The force of attraction between the plates of a charged condenser is :

q^{2} (2 ε_{0} A)

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q^{2} (2 ε_{0} A^{2})

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q^{2} / (2 ε_{0} A)

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none of these

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Solution

The correct option is C $q^{2} / (2 ε_{0} A)$
The electric field due to the charge plate Q is $E = \frac{Q}{2 ϵ_{0} A}$
The force between the charged plates is $F = Q E = \frac{Q^{2}}{2 ϵ_{0} A}$

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The force of attraction between the plates of a charged condenser is :

The correct option is C q2/(2ε0A)The electric field due to the charge plate Q is E=Q2ϵ0AThe force between the charged plates is F=QE=Q22ϵ0A

The correct option is C $q^{2} / (2 ε_{0} A)$
The electric field due to the charge plate Q is $E = \frac{Q}{2 ϵ_{0} A}$
The force between the charged plates is $F = Q E = \frac{Q^{2}}{2 ϵ_{0} A}$