Chapter 24 Capacitance, Dielectrics, Electric Energy Storage
(c) If the separation between the plates doubles, the capacitance is halved. The potential
difference across the plates doesn’t change if the capacitor remains connected to the battery, so
the energy stored is also halved: U 1 CV 2 .
2
10. (c) If the voltage across a capacitor is doubled, the amount of energy it can store is quadrupled:
U 1 CV 2 .
2
11. The dielectric will be pulled into the capacitor by the electrostatic attractive forces between the
charges on the capacitor plates and the polarized charges on the dielectric’s surface. (Note that the
addition of the dielectric decreases the energy of the system.)
12. If the battery remains connected to the capacitor, the energy stored in the electric field of the
capacitor will increase as the dielectric is inserted. Since the energy of the system increases, work
must be done and the dielectric will have to be pushed into the area between the plates. If it is
released, it will be ejected.
13. (a) If the capacitor is isolated, Q remains constant, and U 1 Q2 becomes U ' 1 Q2 and the
2 C 2 KC
stored energy decreases.
(b) If the capacitor remains connected to a battery so V does not change, U 1 CV 2 becomes
2
U ' 1 KCV 2 , and the stored energy increases.
2
14. For dielectrics consisting of polar molecules, one would expect the dielectric constant to decrease
with temperature. As the thermal energy increases, the molecular vibrations will increase in
amplitude, and the polar molecules will be less likely to line up with the electric field.
15. When the dielectric is removed, the capacitance decreases. The potential difference across the plates
remains the same because the capacitor is still connected to the battery. If the potential difference
remains the same and the capacitance decreases, the charge on the plates and the energy stored in the
capacitor must also decrease. (Charges return to the battery.) The electric field between the plates
will stay the same because the potential difference across the plates and the distance between the
plates remain constant.
16. For a given configuration of conductors and dielectrics, C is the proportionality constant between the
voltage between the plates and the charge on the plates.
17. The dielectric constant is the ratio of the capacitance of a capacitor with the dielectric between the
plates to the capacitance without the dielectric. If a conductor were inserted between the plates of a
capacitor such that it filled the gap and touched both plates, the capacitance would drop to zero since
charge would flow from one plate to the other. So, the dielectric constant of a good conductor would
be zero.
Solutions to Problems
1. The capacitance is found from Eq. 24-1.
Q CV C Q 2.8 103C 3.0 106 F 3.0F
V 930 V
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2. We assume the capacitor is fully charged, according to Eq. 24-1.
Q CV 12.6 106 F 12.0 V 1.51104 C
3. The capacitance is found from Eq. 24-1.
Q CV C Q 75 1012 C 3.11012 F 3.1pF
V 24.0 V
4. Let Q1 and V1 be the initial charge and voltage on the capacitor, and let Q2 and V2 be the final
charge and voltage on the capacitor. Use Eq. 24-1 to relate the charges and voltages to the
capacitance.
Q1 CV1 Q2 CV2 Q2 Q1 CV2 CV1 C V2 V1
C Q2 Q1 26 106 C 5.2 107 F 0.52F
V2 V1 50 V
5. After the first capacitor is disconnected from the battery, the total charge must remain constant. The
voltage across each capacitor must be the same when they are connected together, since each
capacitor plate is connected to a corresponding plate on the other capacitor by a constant-potential
connecting wire. Use the total charge and the final potential difference to find the value of the
second capacitor.
QTotal C1V1 Q1 C V1 final Q2 C V2 final
initial final final
QTotal Q1 Q2 C1 C2 Vfinal C1V1 C1 C2 Vfinal
final final initial
V1 125 V 1
C2 C1 initial 1 7.7 106 F 15 V 5.6 105 F 56F
Vfinal
6. The total charge will be conserved, and the final potential difference across the capacitors will be the
same.
Q0 Q1 Q2 ; V1 V2 Q1 Q2 Q0 Q1 Q1 Q0 C1
C1 C2 C2 C1 C2
Q2 Q0 Q1 Q0 Q0 C1 Q2 Q0 C1 C2
C1 C2 C2
Q1 Q0 C1 Q0
C1 C1 C2 C1 C2
V1 V2 V
C1
7. The work to move the charge between the capacitor plates is W qV , where V is the voltage
difference between the plates, assuming that q Q so that the charge on the capacitor does not
change appreciably. The charge is then found from Eq. 24-1. The assumption that q Q is justified.
W qV q Q Q CW 15μF 15J 1.1C
C
q 0.20 mC
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102
Chapter 24 Capacitance, Dielectrics, Electric Energy Storage
8. (a) The total charge on the combination of capacitors is the sum of the charges on the two
individual capacitors, since there is no battery connected to them to supply additional charge,
and there is no neutralization of charge by combining positive and negative charges. The
voltage across each capacitor must be the same after they are connected, since each capacitor
plate is connected to a corresponding plate on the other capacitor by a constant-potential
connecting wire. Use the total charge and the fact of equal potentials to find the charge on each
capacitor and the common potential difference.
Q1 C1V1 Q2 C2V2 Q1 C V1 final Q2 C V2 final
initial initial initial initial final final
QTotal Q1 Q2 Q1 Q2 C1V1 C2V2 C V1 final C V2 final
initial initial final final initial initial
C1V1 C2V2
Vfinal
initial initial 2.70 106 F 475 V 4.00 106 F 525 V
C1 C2 6.70 106 F
504.85 V 505 V V1 V2
Q1 C V1 final 2.70 106 F 504.85 V 1.36 103C
final
Q2 C V2 final 4.00 106 F 504.85 V 2.02 103C
final
(b) By connecting plates of opposite charge, the total charge will be the difference of the charges on
the two individual capacitors. Once the charges have equalized, the two capacitors will again
be at the same potential.
Q1 C1V1 Q2 C2V2 Q1 C V1 final Q2 C V2 final
initial initial initial initial final final
QTotal Q1 Q2 Q1 Q2 C1V1 C2V2 C V1 final C V2 final
initial initial final final initial initial
C1V1 C2V2
Vfinal
initial initial 2.70 106 F 475 V 4.00 106 F 525 V
C1 C2 6.70 106 F
122.01V 120 V V1 V2
Q1 C V1 final 2.70 106 F 122.01V 3.3 104 C
final
Q2 C V2 final 4.00 106 F 122.01V 4.9 104 C
final
9. Use Eq. 24-1.
Q CV ; t Q CV 1200 F 6.0 V 7.2 106 s 1d s 83 d
Q t Q t 86, 400
1.0 103 C s
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
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N
CV 35 1015 F 1.5 V 3.281105 3.3 105 electrons
e
1.60 1019 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 0.01Q 0.01CV 0.01 35 1015 F 1.5 V 1.75s 1.8 s
Q t Q t Q t
0.30 1015 C s
11. Use Eq. 24-2.
C
0 A A Cd 0.40 106 F 2.8 103 m 126.6 m2 130 m2
d 0 8.85 1012 C2 Nm2
If the capacitor plates were square, they would be about 11.2 m on a side.
12. The capacitance per unit length of a coaxial cable is derived in Example 24-2
C
2 0 2 8.85 1012 C2 Nm2 3.5 1011 F m
l ln R Routside inside ln 5.0 mm 1.0 mm
13. Inserting the potential at the surface of a spherical conductor into Eq. 24.1 gives the capacitance of a
conducting sphere. Then inserting the radius of the Earth yields the Earth’s capacitance.
Q Q
V
4 0 r
C
Q 4 0 r 4 8.85 1012 F/m 6.38 106 m 7.10 104 F
14. From the symmetry of the charge distribution, any electric field Ra d –Q
must be radial, away from the cylinder axis, and its magnitude +Q
must be independent of the location around the axis (for a given
radial location). We assume the cylinders have charge of
magnitude Q in a length l. Choose a Gaussian cylinder of Rb
length d and radius R, centered on the capacitor’s axis, with
d l and the Gaussian cylinder far away from both ends of Gaussian cylinder
the capacitor. On the ends of this cylinder, E dA and so of radius R
there is no flux through the ends. On the curved side of the
cylinder, the field has a constant magnitude and E dA . Thus EdA EdA. Write Gauss’s law.
Qencl
0
EdA E Acurved curved E 2 Rd
walls walls
For R Rb , Qencl 0 E 2 Rd 0 0 E 0.
For R Ra , Qencl Q d Q d 0 , and so Qencl 0 E 2 Rd 0 0 E 0.
ll
15. We assume there is a uniform electric field between the capacitor plates, so that V Ed , and then
use Eqs. 24-1 and 24-2.
Qmax A
CVmax 0 d Emax d 0 AEmax 8.85 1012 F/m 6.8 104 m2 3.0 106 V m
1.8 108 C
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104
Chapter 24 Capacitance, Dielectrics, Electric Energy Storage
16. We assume there is a uniform electric field between the capacitor plates, so that V Ed , and then
use Eqs. 24-1 and 24-2.
A Ed
d
Q
CV 0 0 AE 8.85 1012 F/m 21.0 104 m2 4.80 105 V m
8.92 109 C
17. We assume there is a uniform electric field between the capacitor plates, so that V Ed , and then
use Eqs. 24-1 and 24-2.
E Q 92 106 C 5.8 104 V m
Cd 0.80 106 F 2.0 103 m
Q CV CEd
18. (a) The uncharged plate will polarize so that negative positive plate x
charge will be drawn towards the positive capacitor
plate, and positive charge will be drawn towards the E
negative capacitor plate. The same charge will be d
on each face of the plate as on the original capacitor E d xl
plates. The same electric field will be in the gaps as
before the plate was inserted. Use that electric field negative plate
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 face of the sheet, and so d l x is the distance from the other original plate to the
other face of the sheet.
E Q ; V1 Ex Qx ; V2 E d l x Qd l x
0 A0 A 0
A 0
V V1 V2 Qx Qd l x Qd l Q C 0 A
A 0 C
A 0 A 0 d l
A
0 d l
A
0 d
(b) A A Cfinal d d 1
Cinitial 0 d ; Cfinal 0 d l ; Cinitial d l d 0.40d 0.60 1.7
19. (a) The distance between plates is obtained from Eq. 24-2.
C 0A x 0A
xC
Inserting the maximum capacitance gives the minimum plate separation and the minimum
capacitance gives the maximum plate separation.
xmin
0A 8.85 pF/m 25 106 m2 0.22 m
Cmax
1000.0 1012 F
xmax
oA 8.85 pF/m 25 106 m2 0.22 mm 220m
Cmin
1.0 pF
So 0.22 m x 220 m .
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(b) Differentiating the distance equation gives the approximate uncertainty in distance.
x dx C d 0A C 0A C .
dC dC C C2
The minus sign indicates that the capacitance increases as the plate separation decreases. Since
only the magnitude is desired, the minus sign can be dropped. The uncertainty is finally written
in terms of the plate separation using Eq. 24-2.
x 0A C x 2 C
0A
0A 2
x
(c) The percent uncertainty in distance is obtained by dividing the uncertainty by the
separation distance.
xmin C 0.22 m 0.1pF 100%
oA 8.85 pF/m 25 mm2
xmin
xmin
100% 100% 0.01%
0.22 mm 0.1pF 100%
8.85 pF/m 25 mm2
xmax xmax C
oA
xmax
100% 100% 10%
20. The goal is to have an electric field of strength ES at a radial distance of 5.0 Rb from the center of the
cylinder. Knowing the electric field at a specific distance allows us to calculate the linear charge
density on the inner cylinder. From the linear charge density and the capacitance we can find the
potential difference needed to create the field. From the cylindrically symmetric geometry and
Gauss’s law, the field in between the cylinders is given by E 1 . The capacitance of a
20 R
cylindrical capacitor is given in Example 24-2.
E R 5.0Rb 1 ES 2 0 5.0Rb ES Q
2 0 l
5.0Rb
Q CV V Q Q Q ln Ra Rb 2 0 5.0Rb ES ln Ra Rb
C 2 l l
2 2
ln Ra Rb
5.0Rb ES ln Ra 0.100 m
Rb 5.0 1.0 104 m 2.7 106 N C ln 1.0 104 m 9300 V
21. To reduce the net capacitance, another capacitor must be added in series.
1 11 1 1 1 C1 Ceq
Ceq C1 C2 C2 Ceq C1 C1Ceq
C2
C1Ceq 2.9 109 F 1.6 109 F 3.57 109 F 3600 pF
C1 Ceq 2.9 109 F 1.6 109 F
Yes, an existing connection needs to be broken in the process. One of the connections of the original
capacitor to the circuit must be disconnected in order to connect the additional capacitor in series.
