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58. (a) In the process described in the problem, no charge is gained or lost. Thus,
q
= constant. Hence,
q
=
C
1
V
1
=
C
2
V
2
=
⇒
V
2
=
V
1
C
1
C
2
= (200)
µ
150
10
¶
= 3000 V
.
(b) Eq. 2836, with
τ
=
RC
, describes not only the discharging of
q
but also of
V
.Thu
s
,
V
=
V
0
e
−
t/τ
=
⇒
t
=
RC
ln
µ
V
0
V
¶
=
(
300
×
10
9
Ω
)(
10
×
10
−
12
F
)
ln
µ
3000
100
¶
which yields
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Unformatted text preview: to their next task (such as handling the sensitive electronic equipment). (c) We solve V = V e − t/RC for R with the new values V = 1400 V and t = 0 . 30 s. Thus, R = t C ln( V /V ) = . 30 s (10 × 10 − 12 F) ln(1400 / 100) = 1 . 1 × 10 10 Ω ....
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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, Charge

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