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(D) Since the capacitor remains connected to the battery,the potential difference $V$ across the plates remains constant.The capacitance of a parallel

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All of these

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(D) Since the capacitor remains connected to the battery,the potential difference $V$ across the plates remains constant.The capacitance of a parallel plate capacitor is given by $C = \frac{A \varepsilon_0}{d}$. As the distance $d$ between the plates increases,the capacitance $C$ decreases.The charge on the capacitor is given by $q = CV$. Since $C$ decreases and $V$ is constant,the charge $q$ decreases.The energy stored in the capacitor is given by $U = \frac{1}{2}CV^2$. Since $C$ decreases and $V$ is constant,the energy $U$ decreases.Therefore,charge,capacitance,and energy stored all decrease. The correct option is $D$.

$A$ parallel plate capacitor,after being charged,is kept connected to a battery,and the plates are pulled apart with the help of insulating handles. Which of the following quantities will decrease?

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