22. (a) Capacitors in parallel add according to Eq. 24-3.
Ceq C1 C2 C3 C4 C5 C6 6 3.8 106 F 2.28 105 F 22.8F
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Chapter 24 Capacitance, Dielectrics, Electric Energy Storage
(b) Capacitors in series add according to Eq. 24-4.
Ceq 1 1 1 1 1 1 1 3.8 6 F 1 3.8 106 F 6.3 107 F
C1 106 6
C2 C3 C4 C5 C6
0.63F
23. We want a small voltage drop across C1. Since V Q C , if we put
the smallest capacitor in series with the battery, there will be a large
voltage drop across it. Then put the two larger capacitors in parallel, C3 C1
so that their equivalent capacitance is large and therefore will have a
small voltage drop across them. So put C1 and C3 in parallel with V0 C2
each other, and then put that combination in series with C2. See the
diagram. To calculate the voltage across C1, find the equivalent
capacitance and the net charge. That charge is used to find the
voltage drop across C2, and then that voltage is subtracted from the battery voltage to find the
voltage across the parallel combination.
1 1 Qeq
C2
1 C1 C2 C3 Ceq C2 C1 C3 ; Qeq CeqV0 ; V2 Q2 ;
C1 C2 C3 C2
Ceq C2 C1 C3 C2 C1 C3
V1 V0 V2 C2
Qeq CeqV0 C1 C1 C3 V0 C2 1.5F
C2 C2 C2 C3 C2 6.5F
V0 V0 V0 V0 12 V
C2 C1 C3
2.8 V
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 24-2.
Q1 C1V 0 A1 V ; Q2 C2V 0 A2 V ; Q3 C3V 0 A3 V
d1 d2 d3
Qtotal Q1 Q2 Q3 0 A1 V 0 A2 V 0 A3 V 0 A1 0 A2 0 A3 V
d1 d2 d3 d1 d2 d3
Qtotal 0 A1 0 A2 0 A3 V A1 A2 A3
V d1 d2 d3 d1 d2 d3
Cnet 0 0 C1 C2 C3
V 0
25. Capacitors in parallel add linearly, and so adding a capacitor in parallel will increase the net
capacitance without removing the 5.0 F capacitor.
5.0 F C 16F C 11F connected in parallel
26. (a) The two capacitors are in parallel . Both capacitors have their high voltage plates at the same
potential (the middle plate), and both capacitors have their low voltage plates at the same
potential (the outer plates, which are connected).
(b) The capacitance of two capacitors in parallel is the sum of the individual capacitances.
C C1 C2 0A 0A A 1 1 A d1 d2
d1 d2 d1 d1d2
0 d2 0
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(c) Let l d1 d2 constant. Then C 0 Al 0 Al . We see that C as d1 0 or
d1d2
d1 l d1
d1 l (which is d2 0 ). Of course, a real capacitor would break down as the plates got too
close to each other. To find the minimum capacitance, set dC 0 and solve for d1.
d d1
d l 2d1
d1 d1l d12 2
dC
d d1
d 0 Al 0 Al 0 d1 1 l d2
d1l d12 2
Cmin A d1 d2 A l A 4 A 4
d1d2 l d2
0 d1 0 1 l 1 l 0 0 d1
2 2
1 l
2
Cmin 40 A ; Cmax
d1 d2
27. The maximum capacitance is found by connecting the capacitors in parallel.
Cmax C1 C2 C3 3.6 109 F 5.8 109 F 1.00 108 F 1.94 108 F in parallel
The minimum capacitance is found by connecting the capacitors in series.
Cmin 1 1 1 1 1 1 1 1 1.82 109 F in series
C2 C3 3.6 109 F 5.8 109 F 108
C1 1.00 F
28. When the capacitors are connected in series, they each have the same charge as the net capacitance.
(a) Q1 Q2 Qeq CeqV 1 1 1 V 1 0.80 1 F 1 9.0 V
C1 C2 0.50 106 F 106
2.769 106 C
V1 Q1 2.769 106 C 5.538 V 5.5 V V2 Q2 2.769 106 C 3.461V 3.5 V
C1 0.50 106 F C2 0.80 106 F
(b) Q1 Q2 Qeq 2.769 106 C 2.8 106 C
When the capacitors are connected in parallel, they each have the full potential difference.
(c) V1 9.0 V V2 9.0 V Q1 C1V1 0.50 106 F 9.0 V 4.5 106 C
Q2 C2V2 0.80 106 F 9.0 V 7.2 106C
29. (a) From the diagram, we see that C1 and C2 are in series. That combination is in parallel with C3,
and then that combination is in series with C4. Use those combinations to find the equivalent
capacitance. We use subscripts to indicate which capacitors have been combined.
1 11 C12 1 C ; C123 C12 C3 1 C C 3 C ;
2 2 2
C12 C C
1 1 1 21 5 C1234 3 C
5
C1234 C123 C4 3C C 3C
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Chapter 24 Capacitance, Dielectrics, Electric Energy Storage
(b) The charge on the equivalent capacitor C1234 is given by Q1234 C1234V 3 CV . This is the
5
charge on both of the series components of C1234.
Q123 3 CV C V123 123 3 CV123 V123 2 V
5 2 5
Q4 3 CV C4V4 V4 3 V
5 5
The voltage across the equivalent capacitor C123 is the voltage across both of its parallel
components. Note that the sum of the charges across the two parallel components of C123 is the
same as the total charge on the two components, 3 CV .
5
V123
2 V V12 ; Q12 C12V12 1 C 2 V 1 CV
5 2 5 5
V123 V CV
2 V3 ; Q3 C3V3 C 2 V 2
5 5 5
Finally, the charge on the equivalent capacitor C12 is the charge on both of the series
components of C12.
Q12 1 CV Q1 C1V1 V1 1 V ; Q12 1 CV Q2 C1V2 V2 1 V
5 5 5 5
Here are all the results, gathered together.
Q1 Q2 1 CV ; Q3 2 CV ; Q4 3 CV
5 5 5
V1 V2 1 V ; V3 2 V ; V4 3 V
5 5 5
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 12.4C ; V1 Q1 12.4C 0.775V ; V2 Q2 12.4C 0.775V
C1 16.0F C2 16.0F
V3 V1 V2 1.55V ; Q3 C3V3 16.0F 1.55V 24.8C
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.
Q4 Q1 Q3 12.4C 24.8C 37.2C ; V4 Q4 37.2C 1.31V
C4 28.5F
Vab V4 V3 1.31V 1.55V 2.86 V
Here is a summary of all results.
Q1 Q2 12.4C ; Q3 24.8C ; Q4 37.2C
V1 V2 0.775 V ; V3 1.55 V ; V4 1.31V ; Vab 2.86 V
31. When the switch is down the initial charge on C2 is calculated from Eq. 24-1. C1
C2
Q2 C2V0
V0
When the switch is moved up, charge will flow from C2 to C1 until the voltage S
across the two capacitors is equal.
V Q2 Q1 Q2 Q1 C2
C2 C1 C1
The sum of the charges on the two capacitors is equal to the initial charge on C2.
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Q2 Q2 Q1 Q1 C2 Q1 Q1 C2 C1
C1 C1
Inserting the initial charge in terms of the initial voltage gives the final charges.
Q1 C2 C1 C2V0 Q1 C1C2 V0 ; Q2 Q1 C2 C22 C V0
C1 C2 C1 C1 C2
32. (a) From the diagram, we see that C1 and C2 are in parallel, and C1 C3
C3 and C4 are in parallel. Those two combinations are then in
series with each other. Use those combinations to find the C2 c C4
equivalent capacitance. We use subscripts to indicate which a C12 C34 b
a cb
capacitors have been combined.
C12 C1 C2 ; C34 C3 C4 ;
1 11 1 1
C1234 C12 C34 C1 C2 C3 C4
C1234 C1 C2 C3 C4 C1234 b
C1 C2 C3 C4 a
(b) The charge on the equivalent capacitor C1234 is given by Q1234 C1234V . This is the charge on
both of the series components of C1234. Note that V12 V34 V .
C1 C2 C3 C4
Q12 C1234V C12V12
V12 C1234 V C1 C2 C3 C4 V C3 C4 V
C12
C1 C2 C1 C2 C3 C4
C1 C2 C3 C4
Q34 C1234V C34V34
V34 C1234 V C1 C2 C3 C4 V C1 C2 V
C34
C3 C4 C1 C2 C3 C4
The voltage across the equivalent capacitor C12 is the voltage across both of its parallel
components, and the voltage across the equivalent C34 is the voltage across both its parallel
components.
V12 V1 V2 C3 C4 V ;
C1 C2 C3 C4
C1V1 Q1 C1 C3 C4 V ; C2V2 Q2 C2 C3 C4 V
C1 C2 C3 C4 C1 C2 C3 C4
V34 V3 V2 C1 C2 V ;
C1 C2 C3 C4
C3V3 Q3 C3 C1 C2 V ; C4V4 Q4 C4 C1 C2 V
C1 C2 C3 C4 C1 C2 C3 C4
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110
Chapter 24 Capacitance, Dielectrics, Electric Energy Storage
33. (a) The voltage across C3 and C4 must be the same, since they are in parallel.
V3 V4 Q3 Q4 Q4 Q3 C4 23C 16F 46C
C3 C4 C3 8F
The parallel combination of C3 and C4 is in series with the parallel combination of C1 and C2,
and so Q3 Q4 Q1 Q2. That total charge then divides between C1 and C2 in such a way that
V1 V2.
Q1 Q2 Q3 Q4 69C ; V1 V2 Q1 Q4 69C Q1
C1 C4 C4
Q1 C1 69C 8.0F 69C 23C ; Q2 69C 23C 46C
24.0F
C4 C1
Notice the symmetry in the capacitances and the charges.
(b) Use Eq. 24-1.
V1 Q1 23C 2.875V 2.9 V ; V2 V1 2.9 V
C1 8.0F
V3 Q3 23C 2.875V 2.9 V ; V4 V3 2.9 V
C3 8.0F
(c) Vba V1 V3 2.875V 2.875V 5.75V 5.8 V
34. We have CP C1 C2 and 1 1 1. Solve for C1 and C2 in terms of CP and CS.
CS C1 C2
1 11 1 1 CP C1 C1 CP
CS C1 C2 C1 CP C1 C1 CP C1
C1 CP C1
1 CP C12 CPC1 CPCS 0
CS C1 CP C1
C1 CP CP2 4CPCS 35.0F 35.0F2 435.0F 5.5F
2
2
28.2F, 6.8F
C2 CP C1 35.0F 28.2F 6.8F or 35.0F 6.8F 28.2F
So the two values are 28.2F and 6.8F .
35. Since there is no voltage between points a and b, we can imagine there a
Cx
being a connecting wire between points a and b. Then capacitors C1 and C1 d
C2 are in parallel, and so have the same voltage. Also capacitors C3 and c
Cx are in parallel, and so have the same voltage. V
C2
V1 V2 Q1 Q2 ; V3 Vx Q3 Qx C3
C1 C2 C3 Cx b
V0
Since no charge flows through the voltmeter, we could also remove it
from the circuit and have no change in the circuit. In that case,
capacitors C1 and Cx are in series and so have the same charge.
Likewise capacitors C2 and C3 are in series, and so have the same
charge.
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Q1 Qx ; Q2 Q3
Solve this system of equations for Cx.
Q3 Qx Cx C3 Qx C3 Q1 C3 C1 4.8F 8.9F 2.4F
C3 Cx Q3 Q2 C2 18.0F
36. The initial equivalent capacitance is the series combination of the two individual capacitances. Each
individual capacitor will have the same charge as the equivalent capacitance. The sum of the two
initial charges will be the sum of the two final charges, because charge is conserved. The final
potential of both capacitors will be equal.
1 11
Ceq C1 C2
Ceq C1C2 ; Qeq CeqV0 C1C2 V0 3200 pF 1800 pF 12.0 V 13,824 pC
C1 C2 C1 C2
5000 pF
Q1 Q2 2Qeq Q1
Q1 Q2 2Qeq ; V1 V2 final final final
C1 C2
final final final final C2
Q1 2 C1 Qeq 2 3200 pF 13, 824 pC 17,695pC 1.8 108C
C1 C2 5000 pF
final
Q2 2Qeq Q1 2 13,824 pC 17, 695pC 9953pC 1.0 108C
final final
37. (a) The series capacitors add reciprocally, and then the parallel combination is found by adding
linearly.
