connected by a conducting wire, the net charge
qnet = q1 + q2 = −6.0 × 10−9 C will divide equally between the two identical
spheres. Thus, the force is now
−9
k ( q 2) ⎛
N ⋅ m 2 ⎞ ( −6.0 × 10 C )
= ⎜ 8.99 × 109
F = e ne2t
C 2 ⎟⎠ 4( 0.30 m ) 2
r
⎝
2
or F = 9.0 × 10−7 N (repulsion)
2
Problem 5. Calculate the magnitude and direction of the Coulomb force on each of the three
charges in the following figure .
Solution: The forces are as shown in
the sketch at the right.
+6.00 μC
F1
+1.50 μC F1 F2 -2.00 μC
F2
F3 F3
3.00 cm
F1 =
F2 =
keq1 q3
F3 =
keq2 q3
2
13
r
2
23
r
2.00 cm
2
6.00 × 10−6 C )( 1.50 × 10−6 C )
keq1q2 ⎛
9 N ⋅m ⎞ (
=
8.
99
×
10
= 89.9 N
2
⎜⎝
C 2 ⎟⎠
r122
( 3.00 × 10-2 m )
(
)(
)
−6
−6
⎛
N ⋅ m 2 ⎞ 6.00 × 10 C 2.00 × 10 C
= ⎜ 8.99 × 109
= 43.2 N
2
C 2 ⎟⎠
⎝
5.00 × 10-2 m
(
(
)(
)
)
2
1.50 × 10−6 C 2.00 × 10−6 C
⎛
9 N ⋅m ⎞
= ⎜ 8.99 × 10
= 67.4 N
2
C 2 ⎟⎠
⎝
2.00 × 10-2 m
(
)
The net force on the 6μ C charge is F6 = F1 − F2 = 46.7 N (to the left) .
The net force on the 1.5μ C charge is F1.5 = F1 + F3 = 157 N (to the right) .
The net force on the −2μ C charge is F−2 = F2 + F3 = 111 N (to the left) .
Problem 6. Two small metallic spheres, each of mass 0.20 g, are suspended as pendulums by
light strings from a common point as shown in the figure. The spheres are given the same
electric charge, and it is found that the two come to equilibrium when each string is at an
angle of 5.0° with the vertical. If each string is 30.0 cm long, what is the magnitude of the
charge on each sphere?
Solution: Consider the free-body diagram of one of the spheres given
at the right. Here, T is the tension in the string and Fe is the
repulsive electrical force exerted by the other sphere.
⇒ T cos5.0° = m g , or T =
ΣFx = 0
⇒ Fe = T sin 5.0° = m g tan 5.0°
5.0°
+x
cos5.0°
mg
At equilibrium, the distance separating the two spheres is r = 2L sin 5.0° .
Thus, Fe = m g tan 5.0° becomes
q = ( 2L sin 5.0°)
keq2
( 2L sin 5.0°)
2
= m g tan 5.0° and yields
m g tan 5.0°
ke
= ⎡⎣2( 0.300 m ) sin 5.0°⎤⎦
( 0.20 × 10
T
Fe
mg
ΣFy = 0
+y
−3
)(
)
kg 9.80 m s2 tan 5.0°
8.99 × 10 N ⋅ m
9
2
C2
= 7.2 nC
Problem 7. A piece of aluminum foil of mass 5.00 × 10–2 kg is suspended by a string in an
electric field directed vertically upward. If the charge on the foil is 3.00 μC, find the strength
of the field that will reduce the tension in the string to zero.
Solution: If there is zero tension in the string, the electrical force supports the entire
weight of the foil. Thus,
F = qE = m g,
E=
so
−2
2
m g ( 5.00 × 10 kg)( 9.80 m s )
=
= 1.63 × 105 N C
q
3.00 × 10−6 C
Problem 8. A proton accelerates from rest in a uniform electric field of 640 N/C. At some
later time, its speed is 1.20 × 106 m/s. (a) Find the magnitude of the acceleration of the proton.
(b) How long does it take the proton to reach this speed? (c) How far has it moved in this
interval? (d) What is its kinetic energy at the later time?
−19
F qE 1.60 × 10 C ( 640 N C )
Solution: (a) a = =
=
= 6.12 × 1010 m s2
-27
1.673 × 10 kg
m mp
(
(b) t=
(c)
Δv
a
Δx =
=
1.20 × 106 m s
= 1.96 × 10−5 s = 19.6 μs
6.12 × 1010 m s2
v2f − vi2
(d) KE f =
)
2a
(1.20 × 10 m s) − 0 =
=
2( 6.12 × 10 m s )
2
6
10
(
2
)(
11.8 m
)
2
1
1
m pv2f = 1.673 × 10−27 kg 1.20 × 106 m s = 1.20 × 10−15 J
2
2
Problem 9. In the following figure , determine the point (other than infinity) at which the
total electric field is zero.
