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20. We note that the total equivalent capacitance is
C
123
=
[
(
C
3
)
−
1
+ (
C
1
+
C
2
)
−
1
]
−
1
= 6
µ
F.
(a) Thus, the charge that passed point
a
is
C
123
V
batt
= (6
µ
F)(12 V) = 72
µ
C.
Dividing this
by the value
e
= 1.60
×
10
−
19
C gives the number of electrons: 4.5
×
10
14
, which travel to
the left – towards the positive terminal of the battery.
(b) The equivalent capacitance of the parallel pair is
C
12
=
C
1
+
C
2
= 12
µ
F.
Thus, the
voltage across the pair (which is the same as the voltage across
C
1
and
C
2
individually) is
72
µ
C
12
µ
F
= 6 V .
Thus, the charge on
C
1
is (4
µ
F)(6 V) = 24
µ
C, and dividing this by
e
gives the number of
electrons (1.5
×
10
14
) which have passed (upward) though point
b
.
(c) Similarly, the charge on
C
2
is (8
µ
F)(6 V) = 48
µ
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Spring '10 term at Kansas State University.
 Spring '10
 Rebello,NobelS
 Capacitance, Charge

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