Force between the plates of a capacitor is

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Q.

Force between the plates of a capacitor is

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a

$\frac{1}{2} \in_{0} E^{2}$

b

$\frac{Q^{2}}{ϵ_{0} A}$

c

$\frac{1}{2} ϵ_{0} E^{2} A$

d

$\frac{Q}{ϵ_{0} A}$

answer is C.

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Detailed Solution

$U = \frac{1}{2} ϵ_{0} E^{2} (Ax) \Rightarrow F = \frac{- dU}{dx} = - \frac{1}{2} ϵ_{0} E^{2} A ∴ Attractive force F = \frac{1}{2} ϵ_{0} E^{2} A$

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