P26_075.PDF
(
60 KB
)
Pobierz
Chapter 26 - 26.75
75.(a)
C
=
ε
0
A/
(
d
−
b
), the same as part (a) in problem 74.
(b) Now,
U
U
2
C
V
2
=
C
ε
0
A/d
ε
0
A/
(
d
b
)
=
d
−
b
=
=
.
C
−
d
(c) The work done is
W
=∆
U
=
U
−
U
=
1
2
(
C
−
C
)
V
2
=
ε
0
A
2
1
d
−
b
−
1
d
V
2
=
ε
0
AbV
2
2
d
(
d
b
)
.
Since
W>
0the slab must be pushed in.
1
2
CV
2
1
−
Plik z chomika:
kf.mtsw
Inne pliki z tego folderu:
P26_005.PDF
(60 KB)
P26_002.PDF
(52 KB)
P26_003.PDF
(57 KB)
P26_001.PDF
(52 KB)
P26_019.PDF
(64 KB)
Inne foldery tego chomika:
chap01
chap02
chap03
chap04
chap05
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