Solution: If the resultant field is zero, the
+y
contributions from the two charges must
r1 = d
1.0 m
be in opposite directions and also have
+x
equal magnitudes. Choose the line
q1 = –2.5 μC
q2 = 6.0 μC
E2
E1
connecting the charges as the x-axis, with
r2 = 1.0 m + d
the origin at the –2.5 μC charge. Then, the
two contributions will have opposite
directions only in the regions x < 0 and
x > 1.0 m . For the magnitudes to be equal, the point must be nearer the
smaller charge. Thus, the point of zero resultant field is on the x-axis at x < 0 .
ke q1 ke q2
2.5 μ C
6.0 μ C
= 2 or
=
.
2
2
r1
r2
d
(1.0 m + d) 2
Requiring equal magnitudes gives
Thus,
2.5
=d
6.0
(1.0 m + d)
Solving for d yields
d = 1.8 m ,
1.8 m to the leftofthe − 2.5 μ C charge
or
Problem 10. In the Bohr theory of the hydrogen atom, an electron moves in a circular orbit
about a proton, where the radius of the orbit is 0.53 × 10–10 m. (a) Find the electrostatic force
acting on each particle. (b) If this force causes the centripetal acceleration of the electron,
what is the speed of the electron?
Solution: (a) F =
ke q1 q2 kee2
= 2
r2
r
( 8.99 × 10
=
9
(
N ⋅m
)(
C 2 1.60 × 10−19 C
2
( 0.53 × 10
−10
)
m
)
2
2
= 8.2 × 10−8 N
)
(b) F = m eac = m e v2 r , so
r⋅ F
v=
=
me
( 0.53 × 10
−10
)(
m 8.2 × 10−8 N
9.11 × 10
-31
kg
)=
2.2 × 106 m s
Problem 11. A small 2.00-g plastic ball is suspended by a 20.0-cm-long string in a uniform
electric field, as shown in the figure. If the ball is in equilibrium when the string makes a
15.0° angle with the vertical as indicated, what is the net charge on the ball?
Solution: Consider the free-body diagram shown at the right.
ΣFy = 0 ⇒ T cosθ = m g or T =
T
θ
y
mg
cosθ
x
Fe
ΣFx = 0 ⇒ Fe = T sin θ = m g tan θ
mg
Since Fe = qE , we have
qE = m g tan θ , or q =
( 2.00 × 10
q=
−3
m g tan θ
E
)
)(
kg 9.80 m s2 tan15.0°
1.00 × 10 N C
3
= 5.25 × 10−6 C = 5.25 μ C
Problem 12. Two point charges like those in following figure are called an electric dipole.
Show that the electric field at a distant point along the x axis is given by the expression Ex =
4keqa/x3.
Solution: As shown in the sketch, the electric field
at any point on the x-axis consists of two
parts, one due to each of the charges in the
dipole.
E = E+ − E− =
E=
ke q
( x − a)
2
−
ke q ke q
− 2
r+2
r−
y
r–
+q
–q
a
2
r+
x
⎡ 4ax
⎡ ( x + a) 2 − ( x − a) 2 ⎤
⎢
= ke q ⎢
2
2 ⎥ = keq
⎢ x2 − a2
−
+
x
a
x
a
(
)
(
)
( x + a)
⎣
⎦
⎣
ke q
E–
a
(
)
2
⎤
⎥
⎥
⎦
⎡ 4ax ⎤ 4keqa
Thus, if x2 >> a2 , this gives E ≈ keq⎢ 4 ⎥ =
x3
⎣ x ⎦
Problem 13. A proton moves 2.00 cm parallel to a uniform electric field with E = 200 N/C.
(a) How much work is done by the field on the proton? (b) What change occurs in the
potential energy of the proton? (c) Through what potential difference did the proton move?
Solution: (a) The work done is W = F ⋅ scosθ = ( qE) ⋅ scosθ , or
W = ( 1.60 × 10−19 C ) ( 200 N C ) ( 2.00 × 10−2 m ) cos0° = 6.40 × 10−19 J
(b) The change in the electrical potential energy is
ΔPEe = − W = − 6.40 × 10−19 J
(c) The change in the electrical potential is
ΔV =
ΔPEe
q
=
−6.40 × 10−19 J
= − 4.00 V
1.60 × 10-19 C
E+
Problem 14. A potential difference of 90 mV exists between the inner and outer surfaces of a
cell membrane. The inner surface is negative relative to the outer surface. How much work is
required to eject a positive sodium ion (Na+) from the interior of the cell?