Ceq C1 1 1 1 C1 C3 C2 1 C1 C2 C3 1 C1 C2C3
C2 C2C3 C2C3 C2C3 C2 C3
C3
(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 24.0 106 F 35.0 V
8.40 104 C . The equivalent capacitance of the series combination of C2 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 1 C Qeq
Ceq C C 2 3 CeqV 1 24.0 106 F 35.0 V 2.80 104 C Q2 Q3
3
38. From the circuit diagram, we see that C1 is in parallel with the voltage, and so V1 24 V .
Capacitors C2 and C3 both have the same charge, so their voltages are inversely proportional to their
capacitance, and their voltages must total to 24.0 V.
Q2 Q3 C2V2 C3V3 ; V2 V3 V
V2 C2 V2 V V2 C3 V 4.00F 24.0 V 13.7 V
C3 C2 C3 7.00F
V3 V V2 24.0 V 13.7 V 10.3V
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112
Chapter 24 Capacitance, Dielectrics, Electric Energy Storage
39. For an infinitesimal area element of the capacitance a distance y up from the
small end, the distance between the plates is d x d y tan d y.
Since the capacitor plates are square, they are of dimension A A, and the dy
y
area of the infinitesimal strip is dA A dy. The infinitesimal capacitance dC
d
of the strip is calculated, and then the total capacitance is found by adding
together all of the infinitesimal capacitances, in parallel with each other.
C 0 A dC 0 d dA 0 d A dy
d y y
dC A A dy A
C 0 0 d y
0 A ln d y
0
0 A ln A ln d 0 A d A 0 A ln 1 A
d ln d d
We use the approximation from page A-1 that ln 1 x x 1 x2.
2
C 0 A ln 1 A 0 A A 1 A 2 0A 1 A
2 d 2d
d d d
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 Ceq as follows.
(i) From Eq. (1), V1 V V2. Rewrite equations (5) and (6). V1 has been eliminated.
CeqV C4V4 C1V C1V2 5 ; C2V2 C1V C1V2 C3V3 6
(ii) From Eq. (3), V5 V V4. Rewrite equation (4). V5 has been eliminated.
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
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
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) 1
From Eq. (4), V4 C5 C2V2 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) Group all terms by common voltage.
C5Ceq C4Ceq C4C5 C5C1 V C4C2 C5C1 V2 5
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.
C5Ceq C4Ceq C4C5 C5C1
C4C2 C5C1
C5 C1 C3 C3Ceq C3C5 C5 C1 C2 C3 C3C2
Ceq C1C2C3 C1C2C4 C1C2C5 C1C3C5 C1C4C5 C2C3C4 C2C4C5 C3C4C5
C1C3 C1C4 C1C5 C2C3 C2C4 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
4.58.0 17.0 4.5 12.5 8.08.0 4.5 8.0 4.5 4.5 4.5 F
4.5 17.0
8.0 17.0 4.5 12.5
6.0F
41. The stored energy is given by Eq. 24-5.
U 2200 V2 6.8 103J
1 CV 2 1 2.8 109 F
2 2
42. The energy density is given by Eq. 24-6.
u
E1 2 1 8.85 1012 C2 Nm2 150 V m 2 1.0 107 J m3
2
20
43. The energy stored is obtained from Eq. 24-5, with the capacitance of Eq. 24-2.
4.2 104 C 2 0.0013 m
8.85 1012 C2 Nm2 0.080 m2
UQ2Q2d 2 2.0 103 J
2C
20 A
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114
Chapter 24 Capacitance, Dielectrics, Electric Energy Storage
44. (a) The charge is constant, and the tripling of separation reduces the capacitance by a factor of 3.
U2 Q2 C1 0 A 3
U1 C2 d
2C2 A
Q2 3d
0
2C1
(b) The work done is the change in energy stored in the capacitor.
U2 U1 3U1 U1 2U1 2 Q2 Q2 Q2d
2C1 A 0A
0 d
45. The equivalent capacitance is formed by C1 in parallel with the series combination of C2 and C3.
Then use Eq. 24-5 to find the energy stored.
Cnet C1 C2C3 C C2 3 C
C2 C3 2C 2
U
C V1 2 3 CV 2 3 22.6 106 F 10.0 V 2 1.70 103J
4 4
2 net
46. (a) Use Eqs. 24-3 and 24-5.
U C V parallel
12 1 C1 C2 V2 1 0.65 106 F 28 V 2 2.548 104 J 2.5 104 J
2 eq 2 2
(b) Use Eqs. 24-4 and 24-5.
Useries
1 CeqV 2 1 C1C2 V 2 1 0.45 106 F 0.20 106 F 28 V 2
2 2 C1 C2 2 0.65 106 F
5.428 105 J 5.4 105J
(c) The charge can be found from Eq. 24-5.
U
1 QV Q 2U Q parallel 2 2.548 104 J 1.8 105C
2 V
28 V
Q series 2 5.428 105J 3.9 106 C
28 V
47. The capacitance of a cylindrical capacitor is given in Example 24-2 as C 20l .
ln Ra Rb
(a) If the charge is constant, the energy can be calculated by U 1 Q2 .
2 C
1 Q2 20l
U2 2 C2 C1 ln Ra Rb ln 3Ra Rb 1
U1 1 Q2 C2 ln Ra Rb
20l
2 C1 ln 3Ra Rb
The energy comes from the work required to separate the capacitor components.
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
(b) If the voltage is constant, the energy can be calculated by U 1 CV 2.
2
20l
U2 1 C2V 2 C2 ln 3Ra Rb ln Ra Rb
U1 2 C1 ln 3Ra Rb
20l 1
1 C1V 2
2
ln Ra Rb
Since the voltage remained constant, and the capacitance decreased, the amount of charge on
the capacitor components decreased. Charge flowed back into the battery that was maintaining
the constant voltage.
48. (a) Before the capacitors are connected, the only stored energy is in the initially-charged capacitor.
Use Eq. 24-5.
U1 12.0 V2 1.584 104 J 1.58 104 J
1 C1V02 1 2.20 106 F
2 2
(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.
2.20 106 F 2 12.0 V 2
5.70 106 F
Q Q1 C1V0 1 Q2
; Ceq C1 C2 ; U2 C2 1 C12V02 1
2 C1 C2 2
eq
6.114 105J 6.11105J
(c) U U2 U1 6.114 105 J 1.584 104 J 9.73 105 J
49. (a) With the plate inserted, the capacitance is that of two series capacitors of plate separations
d1 x and d2 d l x.
Ci x d x l 1 0A
0 A 0A d l
With the plate removed the capacitance is obtained directly from Eq. 24-2.
Cf 0A
d
Since the voltage remains constant the energy of the capacitor will be given by Eq. 24-5 written
in terms of voltage and capacitance. The work will be the change in energy as the plate is
removed.
W Cf V2
Uf Ui 1 Ci
2
1 0A 0A V 2 0 AlV 2
2 d d l
2d d l
The net work done is negative. Although the person pulling the plate out must do work, charge
is returned to the battery, resulting in a net negative work done.
(b) Since the charge now remains constant, the energy of the capacitor will be given by Eq. 24-5
written in terms of capacitance and charge.
W Q2 1 1 Q2 d d l Q2l
2 20 A
2 C A 0A
f Ci 0
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116
Chapter 24 Capacitance, Dielectrics, Electric Energy Storage
0A 0A V0 2 l 0 AV02l
d l d l
The original charge is Q CV0 V0 and so W 2d l2 .
20 A
50. (a) The charge remains constant, so we express the stored energy as U 1 Q2 1 Q2x , where x
2 C 2 0A
is the separation of the plates. The work required to increase the separation by dx is
dW Fdx, where F is the force on one plate exerted by the other plate. That work results in an
increase in potential energy, dU.
dW Fdx dU 1 Q 2 dx F 1 Q2
2 0A 2 0A
(b) We cannot use F QE Q Q Q Q2 because the electric field is due to both plates,
0 0A 0A
and charge cannot put a force on itself by the field it creates. By the symmetry of the geometry,
the electric field at one plate, due to just the other plate, is 1 E. See Example 24-10.
2
51. (a) The electric field outside the spherical conductor is that of an equivalent point charge at the
center of the sphere, so E 1 Q ,r R. Consider a differential volume of radius dr, and
4 0 r2
volume dV 4 r2dr, as used in Example 22-5. The energy in that volume is dU udV .
Integrate over the region outside the conductor.
1 Q 2 Q2 1 Q2 1
4 0 r2 8 0 R r2 r R
udV E 2dV 4 r2dr dr
U 8 0
dU 1 1 0
2
20 R
Q2
8 0 R
(b) Use Eq. 24-5 with the capacitance of an isolated sphere, from the text immediately after
Example 24-3.
U 1 Q2 1 Q2 Q2
2 C 2 4 0 R 8 0 R
(c) When there is a charge q < Q on the sphere, the potential of the sphere is V 1 q . The
40 R
work required to add a charge dq to the sphere is then dW Vdq 1 q dq. That work
40 R
increase the potential energy by the same amount, so dU dW Vdq 1 q dq. Build up
40 R
the entire charge from 0 to Q, calculating the energy as the charge increases.
Q 1 1 Q Q2
U dU dW Vdq qdq
q dq 8 0 R
0 40 R 40R 0
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
52. In both configurations, the voltage across the combination of capacitors is the same. So use
U 1 CV .
2
C1 C2 V 2 C1C2 V2
C1 C2
UP1CPV 1 ; US 1 CSV 1
2 2 2 2
V2 5 C1C2 V 2 C1 C2 2 5C1C2
C1 C2
UP 5 US
1 C1 C2 1
2 2
C12 3C1C2 C22 0 C1 3C2 9C22 4C22 C2 3 5
2 2
C1 3 5 , 3 5 2.62, 0.382
C2 2 2
53. First find the ratio of energy requirements for a logical operation in the past to the current energy
requirements for a logical operation.
Epast N CV1 2 Cpast Vpast 2 20 5.0 2 220
Epresent N Cpresent Vpresent 1 1.5
2 past
CV1 2
2 present
So past operations would have required 220 times more energy. Since 5 batteries in the past were
required to hold the same energy as a present battery, it would have taken 1100 times as many
batteries in the past. And if it takes 2 batteries for a modern PDA, it would take 2200 batteries to
power the PDA in the past. It would not fit in a pocket or purse. The volume of a present-day
battery is V r2l 0.5cm2 4 cm 3cm3. The volume of 2200 of them would be 6600 cm3 ,
which would require a cube about 20 cm in side length.
54. Use Eq. 24-8 to calculate the capacitance with a dielectric.
C 2
A 4.2 102 m
K0 d 2.2 8.85 1012 C2 N m2 1.8 103 m 1.9 1011 F
55. The change in energy of the capacitor is obtained from Eq. 24-5 in terms of the constant voltage and
the capacitance.
2 KC0V 2 K 1 C0V 2
U
Uf Ui 1 C0V 1 1
2 2 2
The work done by the battery in maintaining a constant voltage is equal to the voltage multiplied by
the change in charge, with the charge given by Eq. 24-1.
Wbattery V Qf Qi V C0V KC0V K 1 C0V 2
The work done in pulling the dielectric out of the capacitor is equal to the difference between the
change in energy of the capacitor and the energy done by the battery.
K 1 C0V 2 K 1 C0V 2
W
U Wbattery 1
2
2 100 V2 1.1 104 J
1 K 1 C0V 3.4 1 8.8 109 F
2
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118
Chapter 24 Capacitance, Dielectrics, Electric Energy Storage
56. We assume the charge and dimensions are the same as in Problem 43. Use Eq. 24-5 with charge and
capacitance.
420 106 C 2 0.0013 m
8.85 1012 C2 N m2 64 104 m2
U Q2
1 Q2 1 KC0 1 Q2d 1 7 289.2 J 290 J
2C 2 2 2
K0 A
57. From Problem 10, we have C 35 1015 F. Use Eq. 24-8 to calculate the area.
C 2
K0 A A Cd 35 1015 F 2.0 109 m 3.164 1013 m2 106 m
d K0 1m
25 8.85 1012 C2 N m2
0.3164 m2 0.32 m2
Half of the area of the cell is used for capacitance, so 1.5cm2 is available for capacitance. Each
capacitor is one “bit.”
