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Question

The force with which the plates of a parallel plate capacitor, having charge Q and area of each plate A, attract each other is

A
zero
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B
Q2ε0A
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C
Q22ε0A
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D
Q22ε0A
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Solution

The correct option is D Q22ε0A
To find the force of attraction between the charged capacitor plates we will use a method called the method of virtual displacement. We simply equate the work W required to make a small change d in the plate separation d to the resulting change U in the stored energy, i.e.,
W=U (i)
If F is the magnitude of the force between the plates, then the work W done to increase the plate separation by d is given by
W=Fd (ii)
Now we know that the energy U of a parallel-plate capacitor of plate area A and capacitance C is
U=Q22C=Q2d2ε0A
Where Q is the charge on the capacitor plates. The increase U in U due to an increase d in d is, therefore, given by
U=Q2d2ε0A (iii)
Equating (ii) and (iii) we get
F=Q22ε0A, which is choice (d)

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