22. Using Equation 2514, the capacitances are
12
2
2
01
1
11
12
2
2
02
2
22
2
2 (8.85 10
C /N m )(0.050 m)
2.53 pF
ln(
/
)
ln(15 mm/5.0 mm)
2
2 (8.85 10
C /N m )(0.090 m)
3.61 pF .
ln(
/
)
ln(10 mm/2.5 mm)
L
C
ba
L
C
ba
πε
π
πε
π
−
−
×⋅
×⋅
==
=
Initially, the total equivalent capacitance is
12
1
2
12
12
1
2
1
2
1
2
1
1
1
(2.53 pF)(3.61 pF)
1.49 pF
2.53 pF 3.61 pF
CC
C
C
C
CC
C
C
C
C
C
+
=+=
⇒
=
=
=
++
,
and the charge on the positive plate of each one is (1.49 pF )(10 V) = 14.9 pC.
Next,
capacitor 2 is modified as described in the problem, with the effect that
12
2
2
02
2
22
2
2 (8.85 10
C /N m )(0.090 m)
2.17 pF .
ln(
/
)
ln(25 mm/2.5 mm)
L
C
ba
πε
π
−
×⋅
′
==
=
′
The new total equivalent capacitance is
12
12
12
(2.53 pF)(2.17 pF)
1.17 pF
2.53 pF 2.17 pF
CC
C
′
==
=
′
++
and the new charge on the positive plate of each one is (1.17 pF)(10 V) = 11.7 pC.
Thus
we see that the charge transferred from the battery (considered in absolute value) as a
result of the modification is 14.9 pC – 11.7 pC = 3.2 pC.
(a) This charge, divided by
e
gives the number of electrons that pass point
P
.
Thus,
12
7
19
3.2 10
C
2.0 10
1.6 10
C
N
−
−
×
==
×
×
.
(b) These electrons move rightwards in the figure (that is, away from the battery) since
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This note was uploaded on 01/25/2011 for the course PHYS 221 taught by Professor Hk during the Spring '10 term at The Petroleum Institute.
 Spring '10
 HK
 Physics, Capacitance

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