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Question

Two place of a parallel plate capacitor of capacity 50μF are charged by a battery to a potential of 100 V. The battery remains connected and the place are separated from each other so that the distance between them is doubled. The energy spent by the battery in doing so, will be :
  1. 25×102J
  2. 12.5×102J
  3. 25×102J
  4. 12.5×102J

A
12.5×102J
B
12.5×102J
C
25×102J
D
25×102J
Solution
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We know that when separation between the plates is doubled the capacitance becomes one half i.e.,C=25μF
The energy spend by the battery is given by
qV=(CV)V=CV2
=25×106×(100)2
=25×102 J

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