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The plates of a parallel plate capacitor of capacity 50 μC are charged to a potential of 100 volts and then separated from each other so that the distance between them is doubled. How much is the energy spent in doing so

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a

25×10−2 J

b

−12.5×10−2J

c

−25×10−2 J

d

12.5×10−2 J

answer is A.

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Detailed Solution

Wext=12C′V′2−12CV2=12C2(2V)2−12CV2=12CV2Wext=12×50×10−6×(100)2=25×10−2 J

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