A parallel plate capacitor is connected to an ideal battery which provides a fixed potential differences. Originally, the energy stored in the capacitor is $U_{0}$ . If the distance between the plates is doubled, then new energy stored in the capacitor will be

4 U_{0}

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2 U_{0}

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U_{0}

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U_{0} / 2

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Solution

The correct option is A $4 U_{0}$
Energy stored in parallel capacitor, $U = \frac{1}{2} ϵ_{o} \frac{A}{d} . E^{2} . d^{2}$
It is seen that energy stored in the capacitor is directly proportional to square of the distance between the plates are doubled, the new energy stored will be $4 U_{o}$

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A parallel plate capacitor is connected to an ideal battery which provides a fixed potential differences. Originally, the energy stored in the capacitor is U0. If the distance between the plates is doubled, then new energy stored in the capacitor will be

The correct option is A 4U0Energy stored in parallel capacitor, U=12ϵoAd.E2.d2It is seen that energy stored in the capacitor is directly proportional to square of the distance between the plates are doubled, the new energy stored will be 4Uo

A parallel plate capacitor is connected to an ideal battery which provides a fixed potential differences. Originally, the energy stored in the capacitor is $U_{0}$ . If the distance between the plates is doubled, then new energy stored in the capacitor will be

The correct option is A $4 U_{0}$
Energy stored in parallel capacitor, $U = \frac{1}{2} ϵ_{o} \frac{A}{d} . E^{2} . d^{2}$
It is seen that energy stored in the capacitor is directly proportional to square of the distance between the plates are doubled, the new energy stored will be $4 U_{o}$