PHYSICS 172
WQ 2010
Solutions to Homework
#4
1. Giancoli Chapter 24, Problem 6
The total charge will be conserved, and the final potential difference across the capacitors
will be the same.
Q0  Q1  Q2 ; V1  V2 
Q2  Q0  Q1  Q0  Q0
V1  V2 
Q1
C1
Q0

Q1
C1

Q2
C2

Q0  Q1
C2

C1
C1  C2

Q1  Q0
C1
C1  C2


 C1  C2 
C2
 Q2  Q0 
C1
C1  C2
C1
 V 
Q0
C1  C2
2. Giancoli Chapter 24, Problem 10
(a) The absolute value of the charge on each plate is given by Eq. 24-1. The plate with
electrons
has a net negative charge.
Q  CV  N  e   CV 
N
CV
 35  10 F 1.5V  3.281  10

15
5
 3.3  105 electrons
e
1.60  1019 C
(b) Since the charge is directly proportional to the potential difference, a 1.0% decrease in
potential
difference corresponds to a 1.0% decrease in charge.
Q  0.01Q ;
t 
Q
Q t

0.01Q
Q t

0.01CV
Q t
3. Giancoli Chapter 24, Problem 18



0.01 35  1015 F 1.5 V 
0.30  10
15
Cs
 1.75s  1.8s
positive plate
E
x
(a) The uncharged plate will polarize so that negative
charge will be drawn towards the positive capacitor
E
d xl
plate, and positive charge will be drawn towards the
negative capacitor plate. The same charge will be on
negative plate
each face of the plate as on the original capacitor
plates. The same electric field will be in the gaps as before the plate was inserted. Use
that electric field to determine the potential difference between the two original plates,
and the new capacitance. Let x be the distance from one original plate to the nearest
d
face of the sheet, and so d  l  x is the distance from the other original plate to the
other face of the sheet.
Q d  l  x

Q
Qx
E

; V1  Ex 
; V2  E  d  l  x  
 0 A 0
A 0
A 0
V  V1  V2 
Qx
A 0

Q d  l  x
A 0

Q d  l 
A 0

Q
C
 C  0
A
d  l 
(b)
Cinitial   0
A
d
; Cfinal   0
A
d  l 
;
0
Cfinal

Cinitial
A
d
d
1
d  l 



 1.7
A
d

l
d

0.40
d
0.60
0
d
4. Giancoli Chapter 24, Problem 24
The capacitors are in parallel, and so the potential is the same for each capacitor, and the
total charge on the capacitors is the sum of the individual charges. We use Eqs. 24-1 and 242.
Q1  C1V   0
A1
d1
V ; Q2  C2V   0
Qtotal  Q1  Q2  Q3   0
A1
d1
V  0
A2
d2
A2
d2
V ; Q3  C3V   0
V  0
A3
d3
A3
d3

A1

d1
V   0
V
 0
A2
d2
 0
A3 
V
d3 
 A1
A
A 
 0   0 2   0 3 V

d1
d2
d3 
 A
Q
A
A 
Cnet  total  
   0 1   0 2   0 3   C1  C2  C3
V
V
d2
d3 
 d1
5. Giancoli Chapter 24, Problem 27
The maximum capacitance is found by connecting the capacitors in parallel.
Cmax  C1  C2  C3  3.6  109 F  5.8  109 F  1.00  108 F  1.94  108 F in parallel
The minimum capacitance is found by connecting the capacitors in series.
1
Cmin
1
1 1 1
1
1
1

  1.82  109 F in series
 
  



9
9
8
 3.6  10 F 5.8  10 F 1.00  10 F 
 C1 C2 C3 
6. Giancoli Chapter 24, Problem 30
C1 and C2 are in series, so they both have the same charge. We then use that charge to find the
voltage across each of C1 and C2. Then their combined voltage is the voltage across C3. The
voltage across C3 is used to find the charge on C3.
Q1  Q2  12.4 C ; V1 
Q1
C1

