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49. (a) The charge on the positive plate of the capacitor is given by
q
=
C
E
(1
−
e
−
t/τ
)
,
where
E
is the emf of the battery,
C
is the capacitance, and
τ
is the time constant. The value of
τ
is
τ
=
RC
=(3
.
00
×
10
6
Ω)(1
.
00
×
10
−
6
F) = 3
.
00 s. At
t
=1
.
00 s,
t/τ
=(1
.
00 s)
/
(3
.
00 s) = 0
.
333
and the rate at which the charge is increasing is
dq
dt
=
C
E
τ
e
−
t/τ
=
(1
.
00
×
10
−
6
)(4
.
00 V)
3
.
00 s
e
−
0
.
333
=9
.
55
×
10
−
7
C
/
s
.
(b) The energy stored in the capacitor is given by
U
C
=
q
2
2
C
.
and its rate of change is
dU
C
dt
=
q
C
dq
dt
.
Now
q
=
C
E
(1
−
e
−
t/τ
)=(1
.
00
×
10
−
6
)(4
.
00 V)(1
−
e
−
0
.
333
)=1
.
13
×
10
−
6
C
,
so
dU
C
dt
=
µ
1
.
13
×
10
−
6
C
1
.
00
×
10
−
6
F
¶
(9
.
55
×
10
−
7
C
/
s) = 1
.
08
×
10
−
6
W
.
(c) The rate at which energy is being dissipated in the resistor is given by
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Capacitance, Charge

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