Chapter 16
Electrical Energy and
Capacitance
Quick Quizzes
1.
(b). The field exerts a force on the electron, causing it to accelerate in the direction opposite
to that of the field. In this process, electrical potential energy is converted into kinetic
energy of the electron. Note that the electron moves to a region of higher potential, but
because the electron has negative charge this corresponds to a decrease in the potential
energy of the electron.
2.
(b), (d). Charged particles always tend to move toward positions of lower potential
energy. The electrical potential energy of a charged particle is PE = qV and, for positivelycharged particles, this increases as V increases. For a negatively-charged particle, the
potential energy decreases as V increases. Thus, a positively-charged particle located at
x = A would move toward the left. A negatively-charged particle would oscillate around
x = B which is a position of minimum potential energy for negative charges.
3.
(d). If the potential is zero at a point located a finite distance from charges, negative
charges must be present in the region to make negative contributions to the potential and
cancel positive contributions made by positive charges in the region.
4.
(c). Both the electric potential and the magnitude of the electric field decrease as the
distance from the charged particle increases. However, the electric flux through the
balloon does not change because it is proportional to the total charge enclosed by the
balloon, which does not change as the balloon increases in size.
5.
(a). From the conservation of energy, the final kinetic energy of either particle will be
given by
(
)
(
)
KE f = KEi + PEi − PE f = 0 + qVi − qVf = − q Vf − Vi = − q ( ∆V )
For the electron, q = − e and ∆V = +1 V giving KE f = − ( − e )( +1 V ) = +1 eV .
For the proton, q = + e and ∆V = −1 V , so KE f = − ( e )( −1 V ) = +1 eV , the same as that of the
electron.
6.
(c). The battery moves negative charge from one plate and puts it on the other. The first
plate is left with excess positive charge whose magnitude equals that of the negative
charge moved to the other plate.
35
36
CHAPTER 16
7.
(a) C decreases.
(d) ∆V increases.
(b) Q stays the same.
(c) E stays the same.
(e) The energy stored increases.
Because the capacitor is removed from the battery, charges on the plates have nowhere to
go. Thus, the charge on the capacitor plates remains the same as the plates are pulled
σ Q A
apart. Because E =
=
, the electric field is constant as the plates are separated.
∈0
∈0
Because ∆V = Ed and E does not change, ∆V increases as d increases. Because the same
charge is stored at a higher potential difference, the capacitance has decreased. Because
Energy stored = Q 2 2C and Q stays the same while C decreases, the energy stored
increases. The extra energy must have been transferred from somewhere, so work was
done. This is consistent with the fact that the plates attract one another, and work must be
done to pull them apart.
8.
(a) C increases.
(d) ∆V remains the same.
(b) Q increases.
(c) E stays the same.
(e) The energy stored increases.
The presence of a dielectric between the plates increases the capacitance by a factor equal to
the dielectric constant. Since the battery holds the potential difference constant while the
capacitance increases, the charge stored ( Q = C∆V ) will increase. Because the potential
difference and the distance between the plates are both constant, the electric field
( E = ∆V d ) will stay the same. The battery maintains a constant potential difference. With
(
∆V constant while capacitance increases, the stored energy Energy stored = 12 C ( ∆V )
2
) will
increase.
9.
(a). Increased random motions associated with an increase in temperature make it more
difficult to maintain a high degree of polarization of the dielectric material. This has the
effect of decreasing the dielectric constant of the material, and in turn, decreasing the
capacitance of the capacitor.
Electrical Energy and Capacitance
37
Answers to Even Numbered Conceptual Questions
2.
Changing the area will change the capacitance and maximum charge but not the
maximum voltage. The question does not allow you to increase the plate separation. You
can increase the maximum operating voltage by inserting a material with higher dielectric
strength between the plates.
4.
Electric potential V is a measure of the potential energy per unit charge. Electrical
potential energy, PE = QV, gives the energy of the total charge Q.
6.
A sharp point on a charged conductor would produce a large electric field in the region
near the point. An electric discharge could most easily take place at the point.
8.
There are eight different combinations that use all three capacitors in the circuit. These
combinations and their equivalent capacitances are:
 1
1
1 
All three capacitors in series - Ceq = 
+
+

 C1 C2 C3 
−1
All three capacitors in parallel - Ceq = C1 + C2 + C3
One capacitor in series with a parallel combination of the other two:
−1
−1
 1
 1

