Gabarito_Listas_Caps25-26

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Chapter 25 
 
12. (a) The potential difference across C1 is V1 = 10.0 V. Thus, 
 
q1 = C1V1 = (10.0 µF)(10.0 V) = 1.00 × 10–4 C. 
 
(b) Let C = 10.0 µF. 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 
 
2
eq
2
1 50 C CC C . C
C C
= + =
+
. 
 
Also, the voltage drop across this combination is 
 
1 1
10 401 50 eq
CV CVV .
C C C . C
= = =
+ +
V . 
 
Since this voltage difference is divided equally between C2 and the one connected in series 
with it, the voltage difference across C2 satisfies V2 = V/2 = V1/5. Thus 
 
( ) 52 2 2
10 0V10 0 F 2 00 10 C.
5
.q C V . .µ − = = = × 
 
 
 
19. (a) After the switches are closed, the potential differences across the capacitors are the 
same and the two capacitors are in parallel. The potential difference from a to b is given by 
Vab = Q/Ceq, where Q is the net charge on the combination and Ceq is the equivalent 
capacitance. The equivalent capacitance is Ceq = C1 + C2 = 4.0 × 10–6 F. The total charge on 
the combination is the net charge on either pair of connected plates. The charge on 
capacitor 1 is 
 
( )( )6 41 1 1.0 10 F 100V 1.0 10 Cq C V − −= = × = × 
 
and the charge on capacitor 2 is 
 
( )( ) ,100.3100100.3 4622 CVFVCq −− ×=×== 
 
so the net charge on the combination is 3.0 × 10–4 C – 1.0 × 10–4 C = 2.0 × 10–4 C. The 
potential difference is 
 
Vab =
×
×
=
−
−
2 0 10
4 0 10
50
4
6
.
.
C
F
V. 
 
(b) The charge on capacitor 1 is now q1 = C1Vab = (1.0 × 10–6 F)(50 V) = 5.0 × 10–5 C. 
 
(c) The charge on capacitor 2 is now q2 = C2Vab = (3.0 × 10–6 F)(50 V) = 1.5 × 10–4 C. 
 
25. We note that the total equivalent capacitance is C123 = [(C3)−1 + (C1 + C2)−1]−1 = 6 µF. 
 
(a) Thus, the charge that passed point a is C123 Vbatt = (6 µF)(12 V) = 72 µC. Dividing this 
by the value e = 1.60 × 10−19 C gives the number of electrons: 4.5 × 1014, which travel to 
the left – towards the positive terminal of the battery. 
 
(b) The equivalent capacitance of the parallel pair is C12 = C1 + C2 = 12 µF. Thus, the 
voltage across the pair (which is the same as the voltage across C1 and C2 individually) is 
 
.6
12
72 V
F
C
=
µ
µ 
 
Thus, the charge on C1 is (4 µF)(6 V) = 24 µC, and dividing this by e gives the number of 
electrons (1.5 × 1014) which have passed (upward) though point b. 
 
(c) Similarly, the charge on C2 is (8 µF)(6 V) = 48 µC, and dividing this by e gives the 
number of electrons (3.0 × 1014) which have passed (upward) though point c. 
 
(d) Finally, since C3 is in series with the battery, its charge is the same 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. 
 
27. (a) In this situation, capacitors 1 and 3 are in series, which means their charges are 
necessarily the same: 
 
( ) ( ) ( )1 3
1 3
1 3
1.00 F 3.00 F 12.0V
9.00 C.
1.00 F+3.00 F
C C Vq q
C C
µ µ
µ
µ µ
= = = =
+
 
 
(b) Capacitors 2 and 4 are also in series: 
 
( ) ( ) ( )2 4
2 4
2 4
2.00 F 4.00 F 12.0V
16.0 C.
2.00 F 4.00 F
C C Vq q
C C
µ µ
µ
µ µ
= = = =
+ +
 
 
(c) 3 1 9.00 C.q q µ= = 
 
(d) q q4 2 16.0 C.µ= = 
 
(e) With switch 2 also closed, the potential difference V1 across C1 must equal the potential 
difference across C2 and is 
 
( )( )3 4
1
1 2 3 4
3.00 F 4.00 F 12.0V
8.40V.
1.00 F 2.00 F 3.00 F 4.00 F
C CV V
C C C C
µ µ
µ µ µ µ
++
= =
+ + + + + +
= 
 
Thus, q1 = C1V1 = (1.00 µF)(8.40 V) = 8.40 µC. 
 