Solution: The work done by the agent moving the charge out of the cell is
W input = −W
field
= − ( −ΔPEe ) = + q( ΔV )
J⎞
⎛
= 1.60 × 10−19 C ⎜ + 90 × 10−3 ⎟ = 1.4 × 10−20 J
⎝
C⎠
(
)
Problem 15. The difference in potential between the accelerating plates of a TV set is about
25 kV. If the distance between these plates is 1.5 cm, find the magnitude of the uniform
electric field in this region.
Solution: E =
ΔV
=
d
25000 J C
= 1.7 × 106 N C
−2
1.5 × 10 m
Problem 16. Suppose an electron is released from rest in a uniform electric field whose
strength is 5.90 × 103 V/m. (a) Through what potential difference will it have passed after
moving 1.00 cm? (b) How fast will the electron be moving after it has traveled 1.00 cm?
Solution: (a)
(
ΔV = Ed = 5.90 × 103 V m
)(1.00 × 10
−2
)
m = 59.0 V
1
m ev2 − 0 = ΔKE = W = −ΔPEe = q( ΔV ) ,
2
(b)
so v =
2 q( ΔV )
me
=
(
)
2 1.60 × 10−19 C ( 59.0 J C )
−31
9.11 × 10
kg
= 4.55 × 106 m s
Problem 17. (a) Find the potential 1.00 cm from a proton. (b) What is the potential difference
between two points that are 1.00 cm and 2.00 cm from a proton?
Solution: (a)
V=
9
2
2
−19
keq ( 8.99 × 10 N ⋅ m C )( 1.60 × 10 C )
=
= 1.44 × 10−7 V
r
1.00 × 10-2 m
(b) ΔV = V 2 − V1 =
⎛ 1 1⎞
keq keq
−
= ( keq) ⎜ − ⎟
r2
r1
⎝ r2 r1 ⎠
(
= 8.99 × 109 N ⋅ m
= − 7.19 × 10−8 V
2
1
1
⎛
⎞
C 2 1.60 × 10−19 C ⎜
−
⎝ 0.0200 m 0.0100 m ⎟⎠
)(
)
Problem 18. Find the potential at point P for the rectangular grouping of charges shown in
the figure.
Solution: Calling q3 the charge at the corner opposite P,
⎛q
V = V1 + V 2 + V 3 = ke ⎜
1
⎝ r1
+
q2
q3 ⎞
+
⎟
r2
r12 + r22 ⎠
⎛
N ⋅ m 2 ⎞ ⎛ 8.00 × 10−6 C 8.00 × 10−6 C
= ⎜ 8.99 × 109
−
+
⎜
C 2 ⎟⎠ ⎜⎝ 0.200 m
0.350 m
⎝
12.0 × 10−6 C
( 0.200) 2 + ( 0.350) 2
⎞
⎟
m ⎟⎠
= 4.22 × 105 V
Problem 19. (a) Find the electric potential, taking zero at infinity, at the upper-right corner
(the corner without a charge) of the rectangle in the following figure . (b) Repeat if the 2.00μC charge is replaced with a charge of –2.00 μC.
Solution: (a) Calling the 2.00 μ C charge q3 ,
V =∑
i
⎛q q
keqi
q3 ⎞
= ke ⎜ 1 + 2 +
⎟
ri
r12 + r22 ⎠
⎝ r1 r2
⎛
N ⋅ m 2 ⎞ ⎛ 8.00 × 10−6 C 4.00 × 10−6 C
= ⎜ 8.99 × 109
+
+
⎜
C 2 ⎟⎠ ⎜⎝ 0.0600 m
0.0300 m
⎝
2.00 × 10−6 C
( 0.0600) 2 + ( 0.0300) 2
V = 2.67 × 106 V
(b) Replacing 2.00 × 10−6 C by − 2.00 × 10−6 C in part (a) yields
V = 2.13 × 106 V
⎞
⎟
m ⎟⎠
Problem 20. (a) How much charge is on each plate of a 4.00-μF capacitor when it is
connected to a 12.0-V battery? (b) If this same capacitor is connected to a 1.50-V battery,
what charge is stored?