1 
1 
1
1 
Ceq = 
+
+
+ 
 , Ceq = 
 , Ceq = 
 C1 + C2 C3 
 C3 + C1 C2 
 C2 + C3 C1 
−1
One capacitor in parallel with a series combination of the other two:
 CC 
 CC 
 CC 
Ceq =  1 2  + C3 , Ceq =  3 1  + C2 , Ceq =  2 3  + C1
 C1 + C2 
 C3 + C1 
 C2 + C3 
10.
Nothing happens to the charge if the wires are disconnected. If the wires are connected to
each other, the charge rapidly recombines, leaving the capacitor uncharged.
12.
All connections of capacitors are not simple combinations of series and parallel circuits. As
an example of such a complex circuit, consider the network of five capacitors C1, C2, C3, C4,
and C5 shown below.
C1
C2
C5
C3
C4
This combination cannot be reduced to a simple equivalent by the techniques of
combining series and parallel capacitors.
38
CHAPTER 16
14.
The material of the dielectric may be able to withstand a larger electric field than air can
withstand before breaking down to pass a spark between the capacitor plates.
16.
(a) i (b) ii
18.
(a) The equation is only valid when the points A and B are located in a region where the
electric field is uniform (that is, constant in both magnitude and direction). (b) No. The
field due to a point charge is not a uniform field. (c) Yes. The field in the region between a
pair of parallel plates is uniform.
Electrical Energy and Capacitance
Answers to Even Numbered Problems
− 6.0 × 10 −4 J
2.
(a)
4.
−3.20 × 10 −19 C
6.
4.3 × 10 6 J
8.
(a)
(b)
1.52 × 10 5 m s
–50 V
(b)
6.49 × 106 m s
10.
40.2 kV
12.
2.2 × 10 2 V
14.
-9.08 J
16.
8.09 × 10 −7 J
18.
7.25 × 10 6 m s
22.
(a)
48.0 µ C
(b)
6.00 µ C
24.
(a)
800 V
(b)
Q f = Qi 2
26.
31.0 Å
28.
1.23 kV
30.
(a)
(b)
1.78 µ F
32.
3.00 pF and 6.00 pF
34.
(a)
(b)
Q4 =144 µ C, Q2 =72.0 µ C, Q24 =Q8 =216 µ C
36.
Yes. Connect a parallel combination of two capacitors in series with another parallel
combination of two capacitors. ∆V = 45.0 V .
38.
30.0 µ F
40.
6.04 µ F
42.
12.9 µ F
44.
(a)
46.
9.79 kg
18.0 µ F
12.0 µ F
0.150 J
(b)
268 V
39
40
CHAPTER 16
48.
(a)
50.
1.04 m
54.
1
1 2
1
C1 = C p ±
C p − C p Cs , C2 = C p ∓
2
4
2
3.00 × 10 3 V m
56.
(a)
(c)
1.8 × 10 4 V
−1.8 × 10 4 V
58.
(a)
C=
60.
κ = 2.33
62.
1.8 × 10 2 µ C on C1 , 89 µ C on C2
64.
121 V
66.
(a)
(b)
42.5 nC
1 2
C p − C p Cs
4
(b)
(d)
−3.6 × 10 4 V
−5.4 × 10 −2 J
(b)
at x = 4.4 mm
ab
ke ( b − a )
0.11 m
(c)
5.31 pC
Electrical Energy and Capacitance
Problem Solutions
16.1
(a) The work done is W = F ⋅ s cosθ = ( qE ) ⋅ s cosθ , or
W = ( 1.60 × 10 −19 C ) ( 200 N C ) ( 2.00 × 10 −2 m ) cos 0° = 6.40 × 10 −19 J
(b) The change in the electrical potential energy is
∆PEe = − W = − 6.40 × 10 −19 J
(c) The change in the electrical potential is
∆V =
16.2
∆PEe −6.40 × 10 −19 J
=
= − 4.00 V
1.60 × 10 -19 C
q
(a) We follow the path from (0,0) to (20 cm,0) to (20 cm,50 cm). The work done on the
charge by the field is
W = W1 + W2 = ( qE ) ⋅ s1 cosθ1 + ( qE ) ⋅ s2 cosθ 2
= ( qE ) ( 0.20 m ) cos 0° + ( 0.50 m ) cos90°
= ( 12 × 10 −6 C ) ( 250 V m ) ( 0.20 m ) + 0  = 6.0 × 10 −4 J
Thus,
(b) ∆V =
16.3
∆PEe = −W = − 6.0 × 10 −4 J
∆PEe − 6.0 × 10 −4 J
=
= −50 J C = − 50 V
12 × 10 -6 C
q
The work done by the agent moving the charge out of the cell is
Winput = −W field = − ( −∆PEe ) = + q ( ∆V )
J

−20
= ( 1.60 × 10 −19 C )  + 90 × 10 −3
 = 1.4 × 10 J
C

16.4
(
)
∆PEe = q ( ∆V ) = q Vf − Vi , so q =
∆PEe
−1.92 × 10 −17 J
=
= − 3.20 × 10 −19 C
Vf − Vi
+60.0 J C
41
42
CHAPTER 16
∆V
25 000 J C
=
= 1.7 × 10 6 N C
1.5 × 10 −2 m
d
16.5
E=
16.6
Since potential difference is work per unit charge ∆V =
W
, the work done is
q
W = q ( ∆V ) = ( 3.6 × 10 5 C ) ( +12 J C ) = 4.3 × 10 6 J
16.7
(a)
E=
∆V
600 J C
=
= 1.13 × 10 5 N C
d
5.33 × 10 −3 m
(b) F = q E = ( 1.60 × 10 −19 C )( 1.13 × 10 5 N C ) = 1.80 × 10 −14 N
(c)
W = F ⋅ s cosθ
= ( 1.80 × 10 −14 N ) ( 5.33 − 2.90 ) × 10 −3 m  cos0° = 4.38 × 10 −17 J
16.8
From conservation of energy,
(a) For the proton,
vf =
(b) For the electron,
vf =
1
mv 2f − 0 = q ( ∆V )
2
or v f =
2 ( 1.60 × 10 −19 C ) ( −120 V )
1.67 × 10 −27 kg
9.11 × 10
kg
m
= 1.52 × 10 5 m s
2 ( −1.60 × 10 −19 C ) ( +120 V )
−31
2 q ( ∆V )
= 6.49 × 106 m s
43
Electrical Energy and Capacitance
16.9
(a) Use conservation of energy
k
( KE + PEs + PEe ) f = ( KE + PEs + PEe )i
or ∆ ( KE ) + ∆ ( PEs ) + ∆ ( PEe ) = 0
Q
x=0
∆ ( KE ) = 0 since the block is at rest at both beginning and end.
∆ ( PEs ) =
1 2
kxmax − 0 ,
2
where xmax is the maximum stretch of the spring.
∆ ( PEe ) = −W = − ( QE ) xmax
1 2
− ( QE ) xmax = 0 , giving
Thus, 0 + kxmax
2
xmax =
−6
5
2 QE 2 ( 50.0 × 10 C )( 5.00 × 10 V m )
=
= 0.500 m
k
100 N m
(b) At equilibrium, ΣF = − Fs + Fe = 0, or − kxeq + QE = 0
xeq =
Therefore,
QE 1
= xmax = 0.250 m
k
2
Note that when the block is released from rest, it overshoots the equilibrium
position and oscillates with simple harmonic motion in the electric field.
16.10
Using ∆y = v0 y t +
0 = v0 y t +
1 2
ay t for the full flight gives
2
− 2 v0 y
1 2
ay t , or ay =
2
t
Then, using vy2 = v02y + 2 ay ( ∆y ) for the upward part of the flight gives
( ∆y )max =
0 − v02y
2 ay
=
(
−v02y
2 −2 v0 y t
)
=
v0 y t
4
=
( 20.1 m s ) ( 4.10 s ) = 20.6 m
4
ur
E
44
CHAPTER 16
From Newton’s second law, ay =
ΣFy
m
=
qE 
− mg − qE

= − g +
 . Equating
m
m

qE  − 2 v0 y

, so the electric field strength is
this to the earlier result gives ay = −  g +
=
m
t


 m   2 v0 y
  2.00 kg   2 ( 20.1 m s )
E =  
− g = 
− 9.80 m s 2  = 1.95 × 10 3 N C

