A parallel plate capacitor is charged fully b | vedclass

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Question

(d)

As plates are moved far away further, capacity of capacitor $\left(C=\frac{\varepsilon_0 A}{d}\right)$ decreases.

As battery remains connect

Vedclass · Accepted Answer

the electrostatic energy stored in the capacitor decreases

Vedclass · Answer

(d)

As plates are moved far away further, capacity of capacitor $\left(C=\frac{\varepsilon_0 A}{d}\right)$ decreases.

As battery remains connected during this activity, potential difference between plates remains same.

So, the charge on plates $(Q=C V)$ decreases.

Also, energy of capacitor $\left(U=\frac{1}{2} C V^2\right)$ decreases as capacitance decreases. So, correct option is $( d )$.

A parallel plate capacitor is charged fully by using a battery. Then, without disconnecting the battery, the plates are moved further apart. Then,

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