106 m 2 1 bit 1 byte
102 cm 0.32m2 8 bits
1.5 cm 2 5.86 107 bytes 59 Mbytes
58. The initial charge on the capacitor is Qinitial C Vinitial . When the mica is inserted, the capacitance
changes to Cfinal KCinitial , and the voltage is unchanged since the capacitor is connected to the same
battery. The final charge on the capacitor is Qfinal CfinalV .
Q Qfinal Qinitial CfinalV C Vinitial K 1 C Vinitial 7 1 3.5 109 F 32 V
6.7 107 C
59. The potential difference is the same on each half of the capacitor, d K1 K2
so it can be treated as two capacitors in parallel. Each parallel
capacitor has half of the total area of the original capacitor.
1 A 1 A A
C 2 2 0 d
C1 C2 K1 0 K 2 0 1 K1 K2
d d 2
60. The intermediate potential at the boundary of the two dielectrics can d K1
be treated as the “low” potential plate of one half and the “high” K2
potential plate of the other half, so we treat it as two capacitors in
series. Each series capacitor has half of the inter-plate distance of
the original capacitor.
111 1 d 1 d d K1 K2 C 20 A K1K2
2 2
C C1 C2 K10 A K20 A 20 A K1K2 d K1 K2
61. The capacitor can be treated as two series capacitors with the same
areas, but different plate separations and dielectrics. Substituting d1 K1
Eq. 24-8 into Eq. 24-4 gives the effective capacitance. d2 K2
C 1 1 1 d1 d2 1 A 0 K1K 2
C1 C2 K1 A0 K2 A0 d1K2 d2K1
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
62. (a) Since the capacitors each have the same charge and the same voltage in the initial situation,
each has the same capacitance of C Q0 . When the dielectric is inserted, the total charge of
V0
2Q0 will not change, but the charge will no longer be divided equally between the two
capacitors. Some charge will move from the capacitor without the dielectric C1 to the
capacitor with the dielectric C2 . Since the capacitors are in parallel, their voltages will be the
same.
V1 V2 Q1 Q2 Q1 2Q0 Q1
C1 C2 C KC
Q1 K 2 1 Q0 2 Q0 0.48Q0 ; Q2 1.52Q0
4.2
(b) V1 V2 Q1 0.48Q0 0.48V0 Q2 1.52Q0
C1 Q0 V0 C2 3.2 Q0 V0
63. (a) We treat this system as two capacitors, one with a dielectric, d ¬
and one without a dielectric. Both capacitors have their high K
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 x
capacitance. Note that x is measured from the right edge of +x
the capacitor, and is positive to the left in the diagram.
C C1 C2 0 ll x K0 lx 0 l2 1 K 1 x
d d l
d
(b) Both “capacitors” have the same potential difference, so use U 1 CV 2.
2
1 V02 l2 1 K 1 x V02
2d l
2
U C1 C2 0
(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 dUcap dUbattery
Fncdx d 1 CV02 d Q Vbattery 0
2
Fnc V1 2 dC V0 dQbattery V1 2 dC V0 dQcap V1 2 dC V02 dC 1 V02 dC
dx dx dx dx dx dx 2 dx
20 20 20
V1 2 l2 K 1 V020l K 1
d 2d
200 l
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120
Chapter 24 Capacitance, Dielectrics, Electric Energy Storage
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 of this attractive force are
V020l K 1 , left .
2d
64. (a) We consider the cylinder as two cylindrical capacitors in parallel. The two “negative plates” are
the (connected) halves of the inner cylinder (half of which is in contact with liquid, and half of
which is in contact with vapor). The two “positive plates” are the (connected) halves of the
outer cylinder (half of which is in contact with liquid, and half of which is in contact with
vapor). Schematically, it is like Figure 24-30 in Problem 59. The capacitance of a cylindrical
capacitor is given in Example 24-2.
2 0 K liq h 2 0 K V l h 20l h
Rb l
ln Ra Rb ln Ra ln Ra Rb
C Cliq
CV Kliq KV KV C
h C ln Ra Rb
l 2 0 l
1 KV
Kliq KV
(b) For the full tank, h 1, and for the empty tank, h 0.
ll
2 0 l h 2 0 lK liq
l
Ra Rb ln Ra Rb
Full: C
ln Kliq KV K V
2
8.85 1012 C2 N m2 2.0 m 1.4 1.5 109 F
ln 5.0 mm 4.5 mm
2 0 l h 2 0 lK V
l
Ra Rb ln Ra Rb
Empty: C K
ln Kliq KV V
2
8.85 1012 C2 N m2 2.0 m 1.0 1.1 109 F
ln 5.0 mm 4.5 mm
65. Consider the dielectric as having a layer of equal and opposite charges at each side of the dielectric.
Then the geometry is like three capacitors in series. One air gap is taken to be d1, and then the other
air gap is d d1 l.
1 1 1 1 d1 l d d1 l 1 l d l
C C1 C2 C3 0A K 0 A 0A 0 A K
C 0A
8.85 1012 C2 N m2 2.50 102 m2 1.72 1010 F
l l 1.00 103 m
K 3.50
d
1.00 103 m
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
66. By leaving the battery connected, the voltage will not change when the dielectric is inserted, but the
amount of charge will change. That will also change the electric field.
(a) Use Eq. 24-2 to find the capacitance.
C0 A 2.50 102 m2 1.106 1010 F 1.11 1010 F
0 d 8.85 1012 C2 N m2 2.00 103 m
(b) Use Eq. 24-1 to find the initial charge on each plate.
Q0 C0V 1.106 1010 F 150 V 1.659 108C 1.66 108 C
In Example 24-12, the charge was constant, so it was simple to calculate the induced charge and then
the electric fields from those charges. But now the voltage is constant, and so we calculate the fields
first, and then calculate the charges. So we are solving the problem parts in a different order.
(d) We follow the same process as in part (f) of Example 24-12.
V E0 d l EDl E0 d l E0 l
K
E0 V 150 V 1.167 105 V m
d l l
K 1.00 103 m
2.00 103 m 1.00 103 m
3.50
1.17 105 V m
(e) ED E0 1.167 105 V m 3.333 104 V m 3.33 104 V m
K
3.50
(h) E0 Q
0 A 0
Q EA0 1.167 105 V m 0.0250 m2 8.85 1012 C2 N m2 2.582 108C
2.58 108 C
Q 1 1 1 1
(c) K 3.50 1.84 108 C
Qind 2.582 108 C
(f) Because the battery voltage does not change, the potential difference between the plates is
unchanged when the dielectric is inserted, and so is V 150 V .
(g) C Q 2.582 108 C 1.72 1010 pF
V 150 V
Notice that the capacitance is the same as in Example 24-12. Since the capacitance is a constant
(function of geometry and material, not charge and voltage), it should be the same value.
67. The capacitance will be given by C Q /V . When a charge Q is placed on one plate and a
charge –Q is placed on the other plate, an electric field will be set up between the two plates.
The electric field in the air-filled region is just the electric field between two charged plates,
E0 Q . The electric field in the dielectric is equal to the electric field in the air,
0 A 0
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122
Chapter 24 Capacitance, Dielectrics, Electric Energy Storage
divided by the dielectric constant: ED E0 Q .
K KA 0
The voltage drop between the two plates is obtained by integrating the electric field between the
two plates. One plate is set at the origin with the dielectric touching this plate. The dielectric
ends at x l . The rest of the distance to x d is then air filled.
d dx l Qdx d Qdx Q l
E 0 KA 0 l A 0 A 0 K
V d l
0
The capacitance is the ratio of the voltage to the charge.
C Q Q 0A
V
Q l d l d l l
A 0 K K
68. Find the energy in each region from the energy density and the volume. The energy density in the
“gap” is given by ugap 1 0 E2 , and the energy density in the dielectric is given by uD E1 2
2 gap
2DD
1 K 0 Egap 2 1 E2 , where Eq. 24-10 is used.
2 K gap
20 K
1 E2 Al
gap
20 K
UD
U total
UD uD Vol D E2
U gap U D ugapVolgap uDVolD gap
E A1 2 d l 1 0 K Al
2
2 0 gap
l
K l d l l 1.00 mm 0.222
K 1.00 mm 3.50 1.00 mm
d l lK
69. There are two uniform electric fields – one in the air, and one in the gap. They are related by Eq. 24-
10. In each region, the potential difference is the field times the distance in the direction of the field
over which the field exists.
V E dair air E dglass glass E dair air Eair d glass
K glass
E Vair K glass
d K dair glass glass
5.80
5.80
90.0 V
3.00 103 m 2.00 103 m
2.69 104 V m
Eglass Eair 2.69 104 V m 4.64 103 V m
K glass
5.80
The charge on the plates can be calculated from the field at the plate, using Eq. 22-5. Use Eq. 24-
11b to calculate the charge on the dielectric.
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Eair plate Qplate
0 0A
Qplate Eair0 A 2.69 104 V m 8.85 1012 C2 N m2 1.45 m2 3.45 107 C
Q 1 1 1 1
Qind K 5.80 2.86 107 C
3.45 107 C
70. (a) The capacitance of a single isolated conducting sphere is given after example 24-3.
C 40r
C 4 1010 F 1m 1012 pF
m 100 cm 1F
r
8.85 1012 C2 N m2 1.11 1.11pF cm
And so C 1.11pF cm r C pF r cm .
(b) We assume that the human body is a sphere of radius 100 cm. Thus the rule C pF r cm
says that the capacitance of the human body is about 100 pF .
(c) A 0.5-cm spark would require a potential difference of about 15,000 V. Use Eq. 24-1.
Q CV 100 pF 15, 000 V 1.5C
71. Use Eq. 24-5 to find the capacitance.
U 1 CV 2 C 2U 2 1200 J 4.3 105 F
2 V2 7500 V2
72. (a) We approximate the configuration as a parallel-plate capacitor, and so use Eq. 24-2 to calculate
the capacitance.
C A r2 4.5in 0.0254 m in2 7.265 1012 F
0 d 0 d 8.85 1012 C2 N m2
0.050 m
7 1012 F
(b) Use Eq. 24-1.
Q CV 7.265 1012 F 9 V 6.539 1011C 7 1011C
(c) The electric field is uniform, and is the voltage divided by the plate separation.
E V 9V 180 V m 200 V m
d 0.050 m
(d) The work done by the battery to charge the plates is equal to the energy stored by the capacitor.
Use Eq. 24-5.
9 V 2 2.942 1010 J 3 1010 J
U
1 CV 2 1 7.265 1012 F
2 2
(e) The electric field will stay the same, because the voltage will stay the same (since the capacitor
is still connected to the battery) and the plate separation will stay the same. The capacitance
changes, and so the charge changes (by Eq. 24-1), and so the work done by the battery changes
(by Eq. 24-5).
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124
Chapter 24 Capacitance, Dielectrics, Electric Energy Storage
73. Since the capacitor is disconnected from the battery, the charge on it cannot change. The
capacitance of the capacitor is increased by a factor of K, the dielectric constant.
C Cinitial 1
V V V final initial 2.2
C KCfinal
Q C V C Vinitial initial
final final initial initial 34.0 V 15 V
initial
74. The energy is given by Eq. 24-5. Calculate the energy difference for the two different amounts of
charge, and then solve for the difference.
U 1 Q2 U 1 Q Q 2 1 Q2 1 Q Q2 Q2 Q 2Q Q
2 C 2 2 C 2C 2C
C
Q
CU 1 Q 17.0 106 F 18.5J 1 13.0 103C 17.7 103C 17.7 mC
Q 2 2
13.0 103C
75. The energy in the capacitor, given by Eq. 24-5, is the heat energy absorbed by the water, given by
Eq. 19-2.
U Qheat 1 CV 2 mcT
2
2mcT 2 3.5 kg 4186 J 95C 22C
C kgC
V 844 V 840 V
3.0 F
76. (a) The capacitance per unit length of a cylindrical capacitor with no dielectric is derived in
Example 24-2, as C 20 . The addition of a dielectric increases the capacitance
l ln R Routside inside
by a factor of K.
C 2 0 K
l ln R Routside inside
(b)
2 0 K 2 8.85 1012 C2 N m2 2.6
C 1.1 1010 F m
ln 9.0 mm 2.5 mm
l ln R Routside inside
77. The potential can be found from the field and the plate separation. Then the capacitance is found
from Eq. 24-1, and the area from Eq. 24-8.