−6
4.10 s
  5.00 × 10 C  
 q  t

Thus,
16.11
( ∆V )max = ( ∆ymax ) E = ( 20.6 m ) (1.95 × 10 3 N C ) = 4.02 × 10 4 V = 40.2 kV
(a) V =
9
2
2
−19
k e q ( 8.99 × 10 N ⋅ m C )( 1.60 × 10 C )
=
= 1.44 × 10 −7 V
-2
r
1.00 × 10 m
(b) ∆V = V2 − V1 =
1 1
ke q ke q
−
= ( ke q )  − 
r2
r1
 r2 r1 


1
1
= ( 8.99 × 109 N ⋅ m 2 C 2 )( 1.60 × 10 −19 C ) 
−

 0.020 0 m 0.010 0 m 
= − 7.19 × 10 −8 V
16.12
q q 
V = V1 + V2 = k e  1 + 2  where r1 = 0.60 m − 0 = 0.60 m , and
 r1 r2 
r2 = 0.60 m − 0.30 m = 0.30 m. Thus,

N ⋅ m 2   3.0 × 10 −9 C 6.0 × 10 −9 C 
2
V =  8.99 × 109
+

 = 2.2 × 10 V
2
C
0.
60
m
0.30
m



45
Electrical Energy and Capacitance
16.13
(a) Calling the 2.00 µ C charge q3 ,
V =∑
i
q q
k e qi
q3
= ke  1 + 2 +
2
 r1 r2
ri
r1 + r22


N ⋅ m2
=  8.99 × 109
C2






  8.00 × 10 −6 C 4.00 × 10 −6 C
+
+

0.030 0 m
  0.060 0 m

2.00 × 10 −6 C
( 0.060 0 ) + ( 0.030 0 )
2
2



m 
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
16.14
(
)
W = q ( ∆V ) = q Vf − Vi , and
Vf = 0 since the 8.00 µ C is infinite distance from other charges.
q q  
N ⋅ m2
Vi = k e  1 + 2  =  8.99 × 109
C2
 r1 r2  

  2.00 × 10 −6 C
+

  0.030 0 m

4.00 × 10 −6 C
( 0.030 0 ) + ( 0.060 0 )
= 1.135 × 10 6 V
Thus, W = ( 8.00 × 10 −6 C )( 0 − 1.135 × 106 V ) = − 9.08 J
16.15
(a) V = ∑
i
k e qi
ri

N ⋅ m 2  5.00 × 10 −9 C 3.00 × 10 −9 C 
=  8.99 × 109
−

 = 103 V
C 2  0.175 m
0.175 m 

2
2



m 
46
CHAPTER 16
(b) PE =
k e qi q2
r12
2
5.00 × 10 −9 C )( − 3.00 × 10 −9 C )

9 N⋅m  (
=  8.99 × 10
= − 3.85 × 10 − 7 J

2
C
0.350
m


The negative sign means that positive work must be done to separate the charges
(that is, bring them up to a state of zero potential energy).
16.16
The potential at distance r = 0.300 m from a charge Q = +9.00 × 10 −9 C is
9
2
2
−9
ke Q ( 8.99 × 10 N ⋅ m C )( 9.00 × 10 C )
=
= + 270 V
V=
0.300 m
r
Thus, the work required to carry a charge q = 3.00 × 10 −9 C from infinity to this location
is
W = qV = ( 3.00 × 10 −9 C ) ( +270 V ) = 8.09 × 10 −7 J
16.17
The Pythagorean theorem gives the distance from the midpoint of the base to the charge
at the apex of the triangle as
r3 =
2
2
( 4.00 cm ) − (1.00 cm ) = 15 cm = 15 × 10 −2 m
Then, the potential at the midpoint of the base is V = ∑ k e qi ri , or
i

N ⋅ m2
V =  8.99 × 109
C2

−9
−9
−9
  ( −7.00 × 10 C ) ( −7.00 × 10 C ) ( +7.00 × 10 C ) 


+
+

0.010 0 m
15 × 10 −2 m 
  0.010 0 m
= −1.10 × 10 4 V = − 11.0 kV
Electrical Energy and Capacitance
16.18
47
Outside the spherical charge distribution, the potential is the same as for a point charge
at the center of the sphere,
V = k e Q r , where Q = 1.00 × 10 −9 C
 1 1
Thus, ∆ ( PEe ) = q ( ∆V ) = − ek e Q  − 
 rf ri 


and from conservation of energy ∆ ( KE ) = −∆ ( PEe ) ,
or

 1 1 
2 k e Qe  1 1 
1
me v 2 − 0 = −  − ek eQ  −   This gives v =
 −  , or
 rf ri  
 rf ri 
2
m

e





N ⋅ m2 
−9
−19
2  8.99 × 10 9
 ( 1.00 × 10 C )( 1.60 × 10 C ) 
2

C 
1
1

v=
−


−31
9.11 × 10 kg
 0.020 0 m 0.030 0 m 
v = 7.25 × 106 m s
16.19
From conservation of energy, ( KE + PEe ) f = ( KE + PEe )i , which gives
0+
2 k Qq 2 k ( 79e )( 2e )
k e Qq 1
= mα vi2 + 0 or rf = e 2 = e
mα vi
mα vi2
2
rf
2