12.4 C
16.0 F
 0.775 V ; V2 
Q2
C2

12.4 C
16.0 F
 0.775 V
V3  V1  V2  1.55 V ; Q3  C3V3  16.0 F 1.55 V   24.8C
From the diagram, C4 must have the same charge as the sum of the charges on C1 and C3.
Then the voltage across the entire combination is the sum of the voltages across C4 and C3.
Q
37.2 C
Q4  Q1  Q3  12.4 C  24.8C  37.2  C ; V4  4 
 1.31V
C4 28.5 F
Vab  V4  V3  1.31V  1.55 V  2.86 V
Here is a summary of all results.
Q1  Q2  12.4 C ; Q3  24.8C ; Q4  37.2 C
V1  V2  0.775 V ; V3  1.55 V ; V4  1.31V ; Vab  2.86 V
7. Giancoli Chapter 24, Problem 37
(a) The series capacitors add reciprocally, and then the parallel combination is found by
adding
linearly.
1
1
1
 1 1 
 C
 C  C3 
C2C3
C 
Ceq  C1      C1   3  2   C1   2
  C1 
C2  C3
 C2 C3 
 C2C3 C2C3 
 C2C3 
(b) For each capacitor, the charge is found by multiplying the capacitance times the voltage.
For
C1 , the full 35.0 V is across the capacitance, so Q1  C1V   24.0  106 F   35.0 V  
8.40  104 C . The equivalent capacitance of the series combination of C 2 and C3 has
the full 35.0 V across it, and the charge on the series combination is the same as the
charge on each of the individual capacitors.
1
1 
C
1
Ceq   


3
C C 2
Qeq  CeqV 
1
3
 24.0  10 F  35.0 V  2.80  10
6
4
C  Q2  Q3
8. Giancoli Chapter 24, Problem 40
No two capacitors are in series or in parallel in the diagram, and so we may not simplify by
that method. Instead use the hint as given in the problem. We consider point a as the higher
voltage. The equivalent capacitance must satisfy Qtot  CeqV .
(a) The potential between a and b can be written in three ways. Alternate but equivalent
expressions are shown in parentheses.
V  V2  V1 ; V  V2  V3  V4 ; V  V5  V4 V2  V3  V5 ; V3  V4  V1 
There are also three independent charge relationships. Alternate but equivalent
expressions are shown in parentheses. Convert the charge expressions to voltage –
capacitance expression.
Qtot  Q2  Q5
; Qtot  Q4  Q1
; Q2  Q1  Q3
Q4  Q3  Q5 
CeqV  C2V2  C5V5 ; CeqV  C4V4  C1V1 ; C2V2  C1V1  C3V3
We have a set of six equations: V  V2  V1 1 ; V  V2  V3  V4  2  ; V  V5  V4  3
CeqV  C2V2  C5V5  4  ; CeqV  C4V4  C1V1  5 ; C2V2  C1V1  C3V3  6 
Solve for C eq as follows.
From Eq. (1), V1  V  V2 . Rewrite equations (5) and (6). V1 has been
(i)
eliminated.
CeqV  C4V4  C1V  C1V2  5 ; C2V2  C1V  C1V2  C3V3  6 
From Eq. (3), V5  V  V4 . Rewrite equation (4). V5 has been eliminated.
(ii)
CeqV  C2V2  C5V  C5V4  4 
(iii)
From Eq. (2), V3  V  V2  V4 . Rewrite equation (6). V3 has been
eliminated.
C2V2  C1V  C1V2  C3V  C3V2  C3V4  6 
 C1  C2  C3 V2  C3V4  C1  C3 V  6
Here is the current set of equations.
CeqV  C2V2  C5V  C5V4  4 
CeqV  C4V4  C1V  C1V2  5
 C1  C2  C3 V2  C3V4   C1  C3 V  6 
(iv)
From Eq. (4), V4 
1
C5
C V
2 2
 C5V  CeqV  . Rewrite equations (5) and
(6).
C5CeqV  C4  C2V2  C5V  CeqV    C5C1V  C5C1V2  5
C5  C1  C2  C3 V2  C3  C2V2  C5V  CeqV    C5 C1  C3 V  6
(v)
C C
5
Group all terms by common voltage.
 C4Ceq  C4C5  C5C1 V   C4C2  C5C1 V2  5
eq
C5  C1  C3   C3Ceq  C3C5  V  C5  C1  C2  C3   C3C2 V2  6 
(vi)
Divide the two equations to eliminate the voltages, and solve for the
equivalent
capacitance.
C C
 C4 Ceq  C4 C5  C5C1 
 C4C2  C5C1 
C5  C1  C3   C3Ceq  C3C5  C5  C1  C2  C3   C3C2 
5
eq
Ceq 