(f) Similarly, q2 = C2V1 = (2.00 µF)(8.40 V) = 16.8 µC. 
 
(g) q3 = C3(V – V1) = (3.00 µF)(12.0 V – 8.40 V) = 10.8 µC. 
 
(h) q4 = C4(V – V1) = (4.00 µF)(12.0 V – 8.40 V) = 14.4 µC. 
 
37. (a) They each store the same charge, so the maximum voltage is across the smallest 
capacitor. With 100 V across 10 µF, then the voltage across the 20 µF capacitor is 50 V and 
the voltage across the 25 µF capacitor is 40 V. Therefore, the voltage across the 
arrangement is 190 V. 
 
(b) Using Eq. 25-21 or Eq. 25-22, we sum the energies on the capacitors and obtain Utotal = 
0.095 J. 
 
46. (a) We use Eq. 25-14: 
 
( )229 N mC
(4.7)(0.15 m)2 0.73 nF.
ln( / ) 2 8.99 10 ln(3.8 cm/3.6 cm)
LC
b a0 ⋅
= πε κ = =
×
 
 
(b) The breakdown potential is (14 kV/mm) (3.8 cm – 3.6 cm) = 28 kV. 
 
50. Let C1 = ε0(A/2)κ1/2d = ε0Aκ1/4d, C2 = ε0(A/2)κ2/d = ε0Aκ2/2d, and C3 = ε0Aκ3/2d. 
Note that C2 and C3 are effectively connected in series, while C1 is effectively connected in 
parallel with the C2-C3 combination. Thus, 
 
( ) ( ) ( )0 2 32 3 0 1 0 2 3
1 1
2 3 2 3 2 3
2 2 2 .
4 2 2 4
A dC C A AC C
C C d d
ε κ κε κ ε κ κκ
κ κ κ
 
= + = + = + + +  κ+
 
 
With , d , 3 21.05 10 mA −= × 33.56 10 m−= × 1 21.0κ = , 2 42.0κ = and 3 58.0κ = , the 
capacitance is, (in SI units) 
 
12 3
11
3
(8.85 10 )(1.05 10 ) 2(42.0)(58.0)21.0 4.55 10 F.
4(3.56) 10 42.0 58.0
C
− −
−
−
× ×  = + × + 
= × 
 
62. (a) The potential across C1 is 10 V, so the charge on it is 
 
q1 = C1V1 = (10.0 µF)(10.0 V) = 100 µC. 
 
(b) Reducing the right portion of the circuit produces an equivalence equal to 6.00 µF, with 
10.0 V across it. Thus, a charge of 60.0 µC is on it -- and consequently also on the bottom 
right capacitor. The bottom right capacitor has, as a result, a potential across it equal to 
 
V
F
C
C
qV 00.6
10
60
===
µ
µ 
 
which leaves 10.0 V − 6.00 V = 4.00 V across the group of capacitors in the upper right 
portion of the circuit. Inspection of the arrangement (and capacitance values) of that group 
reveals that this 4.00 V must be equally divided by C2 and the capacitor directly below it (in 
series with it). Therefore, with 2.00 V across C2 we find 
 
q2 = C2V2 = (10.0 µF)(2.00 V) = 20.0 µC . 
 
69. (a) Since the field is constant and the capacitors are in parallel (each with 600 V across 
them) with identical distances (d = 0.00300 m) between the plates, then the field in A is 
equal to the field in B: 
 
r
E V
d
= = ×2 00 105. .V m 
 
(b) 5| 2.00 10 V m .E = ×
r
| See the note in part (a). 
 