Q = C ( ΔV ) = ( 4.00 × 10−6 F) (12.0 V ) = 48.0 × 10−6 C = 48.0 μ C
Solution: (a)
Q = C ( ΔV ) = ( 4.00 × 10−6 F) (1.50 V ) = 6.00 × 10−6 C = 6.00 μ C
(b)
Problem 21. Consider Earth and a cloud layer 800 m above Earth to be the plates of a
parallel-plate capacitor. (a) If the cloud layer has an area of 1.0 km2 = 1.0 × 106 m2, what is
the capacitance? (b) If an electric field strength greater than 3.0 × 106 N/C causes the air to
break down and conduct charge (lightning), what is the maximum charge the cloud can hold?
6
2
A ⎛
C 2 ⎞ 1.0 × 10 m
Solution: (a) C = ∈0 = ⎜ 8.85 × 10−12
= 1.1 × 10−8 F
d ⎝
N ⋅ m 2 ⎟⎠
( 800 m )
(
)
Q m ax = C ( ΔV ) m ax = C ( Em axd)
(b)
(
)
)(
= 1.11× 10−8 F 3.0 × 106 N C ( 800 m ) = 27 C
Problem 22. The potential difference between a pair of oppositely charged parallel plates is
400 V. (a) If the spacing between the plates is doubled without altering the charge on the
plates, what is the new potential difference between the plates? (b) If the plate spacing is
doubled while the potential difference between the plates is kept constant, what is the ratio of
the final charge on one of the plates to the original charge?
Solution: For a parallel plate capacitor, ΔV =
Qd
Q
Q
=
=
.
C ∈0 ( A d) ∈0 A
(a) Doubling d while holding Q and A constant doubles ΔV to 800 V .
(b) Q =
(∈0 A ) ΔV
d
Thus, doubling d while holding ΔV and A constant will
cut the charge in half, or Q f = Q i 2
Problem 23. The plates of a parallel-plate capacitor are separated by 0.100 mm. If the
material between the plates is air, what plate area is required to provide a capacitance of 2.00
pF?
Solution: C =
∈0 A
d
, so A =
C ⋅d
∈0
( 2.00 × 10
−12
=
)(
F 0.100 × 10−3 m
−12
8.85 × 10
C N ⋅m
2
2
)=
2.26 × 10−5 m
2
Problem 24.An air-filled capacitor consists of two parallel plates, each with an area of 7.60
cm2, separated by a distance of 1.80 mm. If a 20.0-V potential difference is applied to these
plates, calculate (a) the electric field between the plates, (b) the capacitance, and (c) the
charge on each plate.
ΔV
20.0 V
= 1.11× 104 V m = 11.1 kV m directed toward
1.80 × 10-3 m
the negative plate
Solution: (a)
E=
(b) C =
d
=
( 8.85 × 10
−12
∈0 A
=
d
C2 N ⋅m
2
)( 7.60 × 10
−4
m
2
)
1.80 × 10 m
-3
= 3.74 × 10−12 F = 3.74 pF
(c)
Q = C ( ΔV ) = ( 3.74 × 10−12 F) ( 20.0 V ) = 7.47 × 10−11 C = 74.7 pC on one
plate and − 74.7 pC on the other plate.
Problem 25. A parallel-plate capacitor has 2.00-cm2 plates that are separated by 5.00 mm
with air between them. If a 12.0-V battery is connected to this capacitor, how much energy
does it store?
Solution: The capacitance is
C=
∈0 A
d
( 8.85 × 10
=
−12
C2 N ⋅m
2
)( 2.00 × 10
5.00 × 10 m
-3
−4
m
2
) = 3.54 × 10
and the stored energy is
1
2
W = C ( ΔV ) =
2
(
)
1
2
3.54 × 10−13 F (12.0 V ) = 2.55 × 10−11 J
2
−13
F,
Problem 26. Consider the parallel-plate capacitor formed by Earth and a cloud layer. Assume
this capacitor will discharge (that is, lightning occurs) when the electric field strength
between the plates reaches 3.0 × 106 N/C. What is the energy released if the capacitor
discharges completely during a lightning strike?
Solution: The capacitance of this parallel plate capacitor is
C 2 ⎞ ( 1.0 × 10 m
A ⎛
C = ∈0 = ⎜ 8.85 × 10−12
N ⋅ m 2 ⎟⎠
d ⎝
( 800 m )
6
) = 1.1× 10
2
−8
F
With an electric field strength of E = 3.0 × 106 N C and a plate separation of
d = 800 m , the potential difference between plates is
(
ΔV = Ed = 3.0 × 106 V m
) ( 800 m ) = 2.4 × 10
9
V
Thus, the energy available for release in a lightning strike is
1
2
W = C ( ΔV ) =
2
(
)(
1
1.1 × 10−8 F 2.4 × 109 V
2
)
2
= 3.2 × 1010 J