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Q.

The force between the plates of a parallel plate capacitor of capacitance C and distance of separation of the plates d with a potential difference V between the plates, is  CV2yD,what  is the value  of y?

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answer is 2.

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Detailed Solution

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Electric field at the location of one of the plates (which is due to the other plate) is E=σ2ε0

Force on this plate, F=QE=Qσ2ε0=Q22Aε0

But C=Aε0d and Q=CV

F=C2V22Cd=CV22d

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