N ⋅ m2 
−19
2  8.99 × 109
 ( 158 ) ( 1.60 × 10 C )
2
C 
rf = 
= 2.74 × 10 −14 m
2
−27
7
( 6.64 × 10 kg )( 2.00 × 10 m s )
16.20
By definition, the work required to move a charge from one point to any other point on
an equipotential surface is zero. From the definition of work, W = ( F cosθ ) ⋅ s , the work
is zero only if s = 0 or F cosθ = 0 . The displacement s cannot be assumed to be zero in all
cases. Thus, one must require that F cosθ = 0 . The force F is given by F = qE and neither
the charge q nor the field strength E can be assumed to be zero in all cases. Therefore, the
only way the work can be zero in all cases is if cosθ = 0 . But if cosθ = 0 , then θ = 90° or
the force (and hence the electric field) must be perpendicular to the displacement s
(which is tangent to the surface). That is, the field must be perpendicular to the
equipotential surface at all points on that surface.
48
CHAPTER 16
V=
16.21
keQ
r
so
9
2
2
−9
k e Q ( 8.99 × 10 N ⋅ m C )( 8.00 × 10 C ) 71.9 V ⋅ m
=
=
r=
V
V
V
For V = 100 V, 50.0 V, and 25.0 V,
r = 0.719 m, 1.44 m, and 2.88 m
The radii are inversely proportional to the potential.
16.22
16.23
(a)
Q = C ( ∆V ) = ( 4.00 × 10 −6 F ) ( 12.0 V ) = 48.0 × 10 −6 C = 48.0 µ C
(b)
Q = C ( ∆V ) = ( 4.00 × 10 −6 F ) ( 1.50 V ) = 6.00 × 10 −6 C = 6.00 µ C
(a)
(b)
C = ∈0
6
2
A 
C 2  ( 1.0 × 10 m )
=  8.85 × 10 −12
= 1.1 × 10 −8 F
2 
d 
N ⋅ m  ( 800 m )
Qmax = C ( ∆V )max = C ( Emax d )
= ( 1.11 × 10 −8 F )( 3.0 × 10 6 N C ) ( 800 m ) = 27 C
16.24
For a parallel plate capacitor, ∆V =
Qd
Q
Q
=
=
.
C ∈0 ( A d ) ∈0 A
(a) Doubling d while holding Q and A constant doubles ∆V to 800 V .
(∈0 A ) ∆V
Thus, doubling d while holding ∆V and A constant will cut the
d
charge in half, or Q f = Qi 2
(b) Q =
16.25
(a)
∆V
20.0 V
=
= 1.11 × 10 4 V m = 11.1 kV m directed toward the negative
-3
d
1.80 × 10 m
plate
E=
(b) C =
∈0 A ( 8.85 × 10
=
d
−12
C 2 N ⋅ m 2 )( 7.60 × 10 −4 m 2 )
1.80 × 10 -3 m
= 3.74 × 10 −12 F = 3.74 pF
Electrical Energy and Capacitance
(c)
Q = C ( ∆V ) = ( 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.
16.26
∈ A ( 8.85 × 10
∈ A
C = 0 , so d = 0 =
C
d
−12
C 2 N ⋅ m 2 )( 21.0 × 10 −12 m 2 )
60.0 × 10
-15
F
= 3.10 × 10 −9 m
 1Å 
d = ( 3.10 × 10 −9 m )  -10
 = 31.0 Å
 10 m 
16.27
(a)
∆V =
(b) E =
( 400 × 10−12 C )(1.00 × 10−3 m )
Q
Q
Qd
=
=
=
= 90.4 V
C ∈0 A d ∈0 A ( 8.85 × 10 −12 C 2 N ⋅ m 2 )( 5.00 × 10 −4 m 2 )
∆V
90.4 V
=
= 9.04 × 10 4 V m
d
1.00 × 10 -3 m
ΣFy = 0 ⇒ T cos15.0° = mg or T =
16.28
mg
cos15.0°
ur
T
15.0°
ΣFx = 0 ⇒ qE = T sin15.0° = mg tan15.0°
or
E=
ur
ur
F = qE
mg tan15.0°
q
ur
mg
mgd tan15.0°
∆V = Ed =
q
( 350 × 10
∆V =
16.29
−6
kg )( 9.80 m s 2 ) ( 0.040 0 m ) tan15.0°
30.0 × 10 −9 C
(a) For series connection,
= 1.23 × 10 3 V = 1.23 kV
1
1
1
CC
=
+
⇒ Ceq = 1 2
Ceq C1 C2
C1 + C2
 CC
Q=Ceq ( ∆V ) =  1 2
 C1 + C2

 ∆V

 ( 0.050 µ F )( 0.100 µ F ) 
=
 ( 400 V ) = 13.3 µ C on each
 0.050 µ F + 0.100 µ F 
49
50
CHAPTER 16
(b) Q1 =C1 ( ∆V ) = ( 0.050 µ F )( 400 V ) = 20.0 µ C
Q2 =C2 ( ∆V ) = ( 0.100 µ F )( 400 V ) = 40.0 µ C
16.30
(a) For parallel connection,
Ceq =C1 + C2 + C3 = ( 5.00 + 4.00 + 9.00 ) µ F = 18.0 µ F
(b) For series connection,
1
1
1
1
=
+
+
Ceq C1 C2 C3
1
1
1
1
=
+
+
, giving Ceq = 1.78 µ F
Ceq 5.00 µ F 4.00 µ F 9.00 µ F
16.31
(a) Using the rules for combining capacitors in series and in parallel, the circuit is
reduced in steps as shown below. The equivalent capacitor is shown to be a
2.00 µ F capacitor.
4.00 mF
a
3.00 mF
6.00 mF 3.00 mF
b
a
c
b
c
2.00 mF
a
c
2.00 mF
12.0 V
12.0 V
12.0 V
Figure 1
Figure 2
Figure 3
Electrical Energy and Capacitance
(b) From Figure 3:
From Figure 2:
51
Qac =Cac ( ∆V ) ac = ( 2.00 µ F ) ( 12.0 V ) = 24.0 µ C
Qab =Qbc = Qac = 24.0 µ C
Thus, the charge on the 3.00 µ F capacitor is Q3 = 24.0 µ C
Continuing to use Figure 2, ( ∆V ) ab =
( ∆V )3 = ( ∆V )bc =
and
Qab 24.0 µ C
=
= 4.00 V
Cab 6.00 µ F
Qbc 24.0 µ C
=
= 8.00 V
Cbc 3.00 µ F
From Figure 1, ( ∆V ) 4 = ( ∆V )2 = ( ∆V ) ab = 4.00 V
Q4 =C4 ( ∆V ) 4 = ( 4.00 µ F )( 4.00 V ) = 16.0 µ C
and
Q2 =C2 ( ∆V )2 = ( 2.00 µ F ) ( 4.00 V ) = 8.00 µ C
C parallel = C1 + C2 = 9.00 pF ⇒ C1 = 9.00 pF − C2
16.32
1
Cseries
=
(1)
C C
1
1
+
⇒ Cseries = 1 2 = 2.00 pF
C1 C2
C1 + C2
Thus, using equation (1),
Cseries =
( 9.00 pF − C2 ) C2
( 9.00 pF − C2 ) + C2
= 2.00 pF which reduces to
C22 − ( 9.00 pF ) C2 + 18.0 ( pF ) = 0 , or ( C2 − 6.00 pF )( C2 − 3.00 pF ) = 0
2
Therefore, either C2 = 6.00 pF and, from equation (1), C1 = 3.00 pF
or
C2 = 3.00 pF and C1 = 6.00 pF .
We conclude that the two capacitances are 3.00 pF and 6.00 pF .
52
CHAPTER 16
16.33
a
15.0
mF
6.00
mF
3.00
mF
c
b
a
20.0
mF
2.50
mF
6.00
mF
Figure 1
c
b
20.0
mF
Figure 2
c
a
8.50
mF
b
20.0
mF
Figure 3
(a) The equivalent capacitance of the upper branch between points a and c in Figure 1 is
Cs =
( 15.0 µ F )( 3.00 µ F )
15.0 µ F + 3.00 µ F
= 2.50 µ F
Then, using Figure 2, the total capacitance between points a and c is
Cac = 2.50 µ F+6.00 µ F=8.50 µ F
From Figure 3, the total capacitance is
−1


