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A parallel plate capacitor is connected to a battery that maintains a constant potential difference. How will the magnitude of the charge on the plates change if the separation between the plates is doubled?It will be cut in half.It will not change.It will double.It will quadruple.

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A parallel plate capacitor is connected to a battery that maintains a constant potential difference.

As we know,


\begin{gathered} Q=CV \\ Q=(A\epsilon_0V)/(d) \end{gathered}

Here, Q is the charge which is inversely proportional to the separation between the plates.


Q\propto(1)/(d)

That means if the separation between the plates increases the magnitude of the charge on the plates decreases.

Thus, if the separation between the plates is doubled the magnitude of the charge will be cut in half.

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