(c) For the air-filled capacitor, Eq. 25-4 leads to 
 
σ ε= = = × −q
A
E0
6 2177 10
r
. .C m 
 
(d) For the dielectric-filled capacitor, we use Eq. 25-29: 
 
σ κε= = × −0
6 24 60 10
r
E . .C m 
 
(e) Although the discussion in the textbook (§25-8) is in terms of the charge being held 
fixed (while a dielectric is inserted), it is readily adapted to this situation (where 
comparison is made of two capacitors which have the same voltage and are identical except 
for the fact that one has a dielectric). The fact that capacitor B has a relatively large charge 
but only produces the field that A produces (with its smaller charge) is in line with the point 
being made (in the text) with Eq. 25-34 and in the material that follows. Adapting Eq. 25-
35 to this problem, we see that the difference in charge densities between parts (c) and (d) 
is due, in part, to the (negative) layer of charge at the top surface of the dielectric; 
consequently, 
 
( ) ( ) ./1083.21060.41077.1' 2666 mC−−− ×−=×−×=σ 
 
77. The voltage across capacitor 1 is 
 
V q
C1
1
1
30
10
30= = =µ
µ
C
F
V. . 
 
Since V1 = V2, the total charge on capacitor 2 is 
 
CVCq µµ 60)V 2)(F 20(222 === , 
 
which means a total of 90 µC of charge is on the pair of capacitors C1 and C2. This implies 
there is a total of 90 µC of charge also on the C3 and C4 pair. Since C3 = C4, the charge 
divides equally between them, so q3 = q4 = 45 µC. Thus, the voltage across capacitor 3 is 
V q
C3
3
3
45
20
2 3= = =µ
µ
C
F
V. . 
 
Therefore, |VA – VB| = V1 + V3 =5.3 V. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Chapter 26 
 
13. (a) The current resulting from this non-uniform current density is 
 
2 3 2 4 20
0cylinder 0
2 22 (3.40 10 m) (5.50 10 A/m ) 1.33 A
3 3
R
a
Ji J dA r rdr R J
R
π π π −= = ⋅ = = × × =∫ ∫ . 
 
(b) In this case, 
 
2 3 2
0 0cylinder 0
1 11 2 (3.40 10 m) (5.50 10 A/m )
3 3
0.666 A.
R
b
ri J dA J rdr R J
R
π π π − = = − = = × × 
 
=
∫ ∫ 4 2 
 
(c) The result is different from that in part (a) because Jb is higher near the center of the 
cylinder (where the area is smaller for the same radial interval) and lower outward, 
resulting in a lower average current density over the cross section and consequently a lower 
current than that in part (a). So, Ja has its maximum value near the surface of the wire. 
 
20. Since the potential difference V and current i are related by V = iR, where R is the 
resistance of the electrician, the fatal voltage is V = (50 × 10–3 A)(2000 Ω) = 100 V. 
 
24. The cross-sectional area is A = πr2 = π(0.002 m)2. The resistivity from Table 26-1 is 
ρ = 1.69 × 10−8 Ω·m. Thus, with L = 3 m, Ohm’s Law leads to V = iR = iρL/A, or 
 
 12 × 10−6 V = i (1.69 × 10−8 Ω·m)(3.0 m)/ π(0.002 m)2 
 
which yields i = 0.00297 A or roughly 3.0 mA. 
 
31. (a) i = V/R = 35.8 V/935 Ω = 3.83 × 10–2 A. 
 
(b) J = i/A = (3.83 × 10–2 A)/(3.50 × 10–4 m2) = 109 A/m2. 
 
(c) vd = J/ne = (109 A/m2)/[(5.33 × 1022/m3) (1.60 × 10–19 C)] = 1.28 × 10–2 m/s. 
 
(d) E = V/L = 35.8 V/0.158 m = 227 V/m. 
 