1
1
+
Ceq = 
 = 5.96 µ F
 8.50 µ F 20.0 µ F 
(b)
Qab = Qac = Qcb = ( ∆V ) ab Ceq
= ( 15.0 V )( 5.96 µ F ) = 89.5 µ C
Thus, the charge on the 20.0 µ C is
Q20 = Qcb = 89.5 µ C
 89.5 µ C 
 = 10.53 V
 20.0 µ F 
( ∆V )ac = ( ∆V )ab − ( ∆V )bc = 15.0 V − 
Then,
Q6 = ( ∆V ) ac ( 6.00 µ F ) = 63.2 µ C and
Q15 = Q3 = ( ∆V ) ac ( 2.50 µ F ) = 26.3 µ C
Electrical Energy and Capacitance
16.34
(a) The combination reduces to an equivalent capacitance of 12.0 µ F in stages as
shown below.
24.0
mF
36.0 V
4.00
mF
36.0 V
2.00 8.00
mF mF
4.00
mF
Figure 1
2.00
mF
6.00
mF
36.0 V
Figure 2
(b) From Figure 2,
12.00
mF
Figure 3
Q4 = ( 4.00 µ F )( 36.0 V ) = 144 µ C
Q2 = ( 2.00 µ F ) ( 36.0 V ) = 72.0 µ C
and
Q6 = ( 6.00 µ F )( 36.0 V ) = 216 µ C
Then, from Figure 1,
Q24 = Q8 = Q6 = 216 µ C
16.35
a
1.00
mF
24.0 V
a
6.00
mF
5.00
mF
24.0 V
b
8.00
mF
b
Figure 2
The circuit may be reduced in steps as shown above.
Qac = ( 4.00 µ F )( 24.0 V ) = 96.0 µ C
Then, in Figure 2, ( ∆V ) ab =
and
c
c
c
Using the Figure 3,
4.00
mF
24.0 V
12.0
mF
4.00
mF
Figure 1
a
Qac 96.0 µ C
=
= 16.0 V
Cab 6.00 µ F
( ∆V )bc = ( ∆V ) ac − ( ∆V )ab = 24.0 V − 16.0 V = 8.00 V
Figure 3
53
54
CHAPTER 16
Finally, using Figure 1,
Q1 = C1 ( ∆V ) ab = ( 1.00 µ F )( 16.0 V ) = 16.0 µ C
Q5 = ( 5.00 µ F )( ∆V ) ab = 80.0 µ C ,
and
16.36
Q8 = ( 8.00 µ F )( ∆V )bc = 64.0 µ C
Q4 = ( 4.00 µ F )( ∆V )bc = 32.0 µ C
The technician combines two of the capacitors in
parallel making a capacitor of capacitance 200 µ F .
Then she does it again with two more of the
capacitors. Then the two resulting 200 µ F
capacitors are connected in series to yield an
equivalent capacitance of 100 µ F . Because of the
symmetry of the solution, every capacitor in the
combination has the same voltage across it,
A
B
∆V = ( ∆V ) ab 2 = ( 90.0 V ) 2 = 45.0 V
16.37
(a) From Q = C ( ∆V ) , Q25 = ( 25.0 µ F )( 50.0 V ) = 1.25 × 10 3 µ C = 1.25 mC
and
Q40 = ( 40.0 µ F )( 50.0 V ) = 2.00 × 10 3 µ C = 2.00 mC
(b) When the two capacitors are connected in parallel, the equivalent capacitance is
Ceq = C1 + C2 = 25.0 µ F+40.0 µ F = 65.0 µ F .
Since the negative plate of one was connected to the positive plate of the other, the
total charge stored in the parallel combination is
Q = Q40 − Q25 = 2.00 × 10 3 µ C − 1.25 × 10 3 µ C = 750 µ C
The potential difference across each capacitor of the parallel combination is
∆V =
Q 750 µ C
=
= 11.5 V
Ceq 65.0 µ F
and the final charge stored in each capacitor is
′ = C1 ( ∆V ) = ( 25.0 µ F )( 11.5 V ) = 288 µ C
Q25
and
′ = Q − Q25
′ = 750 µ C − 288 µ C = 462 µ C
Q40
Electrical Energy and Capacitance
16.38
55
From Q = C ( ∆V ) , the initial charge of each capacitor is
Q10 = ( 10.0 µ F ) ( 12.0 V ) = 120 µ C and Qx = Cx ( 0 ) = 0
After the capacitors are connected in parallel, the potential difference across each is
∆V ′ = 3.00 V , and the total charge of Q = Q10 + Qx = 120 µ C is divided between the two
capacitors as
′ = ( 10.0 µ F ) ( 3.00 V ) = 30.0 µ C and
Q10
′ = 120 µ C − 30.0 µ C = 90.0 µ C
Qx′ = Q − Q10
Thus, Cx =
16.39
Qx′
90.0 µ C
=
= 30.0 µ F
∆V ′ 3.00 V
From Q = C ( ∆V ) , the initial charge of each capacitor is
Q1 = ( 1.00 µ F )( 10.0 V ) = 10.0 µ C and Q2 = ( 2.00 µ F ) ( 0 ) = 0
After the capacitors are connected in parallel, the potential difference across one is the
same as that across the other. This gives
∆V =
Q1′
Q2′
=
or Q2′ = 2 Q1′
1.00 µ F 2.00 µ F
(1)
From conservation of charge, Q1′ + Q2′ = Q1 + Q2 = 10.0 µ C . Then, substituting from
equation (1), this becomes
Q1′ + 2 Q1′ = 10.0 µ C , giving
Finally, from equation (1),
Q1′ =
10
µC
3
Q2′ =
20
µC
3
56
16.40
CHAPTER 16
The original circuit reduces to a single equivalent capacitor in the steps shown below.
C1
a
a
C1
Cs
C3
C2
a
C3
Cs
a
Cp1
Ceq
C2
C2
C2
C2
C2
b
b
−1
Cp2
b
b
−1
 1