E V ; Q CV CEd
d
C
Q 0.675 106 C 3.758 109 F 3.76 109 F
Ed 9.21 104 V m 1.95 103 m
C
K0 A A Cd 3.758 109 F 1.95 103 m 0.221m2
d K0
3.75 8.85 1012 C2 N m2
78. (a) If N electrons flow onto the plate, the charge on the top plate is Ne, and the positive charge
associated with the capacitor is Q Ne. Since Q CV , we have Ne CV V Ne C ,
showing that V is proportional to N.
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(b) Given V 1mV and we want N 1, solve for the capacitance.
V Ne V eN
CC
C N 1 1.60 1016 F 2 1016 F
e V 1.60 1019 C 1 103 V
(c) Use Eq. 24-8.
C 0K A 0K l2
d d
Cd
l
0K
1.60 1016 F 100 109 m 7.76 107 m 106m 0.8 m
1m
8.85 1012 C2 N m2 3
79. The relative change in energy can be obtained by inserting Eq. 24-8 into Eq. 24-5.
U Q2 A0
C0 1
2C C d 2K
U0 Q2 KA 0
2C0 1 d
2
The dielectric is attracted to the capacitor. As such, the dielectric will gain kinetic energy as it enters
the capacitor. An external force is necessary to stop the dielectric. The negative work done by this
force results in the decrease in energy within the capacitor.
Since the charge remains constant, and the magnitude of the electric field depends on the charge, and
not the separation distance, the electric field will not be affected by the change in distance between
the plates. The electric field between the plates will be reduced by the dielectric constant, as given in
Eq. 24-10.
E E0 / K 1
E0 E0 K
80. (a) Use Eq. 24-2.
C 0 A
d
8.85 1012 C2 N m2 120 106 m2 7.08 107 F 7.1 107 F
1500 m
(b) Use Eq. 24-1.
Q CV 7.08 107 F 3.5 107 V 24.78C 25C
(c) Use Eq. 24-5.
U
1 QV 1 24.78 C 3.5 107 V 4.337 108J 4.3 108J
2 2
81. We treat this as N capacitors in parallel, so that the total capacitance is N times the capacitance of a
single capacitor. The maximum voltage and dielectric strength are used to find the plate separation
of a single capacitor.
d V 100 V m 3.33 106 m ; N l 6.0 103 m 1800
ES 30 106 V d 3.33 106 m
Ceq NC N0K A
d
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126
Chapter 24 Capacitance, Dielectrics, Electric Energy Storage
K
Ceqd 1.0 106 F 3.33 106 m 14.0 103 m 1.244 1.2
N0 A 1800 8.85 1012 C2 N m2 12.0 103 m
82. The total charge doesn’t change when the second capacitor is connected, since the two-capacitor
combination is not connected to a source of charge. The final voltage across the two capacitors must
be the same. Use Eq. 24-1.
Q0 C1V0 Q1 Q2 C1V1 C2V2 C1V1 C2V1
C2 C1 V0 V1 3.5F 12.4 V 5.9 V 3.856 F 3.9F
5.9 V
V1
83. (a) Use Eq. 24-5 to calculate the stored energy.
2.5 104 V 2 25 J
U
1 CV 2 1 8.0 108 F
2 2
(b) The power is the energy converted per unit time.
P Energy 0.15 25 J 9.38 105 W 940 kW
time
4.0 106 s
84. The pressure is the force per unit area on a face of the dielectric. The force is related to the potential
energy stored in the capacitor by Eq. 8-7, F dU , where x is the separation of the capacitor
dx
plates.
U 1 CV 2 1 K A V 2 F dU K0 AV 2 ; P F K0V 2
2 2 x dx 2x2 A 2x2
0
2 1.0 104 m 2 40.0 Pa
3.1 8.85 1012 C2 N m2
V
2x2P 170 V
K0
85. (a) From the diagram, we see that one group of 4 plates is connected together, and the other group
of 4 plates is connected together. This common grouping shows that the capacitors are
connected in parallel .
(b) Since they are connected in parallel, the equivalent capacitance is the sum of the individual
capacitances. The variable area will change the equivalent capacitance.
Ceq 7C 70 A
d
Cmin
70 Amin 7 8.85 1012 C2 Nm2 2.0 104 m2 7.7 1012 F
d 1.6 103 m
Cmax
70 Amax 7 8.85 1012 C2 Nm2 9.0 104 m2 3.5 1011 F
d 1.6 103 m
And so the range is from 7.7 pF to 35pF .
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86. (a) Since the capacitor is charged and then disconnected from the power supply, the charge is
constant. Use Eq. 24-1 to find the new voltage.
Q CV constant C1V1 C2V2 V2 V1 C1 7500 V 8.0 pF 6.0 104 V
C2 1.0 pF
(b) In using this as a high voltage power supply, once it discharges, the voltage drops, and it needs
to be recharged. So it is not a constant source of high voltage. You would also have to be sure
it was designed to not have breakdown of the capacitor material when the voltage gets so high.
Another disadvantage is that it has only a small amount of energy stored: U 1 CV 2
2
6.0 104 V 2 1.8 103J , and so could actually only supply a small amount
1 1.0 1012 C
2
of power unless the discharge time was extremely short.
87. Since the two capacitors are in series, they will both have the same charge on them.
Q1 Q2 Qseries ; 1 V 11
C Q C Cseries
series 12
C2
Q Cseries 1 125 1012 C 175 1012 F 5.15 1012 F
C1V Qseries 175 1012 F 25.0 V 125 1012 C
88. (a) The charge can be determined from Eqs. 24-1 and 24-2.
Q
CV 0 AV 8.85 1012 C2 Nm2 2.0 104 m2 12 V 4.248 1011C
d 5.0 104 m
4.2 1011C
(b) Since the battery is disconnected, no charge can flow to or from the plates. Thus the charge is
constant.
Q 4.2 1011C
(c) The capacitance has changed and the charge has stayed constant, and so the voltage has
changed.
Q CV constant C1V1 C0V0 0 A V1 0 A V0
d1 d0
V1 d1 V0 0.75 mm 12 V 18 V
d0 0.50 mm
(d) The work is the change in stored energy.
W
U 1 QV1 1 QV0 1 Q V1 V0 1 4.248 1011C 6.0 V 1.3 1010 J
2 2 2 2
89. The first capacitor is charged, and so has a certain amount of charge on its plates. Then, when the
switch is moved, the capacitors are not connected to a source of charge, and so the final charge is
equal to the initial charge. Initially treat capacitors C2 and C3 as their equivalent capacitance,
C23 C2C3 2.0F 2.4F 1.091F. The final voltage across C1 and C23 must be the
C2 C3
4.4F
same. The charge on C2 and C3 must be the same. Use Eq. 24-1.
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Chapter 24 Capacitance, Dielectrics, Electric Energy Storage
Q0 C1V0 Q1 Q23 C1V1 C23V23 C1V1 C23V1
V1 C1 C1 V0 1.0F 24 V 11.48 V V1 V23
C23 1.0F 1.091F
Q1 C1V1 1.0F 11.48 V 11.48C
Q23 C23V23 1.091F 11.48 V 12.52C Q2 Q3
V2 Q2 12.52C 6.26 V ; V3 Q3 12.52C 5.22 V
C2 2.0F C3 2.4F
To summarize: Q1 11 C , V1 11V ; Q2 13 C , V2 6.3V ; Q3 13 C , V3 5.2 V
90. The metal conducting strips connecting cylinders b and c mean that b and c
are at the same potential. Due to the positive charge on the inner cylinder +–
–
and the negative charge on the outer cylinder, cylinders b and c will +
polarize according to the first diagram, with negative charge on cylinder c, Cyl. d
Cyl. b
and positive charge on cylinder b. This is then two capacitors in series, as Rc Rd
illustrated in the second diagram. The capacitance per unit length of a Ra
Rb
cylindrical capacitor is derived in Example 24-2.
C1 2 0 l ; C2 2 0 l ; 1 1 1
Cnet C1
ln Ra Rb ln Rc Rd
C2
2 0 l 2 0 l
Rb Rd
Cnet C1C2 ln Ra ln Rc
C1 C2
2 0 l 2 0 l
ln ln
Ra Rb Rc Rd Cyl. c
Cyl. a
ln Rc 2 0 l Rb 2 0 l
Rd ln Ra ln Ra Rc RbRd
C 2 0
ln Ra Rc RbRd
l
91. The force acting on one plate by the other plate is equal to the electric field produced by one charged
plate multiplied by the charge on the second plate.
F EQ Q Q Q2
2 A0 2 A0
The force is attractive since the plates are oppositely charged. Since the force is constant, the work
done in pulling the two plates apart by a distance x is just the force times distance.
W Fx Q2x
2 A0
The change in energy stored between the plates is obtained using Eq. 24-5.
W U Q2 1 1 Q2 2x x Q2x
2 C2 C1 2 0A 0A 20 A
The work done in pulling the plates apart is equal to the increase in energy between the plates.
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92. Since the other values in this problem manifestly have 2 significant figures, we assume that the
capacitance also has 2 significant figures.
(a) The number of electrons is found from the charge on the capacitor.
Q CV Ne
N CV 30 1015 F 1.5 V 2.8 105 e's
e
1.60 1019 C
(b) The thickness is determined from the dielectric strength.
Emax V d min V 1.5 V m 1.5 109 m
d min Emax 1.0 109 V
(c) The area is found from Eq. 24-8.
C
K0 A A Cd 30 1015 F 1.5 109 m 2.0 1013 m
d K0 25 8.85 1012 C2 N m2
93. Use Eq. 24-2 for the capacitance.
C 0 A N m2
d
d 0A 8.85 1012 C2 1F 1.0 104 m2 9 1016 m
C
No , this is not practically achievable. The gap would have to be smaller than the radius of a proton.
94. See the schematic diagram for the arrangement. The two 25 kV 6.3mm glass
“capacitors” are in series, and so have the same charge. Thus
their voltages, which must total 25kV, will be inversely hand
proportional to their capacitances. Let C1 be the glass-filled feet
capacitor, and C2 be the vinyl capacitor. The area of the foot is 1cm vinyl
approximately twice the area of the hand, and since there are floor
two feet on the floor and only one hand on the screen, the area
ratio is Afoot 4 .
Ahand 1
Q C1V1 C2V2 V1 V2 C2
C1
C1 K A0 glass hand ; C2 K A0 vinyl foot
d glass d vinyl
K A0 vinyl foot
C d K A d2 vinyl 34 0.63
C K A K A d 1 0 glass hand 51 1.0
vinyl foot glass 1.5
glass hand vinyl
d glass
V V1 V2 V2 C2 V2 2.5V2 25, 000 V V2 10, 000 V
C1
95. (a) Use Eq. 24-2 to calculate the capacitance.
C0
0A 8.85 1012 C2 N m2 2.0 m2
d 3.0 103 m 5.9 109 F
Use Eq. 24-1 to calculate the charge.
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130
Chapter 24 Capacitance, Dielectrics, Electric Energy Storage
Q0 C0V0 5.9 109 F 45 V 2.655 107 C 2.7 107 C
The electric field is the potential difference divided by the plate separation.
E0 V0 45 V 15000 V m
d 3.0 103 m
Use Eq. 24-5 to calculate the energy stored.
U0
1 C0V02 1 5.9 109 F 45 V 2 6.0 106 J
2 2
(b) Now include the dielectric. The capacitance is multiplied by the dielectric constant.
C KC0 3.2 5.9 109 F 1.888 108 F 1.9 108 F
The voltage doesn’t change. Use Eq. 24-1 to calculate the charge.
Q CV KC0V 3.2 5.9 109 F 45 V 8.496 107 C 8.5 107 C
Since the battery is still connected, the voltage is the same as before, and so the electric field
doesn’t change.
E E0 15000 V m
Use Eq. 24-5 to calculate the energy stored.
U 45 V2
1 CV 2 1 KC0V 2 1 3.2 5.9 109 F 1.9 105 J
2 2 2
96. (a) For a plane conducting surface, the electric field is given by Eq. 22-5.
E Q
0 0A
Qmax ES0 A 3 106 N C 8.85 1012 C2 N m2 150 104 m2
3.98 107 C 4 107 C
(b) The capacitance of an isolated sphere is derived in the text, right after Example 24-3.
