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56. Initially, the total equivalent capacitance is
C
12
=
[
(
C
1
)
−
1
+ (
C
2
)
−
1
]
−
1
= 3.0
µ
F, and the
charge on the positive plate of each one is (3.0
µ
F)(10 V) = 30
µ
C.
Next, the capacitor
(call is
C
1
) is squeezed as described in the problem, with the effect that the new value of
C
1
is 12
µ
F (see Eq. 259). The new total equivalent capacitance then becomes
C
12
=
[
(
C
1
)
−
1
+ (
C
2
)
−
1
]
−
1
= 4.0
µ
F,
and the new charge on the positive plate of each one is (4.0
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 Spring '10
 Rebello,NobelS
 Capacitance, Charge

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