1 
1
1
+
+
Cs = 
 =
 = 3.33 µ F
 5.00 µ F 10.0 µ F 
 C1 C2 
C p1 = Cs + C3 + Cs = 2 ( 3.33 µ F ) + 2.00 µ F = 8.66 µ F
C p 2 = C2 + C2 = 2 ( 10.0 µ F ) = 20.0 µ F
−1
−1
 1


1 
1
1
+
+
Ceq = 
 =
 = 6.04 µ F
 Cp1 Cp 2 
8.66
F
20.0
F
µ
µ




16.41
Refer to the solution of Problem 16.40 given above. The total charge stored between
points a and b is
Qeq = Ceq ( ∆V ) ab = ( 6.04 µ F )( 60.0 V ) = 362 µ C
Then, looking at the third figure, observe that the charges of the series capacitors of that
figure are Qp1 = Qp 2 = Qeq = 362 µ C . Thus, the potential difference across the upper
parallel combination shown in the second figure is
( ∆V ) p 1 =
Qp1
Cp1
=
362 µ C
= 41.8 V
8.66 µ F
Finally, the charge on C3 is
Q3 = C3 ( ∆V ) p1 = ( 2.00 µ F ) ( 41.8 V ) = 83.6 µ C
Electrical Energy and Capacitance
16.42
Recognize that the 7.00 µ F and the 5.00 µ F of the
center branch are connected in series. The total
capacitance of that branch is
−1
4.00
mF
a
1 
 1
Cs = 
+
 = 2.92 µ F
 5.00 7.00 
57
7.00
mF
b
5.00
mF
Then recognize that this capacitor, the 4.00 µ F
capacitor, and the 6.00 µ F capacitor are all connected
in parallel between points a and b. Thus, the equivalent
capacitance between points a and b is
6.00
mF
Ceq = 4.00 µ F + 2.92 µ F+6.00 µ F = 12.9 µ F
16.43
The capacitance is
∈ A ( 8.85 × 10
C= 0 =
d
−12
C2 N ⋅ m 2 )( 2.00 × 10 −4 m 2 )
5.00 × 10 m
-3
= 3.54 × 10 −13 F
and the stored energy is
1
1
2
2
W = C ( ∆V ) = ( 3.54 × 10 −13 F ) ( 12.0 V ) = 2.55 × 10 −11 J
2
2
16.44
(a) When connected in parallel, the energy stored is
1
1
1
2
2
2
W = C1 ( ∆V ) + C2 ( ∆V ) = ( C1 + C2 ) ( ∆V )
2
2
2
=
1
2
( 25.0 + 5.00 ) × 10 −6 F  ( 100 V ) = 0.150 J
2
(b) When connected in series, the equivalent capacitance is
1 
 1
+
Ceq = 

 25.0 5.00 
−1
µ F = 4.17 µ F
From W = 12 Ceq ( ∆V ) , the potential difference required to store the same energy as
2
in part (a) above is
∆V =
2W
=
Ceq
2 ( 0.150 J )
= 268 V
4.17 × 10 −6 F
58
16.45
CHAPTER 16
The capacitance of this parallel plate capacitor is
6
2
A 
C 2  ( 1.0 × 10 m )
−12
C = ∈0 =  8.85 × 10
= 1.1 × 10 −8 F
2 
d 
N ⋅ m  ( 800 m )
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
∆V = Ed = ( 3.0 × 10 6 V m ) ( 800 m ) = 2.4 × 109 V
Thus, the energy available for release in a lightning strike is
2
1
1
2
W = C ( ∆V ) = ( 1.1 × 10 −8 F )( 2.4 × 10 9 V ) = 3.2 × 1010 J
2
2
16.46
The energy transferred to the water is
8
1 1
 ( 50.0 C ) ( 1.00 × 10 V )
= 2.50 × 107 J
W=
Q ( ∆V )  =

100  2
200

Thus, if m is the mass of water boiled away,
W = m c ( ∆T ) + Lv  becomes


J 
6
2.50 × 107 J = m  4186
 ( 100°C − 30.0°C ) + 2.26 × 10 J kg 
kg ⋅ °C 


giving
16.47
2.50 × 107 J
m=
= 9.79 kg
2.55 J kg
The initial capacitance (with air between the plates) is Ci = Q ( ∆V )i , and the final
capacitance (with the dielectric inserted) is C f = Q ( ∆V ) f where Q is the constant
quantity of charge stored on the plates.
Thus, the dielectric constant is κ =
16.48
(a)
E=
Cf
Ci
=
( ∆V )i 100 V
=
= 4.0
( ∆V ) f 25 V
∆V
6.00 V
=
= 3.00 × 10 3 V m
d
2.00 × 10 -3 m
Electrical Energy and Capacitance
59
(b) With air between the plates, the capacitance is
Cair
−4
2
A 
C 2  ( 2.00 × 10 m )
−12
= ∈0 =  8.85 × 10
= 8.85 × 10 −13 F
2 
−3
d 
N ⋅ m  ( 2.00 × 10 m )
and with water (κ = 80 ) between the plates, the capacitance is
C = κ Cair = ( 80 ) ( 8.85 × 10 −13 F ) = 7.08 × 10 −11 F
The stored charge when water is between the plates is
Q = C ( ∆V ) = ( 7.08 × 10 −11 F ) ( 6.00 V ) = 4.25 × 10 −10 C = 42.5 nC
(c) When air is the dielectric between the plates, the stored charge is
Qair = Cair ( ∆V ) = ( 8.85 × 10 −13 F ) ( 6.00 V ) = 5.31 × 10 −12 C = 5.31 pC
16.49
(a) The dielectric constant for Teflon is κ = 2.1 , so the capacitance is
®
C=
κ ∈0 A
d
=
( 2.1) ( 8.85 × 10 −12
C 2 N ⋅ m 2 )( 175 × 10 −4 m 2 )
0.040 0 × 10 -3 m
C = 8.13 × 10 −9 F = 8.13 nF
(b) For Teflon , the dielectric strength is Emax = 60.0 × 10 6 V m , so the maximum voltage
is
®
Vmax = Emax d = ( 60.0 × 106 V m )( 0.040 0 × 10 -3 m
)
Vmax = 2.40 × 10 3 V = 2.40 kV
16.50
Before the capacitor is rolled, the capacitance of this parallel plate capacitor is
C=
κ ∈0 A
d
=
κ ∈0 ( w × L )
d
where A is the surface area of one side of a foil strip. Thus, the required length is
9.50 × 10 −8 F )( 0.025 0 × 10 −3 m )
(
C⋅d
=
= 1.04 m
L=
κ ∈0 w ( 3.70 ) ( 8.85 × 10 −12 C 2 N ⋅ m 2 )( 7.00 × 10 −2 m )
60
16.51
CHAPTER 16
m
(a) V =
ρ
=
1.00 × 10 −12 kg
= 9.09 × 10 −16 m 3
1100 kg m 3
Since V =
4π r 3
 3V 
, the radius is r = 
3
 4π 
 3V 
A = 4π r 2 = 4π 
 4π 
(b) C =
, and the surface area is
 3 ( 9.09 × 10 −16 m 3 ) 