C 40r 4 8.85 1012 C2 N m2 1m 1.11 1010 F 1 1010 F
(c) Use Eq. 24-1, with the maximum charge from part (a) and the capacitance from part (b).
Q CV V Q 3.98 107 C 3586 V 4000 V
C 1.11 1010 F
97. (a) The initial capacitance is obtained directly from Eq. 24-8.
C0 K0 A 3.7 8.85 pF/m 0.21m 0.14 m 32 nF
d
0.030 103 m
(b) Maximum charge will occur when the electric field between the plates is equal to the dielectric
strength. The charge will be equal to the capacitance multiplied by the maximum voltage,
where the maximum voltage is the electric field times the separation distance of the plates.
Qmax C0V C0Ed 32 nF 15 106 V/m 0.030 103 m
14 C
(c) The sheets of foil would be separated by sheets of paper with
alternating sheets connected together on each side. This capacitor
would consist of 100 sheets of paper with 101 sheets of foil.
t 101dAl 100dpaper 1010.040 mm 100 0.030 mm
7.0 mm
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(d) Since the capacitors are in parallel, each capacitor has the same voltage which is equal to the
total voltage. Therefore breakdown will occur when the voltage across a single capacitor
provides an electric field across that capacitor equal to the dielectric strength.
Vmax Emaxd 15 106 V/m 0.030 103 m 450 V
98. From Eq. 24-2, C 0 A. So if C (pF) 1200
d
1000 C = 8606 A 0.12
we plot C vs. A, we should get a 800 R2 = 0.99
straight line with a slope of 0 .
600
d
0 slope 400
d
200
d 0
slope 0
0.00 0.02 0.04 0.06 0.08 0.10
8.85 1012 C2 N m2
8606 1012 F m2 A (m2)
1.03 103 m 1.0 mm
The spreadsheet used for this problem can be found on the Media Manager, with filename
“PSE4_ISM_CH24.XLS,” on tab “Problem 24.98.”
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132
CHAPTER 25: Electric Currents and Resistance
Responses to Questions
1. A battery rating in ampere-hours gives the total amount of charge available in the battery.
2. The chemical reactions within the cell cause electrons to pile up on the negative electrode. If the
terminals of the battery are connected in a circuit, then electrons flow from the negative terminal
because it has an excess of electrons. Once the electrons return to the cell, the electrolyte again
causes them to move to the negative terminal.
3. When a flashlight is operated, the battery energy is being used up.
4. The terminal of the car battery connected to “ground” is actually connected to the metal frame of the
car. This provides a large “sink” or “source” for charge. The metal frame serves as the common
ground for all electrical devices in the car, and all voltages are measured with respect to the car’s
frame.
5. Generally, water is already in the faucet spout, but it will not come out until the faucet valve is
opened. Opening the valve provides the pressure difference needed to force water out of the spout.
The same thing is essentially true when you connect a wire to the terminals of a battery. Electrons
already exist in the wires. The battery provides the potential that causes them to move, producing a
current.
6. Yes. They might have the same resistance if the aluminum wire is thicker. If the lengths of the wires
are the same, then the ratios of resistivity to cross-sectional area must also be the same for the
resistances to be the same. Aluminum has a higher resistivity than copper, so if the cross-sectional
area of the aluminum is also larger by the same proportion, the two wires will have the same
resistance.
7. If the emf in a circuit remains constant and the resistance in the circuit is increased, less current will
flow, and the power dissipated in the circuit will decrease. Both power equations support this result.
If the current in a circuit remains constant and the resistance is increased, then the emf must increase
and the power dissipated in the circuit will increase. Both equations also support this result. There is
no contradiction, because the voltage, current, and resistance are related to each other by V = IR.
8. When a lightbulb burns out, the filament breaks, creating a gap in the circuit so that no current flows.
9. If the resistance of a small immersion heater were increased, it would slow down the heating
process. The emf in the circuit made up of the heater and the wires that connect it to the wall socket
is maintained at a constant rms value. If the resistance in the circuit is increased, less current will
flow, and the power dissipated in the circuit will decrease, slowing the heating process.
10. Resistance is proportional to length and inversely proportional to cross-sectional area.
(a) For the least resistance, you want to connect the wires to maximize area and minimize length.
Therefore, connect them opposite to each other on the faces that are 2a by 3a.
(b) For the greatest resistance, you want to minimize area and maximize length. Therefore, connect
the wires to the faces that are 1a by 2a.
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11. When a light is turned on, the filament is cool, and has a lower resistance than when it is hot. The
current through the filament will be larger, due to the lower resistance. This momentary high current
will heat the wire rapidly, possibly causing the filament to break due to thermal stress or vaporize.
After the light has been on for some time, the filament is at a constant high temperature, with a
higher resistance and a lower current. Since the temperature is constant, there is less thermal stress
on the filament than when the light is first turned on.
12. When connected to the same potential difference, the 100-W bulb will draw more current (P = IV).
The 75-W bulb has the higher resistance (V = IR or P = V²/R).
13. The electric power transferred by the lines is P = IV. If the voltage across the transmission lines is
large, then the current in the lines will be small. The power lost in the transmission lines is P = I²R.
The power dissipated in the lines will be small, because I is small.
14. If the circuit has a 15-A fuse, then it is rated to carry current of no more than 15 A. Replacing the 15-
A fuse with a 25-A fuse will allow the current to increase to a level that is dangerously high for the
wiring, which might result in overheating and possibly a fire.
15. The human eye and brain cannot distinguish the on-off cycle of lights when they are operated at the
normal 60 Hz frequency. At much lower frequencies, such as 5 Hz, the eye and brain are able to
process the on-off cycle of the lights, and they will appear to flicker.
16. The electrons are not “used up” as they pass through the lamp. Their energy is dissipated as light and
heat, but with each cycle of the alternating voltage, their potential energy is raised again. As long as
the electrons keep moving (converting potential energy into kinetic energy, light, and heat) the lamp
will stay lit.
17. Immediately after the toaster is turned on, the Nichrome wire heats up and its resistance increases.
Since the (rms) potential across the element remains constant, the current in the heating element
must decrease.
18. No. Energy is dissipated in a resistor but current, the rate of flow of charge, is not “used up.”
19. In the two wires described, the drift velocities of the electrons will be about the same, but the current
density, and therefore the current, in the wire with twice as many free electrons per atom will be
twice as large as in the other wire.
20. (a) If the length of the wire doubles, its resistance also doubles, and so the current in the wire will
be reduced by a factor of two. Drift velocity is proportional to current, so the drift velocity will
be halved.
(b) If the wire’s radius is doubled, the drift velocity remains the same. (Although, since there are
more charge carriers, the current will quadruple.)
(c) If the potential difference doubles while the resistance remains constant, the drift velocity and
current will also double.
21. If you turn on an electric appliance when you are outside with bare feet, and the appliance shorts out
through you, the current has a direct path to ground through your feet, and you will receive a severe
shock. If you are inside wearing socks and shoes with thick soles, and the appliance shorts out, the
current will not have an easy path to ground through you, and will most likely find an alternate path.
You might receive a mild shock, but not a severe one.
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134
Chapter 25 Electric Currents and Resistance
Solutions to Problems
1. Use the definition of current, Eq. 25-1a.
I Q 1.30 A 1.30C 1 electron 8.131018 electrons s
t s 1.601019C
2. Use the definition of current, Eq. 25-1a.
I Q Q It 6.7 A5.0 h3600s h 1.2105C
t
3. Use the definition of current, Eq. 25-1a.
I
Q 1200 ions 1.601019 C ion 5.51011A
t
3.5 106s
4. Solve Eq. 25-2a for resistance.
R V 120 V 29
I 4.2 A
5. (a) Use Eq. 25-2b to find the current.
V IR I V 240 V 27.91A 28 A
R 8.6
(b) Use the definition of current, Eq. 25-1a.
I Q Q It 27.91A 50 min 60s min 8.4 104C
t
6. (a) Solve Eq. 25-2a for resistance.
R V 120 V 12.63 13
I 9.5A
(b) Use the definition of average current, Eq. 25-1a.
I Q Q It 9.5A 15min 60s min 8600C
t
7. Use Ohm’s Law, Eq. 25-2a, to find the current. Then use the definition of current, Eq. 25-1a, to
calculate the number of electrons per minute.
I V Q 4.5 V 2.8 C 1 electron 60 s 1.1 1021 electrons
R t 1.6 s 1.60 1019 C minute
1 min
8. Find the potential difference from the resistance and the current.
R 2.5 105 m 4.0 102 m 1.0 106
V IR 3100 A 1.0 106 3.1103V
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
9. (a) Use Eq. 25-2b to find the resistance.
R V 12 V 20 (2 sig. fig.)
I 0.60 A
(b) An amount of charge Q loses a potential energy of QV as it passes through the resistor.
The amount of charge is found from Eq. 25-1a.
U QV ItV 0.60 A 60s 12 V 430J
10. (a) If the voltage drops by 15%, and the resistance stays the same, then by Eq. 25-2b, V IR , the
current will also drop by 15%.
Ifinal 0.85Iinitial 0.856.50 A 5.525 A 5.5 A
(b) If the resistance drops by 15% (the same as being multiplied by 0.85), and the voltage stays the
same, then by Eq. 25-2b, the current must be divided by 0.85.
Ifinal Iinitial 6.50 A 7.647 A 7.6 A
0.85 0.85
11. Use Eq. 25-3 to find the diameter, with the area as A r2 d 2 4 .
R
l 4l d 4l 4 1.00 m 5.6 108m 4.7 104 m
A d2 R 0.32
12. Use Eq. 25-3 to calculate the resistance, with the area as A r2 d 2 4 .
l 4l 4 4.5 m
A d2
1.5 103 m
R 4.3 102
1.68 108m 2
13. Use Eq. 25-3 to calculate the resistances, with the area as A r2 d 2 4.
R l 4l .
A d2
Al 4lAl
RAl 1.8 mm 2
RCu 2.0 mm 2 0.64
d 2 l d2 2.65 108m 10.0 m
Al Al Al Cu 1.68 108m 20.0 m
Cu 4lCu l d2
Cu Cu Al
d2
Cu
14. Use Eq. 25-3 to express the resistances, with the area as A r2 d 2 4, and so R l 4l .
A d2
RW RCu 4l Cu 4l
W d 2 d2
W Cu
dW dCu W 2.2 mm 5.6 108m 4.0 mm
1.68 108m
Cu
The diameter of the tungsten should be 4.0 mm.
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136
Chapter 25 Electric Currents and Resistance
15. (a) If the wire obeys Ohm’s law, then V IR or I 1V, showing a linear relationship between I
R
and V. A graph of I vs. V should give a straight line with a slope of 1 and a y-intercept of 0.
R
(b) From the graph and the 0.4
calculated linear fit, we see that
the wire obeys Ohm’s law. 0.3 I = 0.720 V
slope 1 0.2
R I (A)
R 1 A V 0.1
0.720
1.39 0 0.6
0 0.1 0.2 0.3 0.4 0.5
The spreadsheet used for this
problem can be found on the V (V)
Media Manager, with filename “PSE4_ISM_CH25.XLS,” on tab “Problem 25.15b.”
(c) Use Eq. 25-3 to find the resistivity.
3.2 104 m 2 1.39
4 0.11 m
R l
A
AR d 2R 1.0 106m
l 4l
From Table 25-1, the material is nichrome.
16. Use Eq. 25-5 multiplied by l A so that it expresses resistance instead of resistivity.
R R0 1 T T0 1.15R0 1 T T0 1.15
T T0 0.15 0.15 22 C
.0068 C 1
So raise the temperature by 22 C to a final temperature of 42C .
17. Since the resistance is directly proportional to the length, the length of the long piece must be 4.0
times the length of the short piece.
l lshort llong lshort 4.0lshort 5.0lshort lshort 0.20l , llong 0.80l
Make the cut at 20% of the length of the wire .
lshort 0.20l , llong 0.80l Rshort 0.2R 2.0 , Rlong 0.8R 8.0
18. Use Eq. 25-5 for the resistivity.
T Al 0 Al 1 Al T T0 0 W
T T0 1 0 W 1 20C 1 C 1 5.6 108m 1 279.49C 280C
0 Al 0.00429 2.65 108m
Al
19. Use Eq. 25-5 multiplied by l A so that it expresses resistances instead of resistivity.
R R0 1 T T0
T T0 1 R 1 20C 1 140 1 2390C 2400C
R0 12
0.0045C1
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
20. Calculate the voltage drop by combining Ohm’s Law (Eq. 25-2b) with the expression for resistance,
Eq. 25-3.
