A parallel plate capacitor is charged by connecting it to a battery. The battery is disconnected and the plates of the capacitor are pulled apart to make the separation between the plates twice. Again the capacitor is connected to the battery (with same polarity) then

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After reconnection of battery potential difference between the plate will immediately becomes half of the initial potential difference. (Just after disconnecting the battery)

Charge from the battery flows into the capacitor after reconnection

Charge from capacitor flows into the battery after reconnection.

The potential difference between the plates increases when the plates are pulled apart.

answer is B, C.

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Detailed Solution

Initially

When the battery is disconnected and the distance of separation is doubled,

Charge on the capacitor will remain same and the capacitance is halved C₁= C/2, V₁ = 2V, Q₁ = Q

When the battery is again connected, After reconnection of battery potential difference will be ‘V’ , resulting the charge on the capacitor become CV/2.

It implies that charge is flown from the capacitor to the battery.

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