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Question: The plates of a parallel plate capacitor of capacity \(50 \mu C\) are charged to a potential of 100 ...

The plates of a parallel plate capacitor of capacity 50μC50 \mu C 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

A

25×10−2J25 \times 10 ^ { - 2 } J

B

C

D

Answer

25×10−2J25 \times 10 ^ { - 2 } J

Explanation

Solution

=(12)(C2)(2V)2−12CV2=12CV2= \left( \frac { 1 } { 2 } \right) \left( \frac { C } { 2 } \right) ( 2 V ) ^ { 2 } - \frac { 1 } { 2 } C V ^ { 2 } = \frac { 1 } { 2 } C V ^ { 2 }

Wext=12×50×10−6×(100)2=25×10−2JW _ { e x t } = \frac { 1 } { 2 } \times 50 \times 10 ^ { - 6 } \times ( 100 ) ^ { 2 } = 25 \times 10 ^ { - 2 } J