= 4π 
4π


23
= 4.54 × 10 −10 m 2
κ ∈0 A
=
(c)
23
13
d
( 5.00 ) ( 8.85 × 10 −12 C2 N ⋅ m 2 )( 4.54 × 10 −10 m 2 )
100 × 10
−9
m
= 2.01 × 10 −13 F
Q = C ( ∆V ) = ( 2.01 × 10 −13 F )( 100 × 10 -3 V ) = 2.01 × 10 −14 C
and the number of electronic charges is
n=
16.52
Q 2.01 × 10 −14 C
=
= 1.26 × 10 5
-19
e 1.60 × 10 C
Since the capacitors are in parallel, the equivalent capacitance is
Ceq = C1 + C2 + C3 =
or
16.53
Ceq =
∈0 A1 ∈0 A2 ∈0 A3 ∈0 ( A1 + A2 + A3 )
+
+
=
d
d
d
d
∈0 A
where A = A1 + A2 + A3
d
Since the capacitors are in series, the equivalent capacitance is given by
d
d + d2 + d3
d
d
1
1
1
1
=
+
+
= 1 + 2 + 3 = 1
Ceq C1 C2 C3 ∈0 A ∈0 A ∈0 A
∈0 A
or
Ceq =
∈0 A
where d = d1 + d2 + d3
d
Electrical Energy and Capacitance
16.54
For the parallel combination: C p = C1 + C2 which gives
For the series combination:
Thus, we have
C p − C1 =
C2 =
1
1
1
=
+
Cs C1 C2
Cs C1
C1 − Cs
Cs C1
C1 − Cs
We write this result as :
16.55
or
C2 = C p − C1
61
(1)
1
1
1 C1 − Cs
=
−
=
C2 Cs C1
CsC1
and equating this to Equation (1) above gives
or
C p C1 − C p Cs − C12 + Cs C1 = CsC1
C12 − C p C1 + C p Cs = 0
and use the quadratic formula to obtain
1
1 2
C1 = C p ±
C p − C p Cs
2
4
Then, Equation (1) gives
1
C2 = C p ∓
2
1 2
C p − C p Cs
4
The charge stored on the capacitor by the battery is
Q = C ( ∆V )1 = C ( 100 V )
This is also the total charge stored in the parallel combination when this charged
capacitor is connected in parallel with an uncharged 10.0-µ F capacitor. Thus, if ( ∆V )2 is
the resulting voltage across the parallel combination, Q = C p ( ∆V )2 gives
C ( 100 V ) = ( C + 10.0 µ F )( 30.0 V ) or ( 70.0 V ) C = ( 30.0 V )( 10.0 µ F )
and
16.56
 30.0 V 
C=
 ( 10.0 µ F ) = 4.29 µ F
 70.0 V 
(a) The 1.0-µ C is located 0.50 m from point P, so its contribution to the potential at P is
V1 = k e
q1
 1.0 × 10 −6 C 
4
= ( 8.99 × 10 9 N ⋅ m 2 C 2 ) 
 = 1.8 × 10 V
r1
 0.50 m 
(b) The potential at P due to the −2.0-µ C charge located 0.50 m away is
V2 = k e
 − 2.0 × 10 −6 C 
q2
4
= ( 8.99 × 10 9 N ⋅ m 2 C 2 ) 
 = − 3.6 × 10 V
r2
0.50
m


62
CHAPTER 16
(c) The total potential at point P is VP = V1 + V2 = ( +1.8 − 3.6 ) × 10 4 V = − 1.8 × 10 4 V
(d) The work required to move a charge q = 3.0 µ C to point P from infinity is
W = q∆V = q (VP − V∞ ) = ( 3.0 × 10 −6 C )( −1.8 × 10 4 V − 0 ) = − 5.4 × 10 −2 J
16.57
The stages for the reduction of this circuit are shown below.
5.00 mF
3.00 mF
2.00 mF
4.00 mF
9.00 mF
3.00 mF
2.25 mF
3.00 mF
6.25 mF
4.00 mF
6.00 mF
6.00 mF
7.00 mF
48.0 V
12.0 mF
48.0 V
48.0 V
48.0 V
Thus, Ceq = 6.25 µ F
16.58
(a) Due to spherical symmetry, the charge on each of the concentric spherical shells will
be uniformly distributed over that shell. Inside a spherical surface having a uniform
charge distribution, the electric field due to the charge on that surface is zero. Thus,
in this region, the potential due to the charge on that surface is constant and equal
to the potential at the surface. Outside a spherical surface having a uniform charge
kq
distribution, the potential due to the charge on that surface is given by V = e
r
where r is the distance from the center of that surface and q is the charge on that
surface.
In the region between a pair of concentric spherical shells, with the inner shell
having charge + Q and the outer shell having radius b and charge − Q , the total
electric potential is given by
V = Vdue to
inner shell
+ Vdue to
outer shell
=
k eQ k e ( −Q )
1 1
+
= keQ  − 
r
b
r b
Electrical Energy and Capacitance
63
The potential difference between the two shells is therefore,
b−a
1 1
1 1
∆V = V r = a − V r = b = k e Q  −  − k e Q  −  = ke Q 

a b
b b
 ab 
The capacitance of this device is given by
C=
Q
ab
=
∆V
ke ( b − a )
(b) When b >> a , then b − a ≈ b . Thus, in the limit as b → ∞ , the capacitance found
above becomes
C→
16.59
ab
a
= = 4π ∈0 a
ke ( b ) ke
1
2
The energy stored in a charged capacitor is W = C ( ∆V ) . Hence,
2
∆V =
16.60
2 ( 300 J )
2W
=
= 4.47 × 10 3 V = 4.47 kV
C
30.0 × 10 -6 F
From Q = C ( ∆V ) , the capacitance of the capacitor with air between the plates is
C0 =
Q0 150 µ C
=
∆V
∆V
After the dielectric is inserted, the potential difference is held to the original value, but
the charge changes to Q = Q0 + 200 µ C=350 µ C . Thus, the capacitance with the dielectric
slab in place is
C=
Q
350 µ C
=
∆V
∆V
The dielectric constant of the dielectric slab is therefore
κ=
C  350 µ C   ∆V  350
=
= 2.33

