Energy in a Capacitor

Level pending

A parallel plate capacitor is connected to a battery.

While connected to the battery, the area of each plate is increased by a factor of 2.

If the distance between the two plates is decreased by a factor of 4, what happens to the energy stored in the capacitor?

It will increase by a factor of 2 It will decrease by a factor of 2 It will increase by a factor of 8 It will decrease by a factor of 8

Alain Caesar Torre by Alain Caesar Torre , 25, Philippines

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1 solution

Arghyanil Dey
May 1, 2014

C = € A d C=\frac {€A}{d} where A=area of each plate , d=distance between two plates As the capacitance increase by a factor 8 , increase in energy also by a factor 8 [as E = . 5 C V 2 E=.5 CV^{2} ]

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