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(A) The potential difference $V$ across the plates remains constant because the capacitor is connected to a battery.The electric field $E$ between the

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decreases and the charge on the plates also decreases.

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(A) The potential difference $V$ across the plates remains constant because the capacitor is connected to a battery.The electric field $E$ between the plates is given by $E = V/d$,where $d$ is the distance between the plates.As the plates are moved further apart,$d$ increases,so $E = V/d$ decreases.The capacitance $C$ of a parallel plate capacitor is given by $C = \epsilon_0 A / d$.As $d$ increases,the capacitance $C$ decreases.The charge $Q$ on the plates is given by $Q = CV$.Since $C$ decreases and $V$ is constant,the charge $Q$ on the plates also decreases.Therefore,both the electric field intensity and the charge on the plates decrease.

$A$ parallel plate capacitor is connected to a battery,which maintains a constant potential difference. If the plates of the capacitor are moved further apart,the electric field intensity...

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