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Question

A capacitor of capacitance C is charged to a potential difference V from a cell and then disconnected from it. A charge +Q is now given to its positive plate. The potential difference across the capacitor is now

A
V
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B
V+QC
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C
V+Q2C
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D
VQC
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Solution

The correct option is C V+Q2C
Let the area of the plate be A and distance between the plates be d.

Initial charge on the plates,

Q0=CV

As, C=ε0Ad

Q0=Aε0dV .....(1)

Now, when charge Q is added to the positive plate, electric field between the plates will be,

E=Q+Q02Aε0+Q02Aε0



Now new voltage across the plates will be,

V=Ed=Qd+Q0d2Aε0+Q0d2Aε0

V=Qd2Aε0+Q0d2Aε0+Q0d2Aε0

V=Qd2Aε0+Q0dAε0

From equation (1), V=Q0dAε0 and C=ε0Ad, we have

V=Q2C+V

Hence, option (c) is the +correct answer.
Key concept : Effect on potential difference on PPC (Parallel Plate Capacitor) with battery removed after charging and when some more charge is given to its positive plate after removal of battery.

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