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(D) $1$. When the capacitor is charged and disconnected,the charge $Q$ on the plates remains constant. As the plates are drawn apart,the distance $d$

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all of the above

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(D) $1$. When the capacitor is charged and disconnected,the charge $Q$ on the plates remains constant. As the plates are drawn apart,the distance $d$ increases. Since capacitance $C = \frac{\epsilon_0 A}{d}$,the capacitance $C$ decreases.$2$. The potential difference $V = \frac{Q}{C}$. Since $Q$ is constant and $C$ decreases,$V$ increases. Thus,option $A$ is correct.$3$. The electric field $E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A \epsilon_0}$. Since $Q$ and $A$ are constant,$E$ remains constant while the plates are drawn apart. Thus,option $C$ is correct.$4$. When reconnected to the source of $E.M.F.$ $V_0$,the potential difference across the capacitor must return to $V_0$. Since the potential $V$ had increased to a value greater than $V_0$ (because $V = \frac{Q}{C}$ and $C$ decreased),charge must flow from the capacitor back into the source to restore the potential to $V_0$. Thus,option $B$ is correct.$5$. Since $A, B,$ and $C$ are correct,the correct answer is $D$.

$A$ parallel plate capacitor is charged and then disconnected from the source of steady $E.M.F.$ The plates are then drawn apart farther. Again it is connected to the same source. Then:

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