Physics 133
Homework Chapter 25
Q1, Q3, P14, P23, P46, P75
Q1)
C = q/V, which is equal to the slope of the curve. The higher the slope, the higher
the capacitance. In addition, C α A/d, so the higher the ratio of A/d, the higher the
capacitance. Matching the highest slopes to the highest A/d ratios, it is apparent
that C2 belongs to curve a, C1 belongs to curve b and C3 belongs to curve c.
Q3)
a) For capacitors to be in series, only one end of that capacitor must connect to
only one end of the other capacitor, with nothing else connected to that point.
Therefore, this does not apply to capacitors C1 and C3 and they are not in series.
b)
For capacitors to be in parallel, each end of each capacitor must connect to each
end of the other capacitor. This does apply to the case of capacitors C1 and C2
and the are, therefore, in parallel.
c)
Examining each configuration, it is apparent that each case involves capacitors C1
and C2 in parallel and this parallel combination in series with capacitor C3.
Therefore, each circuit has the same equivalent capacitance,
1
C equivalent
P14)
 1
1 
 .
 

C

C
C
2
3 
 1
(a) The potential difference across C1 is V1 = 10.0 V. Thus,
q1 = C1V1 = (10.0 F)(10.0 V) = 1.00  10–4 C.
(b) We first consider the three-capacitor combination consisting of C2 and its two closest
neighbors, each of capacitance C. The equivalent capacitance of this combination is
1
Cequivalent
 1 1
 C  
   15F .
C
 2 C
This equivalent capacitance is in series with another 10 F capacitor. So the series
capacitance of these is,
1
C series
1

 1
1
1
1 

   

  6 F .
C

 15 F 10 F 
 equivalent C 
Since the 10 volt battery is across this series capacitance, the charge on each of these
capacitors in series is just,
q  C seriesV  6 F (10volts ) 1  60 C .
Therefore, the voltage across the original three-capacitor combination involving C2 is,
V 
q
Cequivalent

60 C
 4volts.
15 F
This 4 volt potential is across the C2 capacitor and the 10 F capacitor in series with it,
with their series capacitance of 5 F. Therefore, the charge on each of these capacitors is
determined to be,
q  CV  5F 4volts   20C  2.00  10 5 C.
P23)
F.
We note that the total equivalent capacitance is C123 = [(C3)1 + (C1 + C2)1]1 = 6
(a) Thus, the charge that passed point a is C123 Vbatt = (6 F)(12 V) = 72 C. Dividing this
by the value e = 1.60  1019 C gives the number of electrons: 4.5  1014, which travel to
the left, toward the positive terminal of the battery.
(b) The equivalent capacitance of the parallel pair is C12 = C1 + C2 = 12 F. Thus, the
voltage across the pair (which is the same as the voltage across C1 and C2 individually) is
72 C
= 6 V.
12 F
Thus, the charge on C1 is q1  (4 F)(6 V) = 24 C, and dividing this by e gives
N1  q1 / e  1.5 1014 , the number of electrons that have passed (upward) through point b.
(c) Similarly, the charge on C2 is q2  (8 F)(6 V) = 48 C, and dividing this by e gives
N 2  q2 / e  3.0 1014 , the number of electrons which have passed (upward) through
point c.
(d) Finally, since C3 is in series with the battery, its charge is the same charge that passed
through the battery (the same as passed through the switch). Thus, 4.5  1014 electrons
passed rightward though point d. By leaving the rightmost plate of C3, that plate is then
the positive plate of the fully charged capacitor, making its leftmost plate (the one closest
to the negative terminal of the battery) the negative plate, as it should be.
(e) As stated in (b), the electrons travel up through point b.
(f) As stated in (c), the electrons travel up through point c.
P46) Each capacitor has 12.0 V across it, so Eq. 25-1 yields the charge values once we
know C1 and C2. From Eq. 25-9,
C2 =
and from Eq. 25-27,
C1 =
This leads to
0 A
d
 0 A
d
= 2.21  1011 F ,
= 6.64  1011 F .
q1 = C1V1 = 8.00  1010 C, q2 = C2V2 = 2.66  1010 C.
The addition of these gives the desired result: qtot = 1.06  109 C. Alternatively, the
circuit could be reduced to find the qtot.
P75) We cannot expect simple energy conservation to hold since energy is presumably
dissipated either as heat in the hookup wires or as radio waves while the charge oscillates
in the course of the system “settling down” to its final state (of having 40 V across the
parallel pair of capacitors C and 60 F). We do expect charge to be conserved. Thus, if
Q is the charge originally stored on C and q1, q2 are the charges on the parallel pair after
“settling down,” then
Q  q1  q2

C 100 V   C  40 V    60  F  40 V 
which leads to the solution C = 40 F.