V
IR I l I 4l 4 1.68 108m 26 m 2.5 V
A d2
12 A 1.628 103 m 2
21. The wires have the same resistance and the same resistivity.
R Rlong short 4 2l 4lshort dlong
llong lshort 2 short dshort 2
A1 A2 2
d dlong short
22. In each case calculate the resistance by using Eq. 25-3 for resistance.
(a)
Rx lx 3.0 105m 1.0 102 m 3.75 104 3.8 104
Ayz 2.0 102 m 4.0 102 m
(b)
Ry ly 3.0 105m 2.0 102 m 1.5 103
Axz 1.0 102 m 4.0 102 m
(c)
Rz lz 3.0 105m 4.0 102 m 6.0 103
Axy 1.0 102 m 2.0 102 m
23. The original resistance is R0 V I0 , and the high temperature resistance is R V I , where the two
voltages are the same. The two resistances are related by Eq. 25-5, multiplied by l A so that it
expresses resistance instead of resistivity.
R R0 1 T T0 T T0 1 R 1 T0 1 V I 1 T0 1 I0 1
R0 V I0 I
20.0C 1 C1 0.4212 A 1 44.1C
0.00429 0.3818 A
24. For the cylindrical wire, its (constant) volume is given by V l0 A0 lA, and so AV . Combine
l
this relationship with Eq. 25-3. We assume that l l0.
R0 l0 l02 ; R l l2 ; dR 2 l
A0 V0 A V0 d l V0
R dR l 2 l l l V0R l V0R R 1 R
d l V0 2l l 2l2 2 l2 2R
V0
This is true for any initial conditions, and so l 1 R
l0 2 R0
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138
Chapter 25 Electric Currents and Resistance
25. The resistance depends on the length and area as R l A. Cutting the wire and running the wires
side by side will halve the length and double the area.
R2 1 l 1 l R1
2 4 A
41
2A
26. The total resistance is to be 3700 ohms Rtotal at all temperatures. Write each resistance in terms of
Eq.25-5 (with T0 0o C ), multiplied by l A to express resistance instead of resistivity.
Rtotal R0C 1 CT R0N 1 NT R0C R0CCT R0N R0NNT
R0C R0N R0CC R0NN T
For the above to be true, the terms with a temperature dependence must cancel, and the terms
without a temperature dependence must add to .Rtotal Thus we have two equations in two unknowns.
0 R0CC R0NN T R0N R0CC
N
Rtotal
R0C R0 N R0C R0CC R0C N C
N N
0.0004 Co 1
Co 1 0.0005
R0C
Rtotal N 3700 0.0004 Co 1 1644 1600
N C
R0N Rtotal R0C 3700 1644 2056 2100
27. We choose a spherical shell of radius r and thickness dr as a differential element. The area of this
element is 4 r2. Use Eq. 25-3, but for an infinitesimal resistance. Then integrate over the radius of
the sphere.
l dR dl dr R r2 dr 1 1 r2 1 1 1
A A 4 r2 r1 4 r2 4 r r1
R dR 4 r1 r2
28. (a) Let the values at the lower temperature be indicated by a subscript “0”. Thus R0 0 l0
A0
0 4l0 . The change in temperature results in new values for the resistivity, the length, and
d02
the diameter. Let represent the temperature coefficient for the resistivity, and T represent
the thermal coefficient of expansion, which will affect the length and diameter.
R l 4l 0 1 T T0 4l0 1 T T T0 0 4l0 1 T T0
A d2 1 T T T0 2 d02 1 T T T0
d0
R0 1 T T0 R1 T T T0 R0 1 T T0
1 T T T0
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
T0 R R0 140 12
0.0045C1 140
T R0 RT
20C 12 5.5 106 C1
20C 2405C 2425C 2400C
(b) The net effect of thermal expansion is that both the length and diameter increase, which lowers
the resistance.
0 4l
R 0
R0 2 2
d2 ld 0 l0 1 T T T0 d 0 1
l0
4l0 l0d 2 d0 1 T T T0 2 1 T T T0
d 2
0
1 2405C 0.9869
5.5 106 C1
1
% change R R0 100 R 1100 1.31 1.3 %
R0 R0
The net effect of resistivity change is that the resistance increases.
4l0
2 0 1 T T0
0
0
R d
R0
4l0 0 1 T T0 1 0.0045 C1 2405C
d02
0
11.82
% change R R0 100 R 1100 1082 1100%
R0 R0
29. (a) Calculate each resistance separately using Eq. 25-3, and then add the resistances together to find
the total resistance.
RCu
Cu l 4Cu l 4 1.68 108m 5.0 m 0.054567
A d2
1.4 103 m 2
RAl
Al l 4 Al l 4 2.65 108m 5.0 m 0.086074
A d2
1.4 103 m 2
Rtotal RCu RAl 0.054567 0.086074 0.140641 0.14
(b) The current through the wire is the voltage divided by the total resistance.
I V 85 103 V 0.60438 A 0.60 A
Rtotal 0.140641
(c) For each segment of wire, Ohm’s law is true. Both wires have the current found in (b) above.
VCu IRCu 0.60438A 0.054567 0.033V
VAl IRAl 0.60438A 0.086074 0.052 V
Notice that the total voltage is 85 mV.
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140
Chapter 25 Electric Currents and Resistance
30. (a) Divide the cylinder up into concentric cylindrical shells of
radius r, thickness dr, and length l. See the diagram. The dr
r1
resistance of one of those shells, from Eq. 25-3, is found. Note
that the “length” in Eq. 25-3 is in the direction of the current flow,
so we must substitute in dr for the “length” in Eq. 25-3. The area
is the surface area of the thin cylindrical shell. Then integrate over r
the range of radii to find the total resistance.
R "l" dR dr ; r2
A 2 rl
r2 dr ln r2
dR
R
r1 2 rl 2 l r1
(b) Use the data given to calculate the resistance from the above
formula.
R ln r2 15 105m ln 1.8 mm 5.8 104
2 l r1 1.0 mm
2 0.024 m
(c) For resistance along the axis, we again use Eq. 25-3, but the current is flowing in the direction
of length l. The area is the cross-sectional area of the face of the hollow cylinder.
R
l l 15 105m 0.024 m 0.51
A r22 r12
1.8 103 m 2 1.0 103m 2
31. Use Eq. 25-6 to find the power from the voltage and the current.
P IV 0.27 A 3.0 V 0.81W
32. Use Eq. 25-7b to find the resistance from the voltage and the power.
P V2 R V2 240 V2 17
R P 3300 W
33. Use Eq. 25-7b to find the voltage from the power and the resistance.
P V 2 V RP 3300 0.25W 29 V
R
34. Use Eq. 25-7b to find the resistance, and Eq. 25-6 to find the current.
(a) P V2 R V2 110 V2 161.3 160
R P 75W
P IV I P 75W 0.6818 A 0.68 A
V 110 V
(b) P V2 R V2 110 V2 27.5 28
R P 440 W
P IV I P 440 W 4.0 A
V 110 V
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141
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
35. (a) From Eq. 25-6, if power P is delivered to the transmission line at voltage V, there must be a
current I P V . As this current is carried by the transmission line, there will be power losses
of I 2R due to the resistance of the wire. This power loss can be expressed as P I 2R
P2R V 2 . Equivalently, there is a voltage drop across the transmission lines of V IR.
Thus the voltage available to the users is V V , and so the power available to the users is
P V V I VI V I VI I 2R P I 2R. The power loss is P P P P P I 2R
I 2R P2R V 2 .
(b) Since P 1 , V should be as large as possible to minimize P.
V2
36. (a) Since P V 2 R V 2 says that the resistance is inversely proportional to the power for a
RP
constant voltage, we predict that the 850 W setting has the higher resistance.
(b) R V2 120 V2 17
P 850 W
(c) R V2 120 V2 12
P 1250 W
37. (a) Use Eq. 25-6 to find the current.
P IV I P 95W 0.83A
V 115V
(b) Use Eq. 25-7b to find the resistance.
P V2 R V2 115 V2 140
R P 95W
38. The power (and thus the brightness) of the bulb is proportional to the square of the voltage,
according to Eq. 25-7b, P V 2 . Since the resistance is assumed to be constant, if the voltage is cut
R
in half from 240 V to 120V, the power will be reduced by a factor of 4. Thus the bulb will appear
only about 1/4 as bright in the United States as in Europe.
39. To find the kWh of energy, multiply the kilowatts of power consumption by the number of hours in
operation.
Energy P in kW t in h 550 W 1kW 6.0 min 1h 0.055 kWh
1000 W 60 min
To find the cost of the energy used in a month, multiply times 4 days per week of usage, times 4
weeks per month, times the cost per kWh.
Cost 0.055 kWh 4d 4 week 9.0 cents 7.9 cents month
d 1week 1month kWh
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142
Chapter 25 Electric Currents and Resistance
40. To find the cost of the energy, multiply the kilowatts of power consumption by the number of hours
in operation times the cost per kWh.
Cost 25 W 1kW 365 day 24 h $0.095 $ 21
1000 W 1day kWh
41. The Ah rating is the amount of charge that the battery can deliver. The potential energy of the
charge is the charge times the voltage.
U QV 75 Ah 3600 s 12 V 3.2 106J 0.90 kWh
1h
42. (a) Calculate the resistance from Eq. 25-2b and the power from Eq. 25-6.
R V 3.0 V 7.895 7.9 P IV 0.38A 3.0 V 1.14 W 1.1W
I 0.38 A
(b) If four D-cells are used, the voltage will be doubled to 6.0 V. Assuming that the resistance of
the bulb stays the same (by ignoring heating effects in the filament), the power that the bulb
would need to dissipate is given by Eq. 25-7b, P V 2 . A doubling of the voltage means the
R
power is increased by a factor of 4 . This should not be tried because the bulb is probably not
rated for such a high wattage. The filament in the bulb would probably burn out, and the glass
bulb might even explode if the filament burns violently.
43. Each bulb will draw an amount of current found from Eq. 25-6.
P IV I bulb P
V
The number of bulbs to draw 15 A is the total current divided by the current per bulb.
I total nI bulb nP n VItotal 120 V 15 A 24 bulbs
V
P 75 W
44. Find the power dissipated in the cord by Eq. 25-7a, using Eq. 25-3 for the resistance.
l l 4l 4 5.4 m
A d2 d2
0.129 102 m 2
P
I2R I2 I 2 I 2 15.0 A2 1.68 108m
4
15.62 W 16 W
45. Find the current used to deliver the power in each case, and then find the power dissipated in the
resistance at the given current.
P IV I P P =dissipated I 2R P2 R
V V2
7.5 105 W 2
1.2 104 V 2
P dissipated 3.0 11719 W
12,000 V
7.5 105 W 2
5 104 V 2
P dissipated 3.0 675 W difference 11719 W 675 W 1.1104 W
50,000 V
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143
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
46. (a) By conservation of energy and the efficiency claim, 75% of the electrical power dissipated by
the heater must be the rate at which energy is absorbed by the water.
0.75 Pemitted by absorbed 0.75 IV Qheat water mcT
electromagnet by water tt
I mcT 0.120 kg 4186 J kg 95C 25C 8.139 A 8.1A
0.75Vt 0.75 12 V 480s
(b) Use Ohm’s law to find the resistance of the heater.
V IR R V 12 V 1.5
I 8.139 A
47. The water temperature rises by absorbing the heat energy that the electromagnet dissipates. Express
both energies in terms of power, which is energy per unit time.
P Pelectric to heat IV Qheat water mcT
tt
water
m IV 17.5A 240 V 0.154 kg s 0.15 kg s
t cT 4186 J kgC 6.50C
This is 154 mL s.
48. For the wire to stay a constant temperature, the power generated in the resistor is to be dissipated by
radiation. Use Eq. 25-7a and 19-18, both expressions of power (energy per unit time). We assume
that the dimensions requested and dimensions given are those at the higher temperature, and do not
take any thermal expansion effects into account. We also use Eq. 25-3 for resistance.