C1C2 C3  C1C2 C4  C1C2 C5  C1C3C5  C1C4 C5  C2 C3C4  C2 C4 C5  C3C4 C5
C1C3  C1C4  C1C5  C2 C3  C2 C4  C2C5  C3C4  C3C5
(b) Evaluate with the given data. Since all capacitances are in  F, and the expression
involves capacitance cubed terms divided by capacitance squared terms, the result will
be in  F.
Ceq 


C1C2C3  C1C2C4  C1C2C5  C1C3C5  C1C4C5  C2C3C4  C2C4C5  C3C4C5
C1C3  C1C4  C1C5  C2C3  C2C4  C2C5  C3C4  C3C5
C1 C2  C3  C4  C5   C5  C3  C4    C4  C2C3  C2C5  C3C5 
C1  C3  C4  C5   C2  C3  C4  C5   C3  C4  C5 
 4.5 8.0 17.0    4.512.5  8.0  8.0  4.5  8.0   4.5   4.5 4.5
F
 4.517.0   8.0 17.0    4.512.5
 6.0  F
9. Giancoli Chapter 24, Problem 48
(a) Before the capacitors are connected, the only stored energy is in the initially-charged
capacitor.
Use Eq. 24-5.
U1  12 C1V02 
1
2
 2.20  10 F  12.0 V 
6
2
 1.584  104 J  1.58  10 4 J
(b) The total charge available is the charge on the initial capacitor. The capacitance changes
to the
equivalent capacitance of the two capacitors in parallel.
Q  Q1  C1V0 ; Ceq  C1  C2 ; U 2 
1
2
Q2
Ceq

1
2
C12V02
C1  C2
 2.20  10 F  12.0 V 
 5.70  10 F 
6

1
2
2
2
6
 6.114  105 J  6.11  10 5 J
(c) U  U 2  U1  6.114  10 5 J  1.584  10 4 J  9.73  10 5 J
¬
d
10. Giancoli Chapter 24, Problem 63
K
x
(a) We treat this system as two capacitors, one with a dielectric,
x
+
and one without a dielectric. Both capacitors have their high
voltage plates in contact and their low voltage plates in contact, so they are in parallel.
Use Eq. 24-2 and 24-8 for the capacitance. Note that x is measured from the right edge
of the capacitor, and is positive to the left in the diagram.
C  C1  C2   0
l l  x
d
l2 
x
 K 0
 0
1   K  1

d
d 
l 
lx
(b) Both “capacitors” have the same potential difference, so use U  12 CV 2 .
U
1
2
 C1  C2 V02 
0
l2 
x
1   K  1 V02
2d 
l 
(c) We must be careful here. When the voltage across a capacitor is constant and a dielectric
is inserted, charge flows from the battery to the capacitor. So the battery will lose energy
and the capacitor gain energy as the dielectric is inserted. As in Example 24-10, we assume
that work is done by an external agent Wnc  in such a way that the dielectric has no kinetic
energy. Then the work-energy principle (Chapter 8) can be expressed as Wnc  U or
dWnc  dU . This is analogous to moving an object vertically at constant speed. To increase
(decrease) the gravitational potential energy, positive (negative) work must be done by an
outside, non-gravitational source.
In this problem, the potential energy of the voltage source and the potential energy of the
capacitor both change as x changes. Also note that the change in charge stored on the
capacitor is the opposite of the change in charge stored in the voltage supply.
dWnc  dU  dU cap  dU battery  Fnc dx  d
Fnc  12 V02
dC
dx
 V0
dQbattery
dx
 12 V02
dC
dx
 V0

1
2

CV02  d QbatteryV0  
dQcap
dx
 12 V02
dC
dx
 V02
dC
dx
  12 V02
dC
dx
l   K  1 
V  l

 K  1


d  l
2d

Note that this force is in the opposite direction of dx, and so is to the right. Since this
force is being applied to keep the dielectric from accelerating, there must be a force of
equal magnitude to the left pulling on the dielectric. This force is due to the attraction of
the charged plates and the induced charge on the dielectric. The magnitude and direction
  12 V02 0
2
of this attractive force are
2
0 0
V02 0 l
2d
 K  1 , left .