P26_078.PDF

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Chapter 26 - 26.78
78. (a) Initially, the capacitance is
C 0 = ε 0 A
d
8 . 85
×
10 − 12 C 2
N · m 2 (0 . 12 m 2 )
=
=89pF .
1 . 2
×
10 − 2 m
(b) Working through Sample Problem 26-6 algebraically, we find:
C =
ε 0 Aκ
b )+ b =
8 . 85
×
10 − 12 C 2
N
m 2 (0 . 12 m 2 )(4 . 8)
10 − 3 m) = 120 pF .
κ ( d
−
(4 . 8)(1 . 2
−
0 . 40)(10 − 2 m) + (4 . 0
×
(c) Before the insertion, q = C 0 V (89 pF)(120 V) = 11 nC . Since the battery is disconnected, q will
remain the same after the insertion of the slab.
(d) E = q/ε 0 A =11
×
10 − 9 C / 8 . 85
×
10 − 12 C 2
N
·
m 2 0 . 12 m 2 =10kV / m .
(e) E = E/κ =(10kV / m) / 4 . 8=2 . 1kV / m .
(f) V = E ( d
b )+ E b =(10kV / m)(0 . 012 m
−
0 . 0040 m) + (2 . 1kV / m)(0 . 40
×
10 − 3 m) = 88 V.
(g) The work done is
W ext =∆ U = q 2
2
1
C −
1
89
1
C 0
=
(11
×
10 − 9 C) 2
2
10 − 12 F −
1
×
120
×
10 − 12 F
=
−
1 . 7
×
10 − 7 J .
·
−
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