I 2 4l
d2
I 2R A
T T4 4 d l T T4 4
high low high low
1/ 3 1/ 3
d
4I 2 4 15.0 A2 5.6 108 m
2 T T4 4
2 1.0 5.67 108 W m2 K4
high low 3100 K4 293K4
9.92 105 m 0.099 mm
49. Use Ohm’s law and the relationship between peak and rms values.
I peak 2Irms 2 Vrms 2 220 V 0.12 A
R 2700
50. Find the peak current from Ohm’s law, and then find the rms current from Eq. 25-9a.
I peak Vpeak 180 V 0.47368 A 0.47 A I rms I peak 2 0.47368 A 2 0.33A
R 380
51. (a) When everything electrical is turned off, no current will be flowing into the house, even though
a voltage is being supplied. Since for a given voltage, the more resistance, the lower the
current, a zero current corresponds to an infinite resistance.
(b) Use Eq. 25-7a to calculate the resistance.
P V2 R V2 = 120 V2 96
R P 2 75W
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144
Chapter 25 Electric Currents and Resistance
52. The power and current can be used to find the peak voltage, and then the rms voltage can be found
from the peak voltage.
P I Vrms rms I peak Vrms Vrms 2P 2 1500 W 390 V
2 5.4 A
I peak
53. Use the average power and rms voltage to calculate the peak voltage and peak current.
(a) Vpeak 2Vrms 2 660 V 933.4 V 930 V
(b) I peak
P I Vrms rms 2 Vrms I peak 2P 2 1800 W 3.9 A
Vrms 660 V
54. (a) We assume that the 2.5 hp is the average power, so the maximum power is twice that, or 5.0 hp,
as seen in Figure 25-22.
5.0 hp 746 W 3730 W 3700 W
1hp
(b) Use the average power and the rms voltage to find the peak current.
P I Vrms rms I peak Vrms I peak 2P 2 1 3730 W 11 A
2 Vrms 2
240 V
55. (a) The average power used can be found from the resistance and the rms voltage by Eq. 25-10c.
P V2 240 V2 1309 W 1300 W
rms
R 44
(b) The maximum power is twice the average power, and the minimum power is 0.
Pmax 2P 2 1309 W 2600 W Pmin 0 W
56. (a) Find Vrms. Use an integral from Appendix B-4, page A-7.
1 / 2 sin 4 t T 1/ 2 1/ 2
V02 T
1 T V0 2 t 2 dt T t V02 V0
Vrms 0 sin T 2 2 2
T 8
0
T
(b) Find Vrms.
1 T 1 / 2 1 T /2 1 T 1 / 2 V02 T 01/ 2 V0
T T T T /2 T 2 2
V 2dt V02dt 2
0 0
Vrms 0 dt
57. (a) We follow the derivation in Example 25-14. Start with Eq. 25-14, in absolute value.
j nevd vd j I N 1 mole I 4I m
ne neA m 1 mole D N De d 2
e 1 d 2
2
vd 6.02 1023
4 2.3 106 A 63.5 103 kg 0.65 103 m 2 5.1 1010 m s
8.9 103 kg m3 1.60 1019 C
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
(b) Calculate the current density from Eq. 25-11.
j
II 4I 4 2.3 106 A 2 6.931A m2 6.9 A m2
A r2 d2 6.5 104 m
(c) The electric field is calculated from Eq. 25-17.
j 1 E E j 1.68 108 m 6.931A m2 1.2 107 V m
58. (a) Use Ohm’s law to find the resistance.
V IR R V 0.0220 V 0.02933 0.029
I 0.75 A
(b) Find the resistivity from Eq. 25-3.
R l
A
2
RA R r2 0.02933 1.0 103 m
l l 5.80 m 1.589 108 m 1.6 108 m
(c) Use Eq. 25-11 to find the current density.
j I I 0.75 2.387 105 A m2 2.4 105 A m2
A r2
0.0010 m2
(d) Use Eq. 25-17 to find the electric field.
j 1E
E j 1.589 108 m 2.387 105 A m2 3.793 103 V m 3.8 103 V m
(e) Find the number of electrons per unit volume from the absolute value of Eq. 25-14.
n j 2.387 105 A m2 8.8 1028 e m3
vde 1.7 105 m s 1.60 1019 C
j nevd
59. We are given a charge density and a speed (like the drift speed) for both types of ions. From that we
can use Eq. 25-13 (without the negative sign) to determine the current per unit area. Both currents
are in the same direction in terms of conventional current – positive charge moving north has the
same effect as negative charge moving south – and so they can be added.
I neAvd
I
A
nevd He nevd O 2.8 1012 ions m3 2 1.60 1019 C ion 2.0 106 m s
7.0 1011 ions m3 1.60 1019 C ion 6.2 106 m s
2.486 A m2 2.5 A m2 , North
60. The magnitude of the electric field is the voltage change per unit meter.
E V 70 103 V 7.0 106 V m
x 1.0 108 m
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146
Chapter 25 Electric Currents and Resistance
61. The speed is the change in position per unit time.
v x 7.20 102 m 3.40 102 m 35 m s
t 0.0063s 0.0052 s
Two measurements are needed because there may be a time delay from the stimulation of the nerve
to the generation of the action potential.
62. The power is the work done per unit time. The work done to move a charge through a potential
difference is the charge times the potential difference. The charge density must be multiplied by the
surface area of the cell (the surface area of an open tube, length times circumference) to find the
actual charge moved.
PW QV QV
t tt
mol ions 1.6 C
3 107 m2 s 6.02 1023 mol 1019 ion 0.10 m 20 106 m 0.030 V
5.4 109 W
63. The energy supplied by the battery is the energy consumed by the lights.
E Esupplied consumed QV Pt
t QV 85Ah 3600s h 12V 39913s 1h 11.09 h 11 h
P 3600
92 W s
64. The ampere-hour is a unit of charge.
1.00Ah 1C s 3600 s 3600 C
1A 1h
65. Use Eqs. 25-3 and 25-7b.
R l l 4l ; P V2 V2
A r2 d2 R 4l
d2
1.5 V2 5.0 104 m 2
4 1.68 108 m 15 W
l
V 2 d 2 1.753 m 1.8 m
4P
If the voltage increases by a factor of 6 without the resistance changing, the power will increase by a
factor of 36. The blanket would theoretically be able to deliver 540 W of power, which might make
the material catch on fire or burn the occupant.
66. Use Eq. 25-6 to calculate the current.
P IV I P 746 W 6.22 A
V 120 V
67. From Eq. 25-2b, if R V I , then G I V
G I 0.48 A 0.16 S
V 3.0 V
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147
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
68. Use Eq. 25-7b to express the resistance in terms of the power, and Eq. 25-3 to express the resistance
in terms of the wire geometry.
P V2 R V2 R l l 2 4 l
RP A r d2
4
l V2 d 4 lP 4 9.71108 m 3.5 m 1500 W 2.3 104 m
d2 V 2 110 V2
P
69. (a) Calculate the total kWh used per day, and then multiply by the number of days and the cost per
kWh.
1.8 kW 2.0 h d 4 0.1kW 6.0 h d 3.0 kW 1.0 h d 2.0 kWh d
11.0 kWh d
Cost 11.0 kWh d 30 d $0.105 $34.65 $35 per month
kWh
(b) The energy required by the household is 35% of the energy that needs to be supplied by the
power plant.
Household Energy 0.35coal mass coal energy per mass
Household Energy 11.0 kWh d 365 d 1000 W 3600 s
kW 1h
coal mass 0.35 coal energy per mass
kcal 4186 J
0.35 7500 kg 1 kcal
1315 kg 1300 kg of coal
70. To deliver 15 MW of power at 120 V requires a current of I P 15 106 W 1.25 105 A .
V 120 V
Calculate the power dissipated in the resistors using the current and the resistance.
L L L 2 2 1.0 m
A r2 d
5.0 103 m 2
P
I2R I 2 I 2 4 I 2 2 4 1.25 105 A 1.68 108 m
2.674 107 W
Cost Power time rate per kWh 1 kW $0.090
2.674 107 W 1000 W 1h kWh
$2407 $2, 400 per hour per meter
71. (a) Use Eq. 25-7b to relate the power to the voltage for a constant resistance.
P V2 P105 105 V2 R 105 V2 0.805 or a 19.5% decrease
R P117 117 V2 R 117 V2
(b) The lower power output means that the resistor is generating less heat, and so the resistor’s
temperature would be lower. The lower temperature results in a lower value of the resistance,
which would increase the power output at the lower voltages. Thus the decrease would be
smaller than the value given in the first part of the problem.
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148
Chapter 25 Electric Currents and Resistance
72. Assume that we have a meter of wire, carrying 35 A of current, and dissipating 1.5 W of heat. The
power dissipated is PR I2R , and the resistance is R l .
A
PR I2R I2 l I 2 l I 2 4l
A r2 d2
d I 2 4l 2I l 2 35 A
PR PR
1.68 108 m 1.0 m 4.2 103 m
1.5 W
73. (a) The resistance at the operating temperature can be calculated directly from Eq. 25-7.
P V2 R V2 120 V2 190
R P 75 W
(b) The resistance at room temperature is found by converting Eq. 25-5 into an equation for
resistances and solving for R0.
R R0 1 T T0
R 192
R0 15
1 T T0 1 0.0045 K1 3000 K 293 K
74. (a) The angular frequency is 210 rad s .
f 210 rad s 33.42 Hz 33 Hz
2 2
(b) The maximum current is 1.80 A.
I rms I max 1.80 A 1.27 A
2
2
(c) For a resistor, V IR .
V IR 1.80 A sin 210 t 24.0 43.2 sin 210 t V
75. (a) The power delivered to the interior is 65% of the power drawn from the source.
Pinterior 0.65Psource Psource Pinterior 950 W 1462 W 1500 W
0.65 0.65
(b) The current drawn is current from the source, and so the source power is used to calculate the
current.
P IVsource I Psource 1462 W 12.18 A 12 A
source Vsource 120 V
76. The volume of wire is unchanged by the stretching. The volume is equal to the length of the wire
times its cross-sectional area, and since the length was increased by a factor of 1.20, the area was
decreased by a factor of 1.20. Use Eq. 25-3.
R0 l0 l 1.20l0 A A0 R l 1.20l 0 1.202 l0 1.44R0 1.44
A0 1.20 A A0 A0
1.20
© 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
149
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
77. The long, thick conductor is labeled as conductor number 1, and the short, thin conductor is labeled
as number 2. The power transformed by a resistor is given by Eq. 25-7b, P V 2 R , and both have
the same voltage applied.
R1 l1 R2 l2 l1 2l2 A1 4 A2 diameter1 2diameter2
A1 A2
P1 V12 R1 R2 l2 A2 l2 A1 142 P1 : P2 2 : 1
P2 V22 R2 R1 l1 A1 l1 A2 2
78. The heater must heat 108 m3 of air per hour from 5o C to 20o C , and also replace the heat being lost
at a rate of 850 kcal/h. Use Eq. 19-2 to calculate the energy needed to heat the air. The density of
air is found in Table 13-1.
Q mcT Q m cT 108 m3 1.29 kg 0.17 kcal 15C 355 kcal
t t h m3 kgC h
Power required 355 kcal 850 kcal 1205 kcal 4186 J 1h 1401W 1400 W
h h h kcal 3600 s
79. (a) Use Eq. 25-7b.
P V2 R V2 240 V2 20.57 21
R P 2800 W
(b) Only 75% of the heat from the oven is used to heat the water. Use Eq. 19-2.
0.75 Poven t Heat absorbed by water mcT
mcT 0.120 L 1 kg 4186 J kgC 85C
1L
t 0.75 Poven 20.33s 20 s 2 sig. fig.
0.752800 W
(c) 11cents 2.8 kW 20.33s 1h s 0.17 cents
kWh 3600
80. (a) The horsepower required is the power dissipated by the frictional force, since we are neglecting
the energy used for acceleration.
P Fv 240 N45 km hr 1m s hr 3000W 1 hp 4.0 hp
3.6 km 746 W
(b) The charge available by each battery is Q 95Ah 95 C s3600 s 3.42 105 C , and so the
total charge available is 24 times that. The potential energy of that charge is the charge times
the voltage. That energy must be delivered (batteries discharged) in a certain amount of time to
produce the 3000 W necessary. The speed of the car times the discharge time is the range of the
car between recharges.
P U QV t QV d
tt Pv
d
vt v QV v QV QV 24 3.42 105C 12 V 410 km
P Fv F 240 N
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