45. During charging, the charge on the positive plate of the capacitor is given by
qC
e
t
=−
−
ε
τ
1
c
h
,
where
C
is the capacitance,
is applied emf, and
=
RC
is the capacitive time constant.
The equilibrium charge is
q
eq
=
C
. We require
q
= 0.99
q
eq
= 0.99
C
, so
099 1
..
−
e
t
Thus,
e
t
−
=
001
. . Taking the natural logarithm of both sides, we obtain
t
/
= – ln 0.01 =
4.61 or
t
= 4.61
.
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View Full Document51. (a) The initial energy stored in a capacitor is given by
2
0
/2 ,
C
UqC
=
where
C
is the
capacitance and
q
0
is the initial charge on one plate. Thus
qC
U
C
0
63
22
1
0
1
0
0
5
0
1
0
1
0
==
×
=
×
−−
..
.
FJ
C
.
c
hb
g
(b) The charge as a function of time is given by
qq
e
t
=
−
0
τ
, where
is the capacitive time
constant. The current is the derivative of the charge
i
dq
dt
q
e
t
=−
=
−
0
,
and the initial current is
i
0
=
q
0
/
. The time constant is
RC
()
66
1.0 10 F 1.0 10
1.0s
−
××
Ω
=
.
Thus
i
0
33
10 10
10
=×
=
×
.
Cs
A
c
h
b
g
.
(c) We substitute
0
t
e
−
=
into
V
C
= q
/
C
to obtain
3
1.0 s
3
1.0
0
6
1.0 10 C
1.0 10 V
,
1.0 10 F
tt
t
C
q
Ve
e
e
C
−
−− −
−
§·
×
=
×
¨¸
×
©¹
where
t
is measured in seconds.
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 Spring '08
 STEPHENIRONS
 Physics, Capacitance, Charge, Electric charge, 10 kg, 6 M, 1.0 W, 0.50 J

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