=
C0  ∆V   150 µ C  150
64
16.61
CHAPTER 16
The charges initially stored on the capacitors are
Q1 = C1 ( ∆V )i = ( 6.0 µ F )( 250 V ) = 1.5 × 10 3 µ C
and
Q2 = C2 ( ∆V )i = ( 2.0 µ F )( 250 V ) = 5.0 × 10 2 µ C
When the capacitors are connected in parallel, with the negative plate of one connected
to the positive plate of the other, the net stored charge is
Q = Q1 − Q2 = 1.5 × 10 3 µ C − 5.0 × 10 2 µ C=1.0 × 10 3 µ C
The equivalent capacitance of the parallel combination is Ceq = C1 + C2 = 8.0 µ F . Thus,
the final potential difference across each of the capacitors is
( ∆V )′ =
Q 1.0 × 10 3 µ C
=
= 125 V
Ceq
8.0 µ F
and the final charge on each capacitor is
Q1′ = C1 ( ∆V )′ = ( 6.0 µ F )( 125 V ) = 750 µ C = 0.75 mC
and
16.62
Q2′ = C2 ( ∆V )′ = ( 2.0 µ F )( 125 V ) = 250 µ C = 0.25 mC
When connected in series, the equivalent capacitance is
−1
−1
 1
 1
1 
1 
4
+
+
Ceq = 
 =
 = µF
3
 4.0 µ F 2.0 µ F 
 C1 C2 
and the charge stored on each capacitor is
400
4

Q1 = Q2 = Qeq = Ceq ( ∆V )i =  µ F  ( 100 V ) =
µC
3
3

When the capacitors are reconnected in parallel, with the positive plate of one connected
to the positive plate of the other, the new equivalent capacitance is Ceq′ = C1 + C2 = 6.0 µ F
and the net stored charge is Q′ = Q1 + Q2 = 800 3 µ C . Therefore, the final potential
difference across each of the capacitors is
( ∆V )′ =
Q′ 800 3 µ C
=
= 44.4 V
6.0 µ F
Ceq′
Electrical Energy and Capacitance
65
The final charge on each of the capacitors is
Q1′ = C1 ( ∆V )′ = ( 4.0 µ F )( 44.4 V ) = 1.8 × 10 2 µ C
and
16.63
Q2′ = C2 ( ∆V )′ = ( 2.0 µ F )( 44.4 V ) = 89 µ C
(a) V = V1 + V2 + V3 =
keQ 2 keQ keQ
−
+
x+d
x
x−d
 x ( x − d ) − 2 ( x 2 − d2 ) + x ( x + d ) 

= keQ 
x ( x2 − d2 )


which simplifies to V =
2 k e Qd 2
x ( x2 − d2 )
=
2 k e Qd 2
x 3 − xd 2
(b) When x >> d , then x 2 − d 2 ≈ x 2
and V =
16.64
2 k e Qd 2
x ( x 2 − d2 )
becomes V ≈
2 k e Qd 2
x3
The energy required to melt the lead sample is
W = m cPb ( ∆T ) + L f 
= ( 6.00 × 10 −6 kg ) ( 128 J kg ⋅ °C ) ( 327.3°C − 20.0°C ) + 24.5 × 10 3 J kg 
= 0.383 J
1
2
The energy stored in a capacitor is W = C ( ∆V ) , so the required potential difference is
2
∆V =
2 ( 0.383 J )
2W
=
= 121 V
C
52.0 × 10 -6 F
66
16.65
CHAPTER 16
The capacitance of a parallel plate capacitor is C =
κ ∈0 A
d
Thus, κ ∈0 A = C ⋅ d , and the given force equation may be rewritten as
(Q C ) C = C ( ∆V )
Q2
Q2
=
=
F=
2 κ ∈0 A 2 C ⋅ d
2d
2d
2
2
With the given data values, the force is
2
( 20 × 10−6 F ) (100 V ) = 50 N
C ( ∆V )
F=
=
2d
2 ( 2.0 × 10 −3 m )
2
16.66
The electric field between the plates is directed downward with magnitude
Ey =
∆V
100 V
=
= 5.00 × 10 4 N m
d
2.00 × 10 -3 m
Since the gravitational force experienced by the electron is negligible in comparison to
the electrical force acting on it, the vertical acceleration is
ay =
Fy
me
=
qEy
me
=
( −1.60 × 10
−19
C )( −5.00 × 10 4 N m )
9.11 × 10
−31
kg
= + 8.78 × 1015 m s 2
(a) At the closest approach to the bottom plate, vy = 0 . Thus, the vertical displacement
from point O is found from vy2 = v02y + 2 ay ( ∆y ) as
∆y =
0 − ( v0 sin θ 0 )
2 ay
2
=
−  − ( 5.6 × 106 m s ) sin 45° 
2 ( 8.78 × 1015 m s 2 )
2
= − 0.89 mm
The minimum distance above the bottom plate is then
d=
D
+ ∆y = 1.00 mm − 0.89 mm = 0.11 mm
2
Electrical Energy and Capacitance
(b) The time for the electron to go from point O to the upper plate is found from
1
∆y = v0 y t + ay t 2 as
2
m
1
 

15 m  2
+1.00 × 10 −3 m =  −  5.6 × 10 6
 sin 45°  t +  8.78 × 10
t
s 
2
s2 
 

Solving for t gives a positive solution of t = 1.11 × 10 −9 s . The horizontal
displacement from point O at this time is
∆x = v0 x t = ( 5.6 × 10 6 m s ) cos 45° ( 1.11 × 10 −9 s ) = 4.4 mm
67
68
CHAPTER 16