35. (a) The current i is shown in Fig. 26-29 entering the truncated cone at the left end and 
leaving at the right. This is our choice of positive x direction. We make the assumption that 
the current density J at each value of x may be found by taking the ratio i/A where A = πr2 
is the cone’s cross-section area at that particular value of x. The direction of 
r
J is identical 
to that shown in the figure for i (our +x direction). Using Eq. 26-11, we then find an 
expression for the electric field at each value of x, and next find the potential difference V 
by integrating the field along the x axis, in accordance with the ideas of Chapter 25. Finally, 
the resistance of the cone is given by R = V/i. Thus, 
 
J i
r
E
= =
π 2 ρ
 
 
where we must deduce how r depends on x in order to proceed. We note that the radius 
increases linearly with x, so (with c1 and c2 to be determined later) we may write 
 
r c c x= +1 2 . 
 
Choosing the origin at the left end of the truncated cone, the coefficient c1 is chosen so that 
r = a (when x = 0); therefore, c1 = a. Also, the coefficient c2 must be chosen so that (at the 
right end of the truncated cone) we have r = b (when x = L); therefore, c2 = 
(b – a)/L. Our expression, then, becomes 
 
.x
L
abar 




 −+= 
 
Substituting this into our previous statement and solving for the field, we find 
 
.
2−





 −+= x
L
abaiE
π
ρ 
 
Consequently, the potential difference between the faces of the cone is 
 
2 1
0 0
0
1 1 .
L
L Li b a i L b aV E dx a x dx a x
L b a L
i L i L b a i L
b a a b b a ab ab
ρ ρ
ρ ρ ρ
− −− −   = − = − + = +   π π −   
− = − = = π − π − π 
∫ ∫
 
 
The resistance is therefore 
 
2
5
3 3
(731 m)(1.94 10 m) 9.81 10 
(2.00 10 m)(2.30 10 m)
V LR
i ab
ρ
π
−
− −
Ω⋅ ×
= = = = × Ω
π × ×
 
 
Note that if b = a, then R = ρL/πa2 = ρL/A, where A = πa2 is the cross-sectional area of the 
cylinder. 
 
40. (a) Referring to Fig. 26-34, the electric field would point down (towards the bottom of 
the page) in the strip, which means the current density vector would point down, too (by 
Eq. 26-11). This implies (since electrons are negatively charged) that the conduction-
electrons would be “drifting” upward in the strip. 
 
(b) Eq. 24-6 immediately gives 12 eV, or (using e = 1.60 × 10−19 C) 1.9 × 10−18 J for the 
work done by the field (which equals, in magnitude, the potential energy change of the 
electron). 
 
(c) Since the electrons don’t (on average) gain kinetic energy as a result of this work done, 
it is generally dissipated as heat. The answer is as in part (b): 12 eV or 1.9 × 10−18 J. 
 
44. The slope of the graph is P = 5.0 × 10−4 W. Using this in the P = V2/R relation leads to 
V = 0.10 Vs. 
 
48. The slopes of the lines yield P1 = 8 mW and P2 = 4 mW. Their sum (by energy 
conservation) must be equal to that supplied by the battery: Pbatt = ( 8 + 4 ) mW = 12 mW. 
 
53. (a) We use Eq. 26-16 to compute the resistances in SI units: 
 
6
2 2
1(2 10 ) 2.5 .CC C
C
LR
r
ρ −= = × =
π π(0.0005)
Ω
V.
 
 
The voltage follows from Ohm’s law: 
 
1 2| | 5.1C CV V V iR− = = = 
 
(b) Similarly, 
6
2 2
1(1 10 ) 5.1DD D
D
LR
r
ρ −= = × =
π π(0.00025)
Ω
V
 
 
and | | . 2 3 10D DV V V iR− = = =
 
(c) The power is calculated from Eq. 26-27: . 2 10WC CP i R= =
 
(d) Similarly, . 2 20W D DP i R= =
 
70. (a) The current is 4.2 × 1018 e divided by 1 second. Using e = 1.60 × 10−19 C we obtain 
0.67 A for the current. 
 
(b) Since the electric field points away from the positive terminal (high potential) and 
towards the negative terminal (low potential), then the current density vector (by Eq. 26-11) 
must also point towards the negative terminal.