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(C) Let the initial capacitance be $C = \frac{\epsilon_0 A}{d}$, charge be $Q = CV$, and electric field be $E = \frac{V}{d}$.When the separation $d$ i

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becomes four times

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(C) Let the initial capacitance be $C = \frac{\epsilon_0 A}{d}$, charge be $Q = CV$, and electric field be $E = \frac{V}{d}$.When the separation $d$ is halved to $d' = \frac{d}{2}$, the new capacitance becomes $C' = \frac{\epsilon_0 A}{d'} = 2C$.The potential difference $V$ remains constant as the battery is connected.The new charge becomes $Q' = C'V = 2CV = 2Q$.The new electric field becomes $E' = \frac{V}{d'} = \frac{V}{d/2} = 2E$.The force of attraction between the plates of a capacitor is given by $F = \frac{QE}{2}$.Substituting the new values, the new force $F'$ is:$F' = \frac{Q' E'}{2} = \frac{(2Q)(2E)}{2} = 4 \left( \frac{QE}{2} \right) = 4F$.Therefore, the force of attraction becomes four times the initial force.

$A$ parallel plate capacitor is connected to a battery of constant potential difference $V$. What happens to the force of attraction between the plates when the separation between them is halved?

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