This preview shows page 1. Sign up to view the full content.
69. (a) The charge on the positive plate of the capacitor is given by
qC
e
t
=−
−
ε
τ
1
c
h
,
where
is the emf of the battery,
C
is the capacitance, and
is the time constant. The
value of
is
=
RC
= (3.00
×
10
6
Ω
)(1.00
×
10
–6
F) = 3.00 s.
At
t
= 1.00 s,
t
/
= (1.00 s)/(3.00 s) = 0.333 and the rate at which the charge is increasing
is
()
( )
6
0.333
7
1.00 10
F 4.00V
9.55 10 C s.
3.00s
t
dq
C
ee
dt
−
−−
−
×
==
=
×
(b) The energy stored in the capacitor is given by
2
,
2
C
q
U
C
=
and its rate of change is
dU
dt
q
C
dq
dt
C
=
.
Now
e
e
t
=−=×
−
=
×
− −
1
100 10
4 00
1
113 10
6
0 333
6
c
h
c
h
bg
c
h
..
.
.
VC
,
so
6
76
6
1.13 10 C
9.55 10 C s
1.08 10 W.
1.00 10 F
C
dU
qdq
dt
C dt
−
−
⎛⎞
×
×
=
×
⎜⎟
×
⎝⎠
(c) The rate at which energy is being dissipated in the resistor is given by
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Capacitance, Charge

Click to edit the document details