A parallel plate air capacitor has capacity $^{'} C^{'}$ , distance of separation between plate is $^{'} d^{'}$ and potential difference $^{'} V^{'}$ is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is:

\frac{C^{2} V^{2}}{2 d^{2}}

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\frac{C^{2} V^{2}}{2 d}

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\frac{C V^{2}}{2 d}

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\frac{C^{2} V}{d}

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Solution

The correct option is C $\frac{C V^{2}}{2 d}$
Force of attraction between the plates is $F = \frac{Q^{2}}{2 A ϵ_{0}}$

or $F = \frac{C^{2} V^{2}}{2 A ϵ_{0}}$ where $Q = C V$

also $C = \frac{A ϵ_{0}}{d} \Rightarrow A ϵ_{0} = C d$

now, $F = \frac{C^{2} V^{2}}{2 C d} = \frac{C V^{2}}{2 d}$

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A parallel plate air capacitor has capacity ′C′, distance of separation between plate is ′d′ and potential difference ′V′ is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is:

The correct option is C CV22dForce of attraction between the plates is F=Q22Aϵ0or F=C2V22Aϵ0 where Q=CV also C=Aϵ0d⇒Aϵ0=Cdnow, F=C2V22Cd=CV22d

A parallel plate air capacitor has capacity $^{'} C^{'}$ , distance of separation between plate is $^{'} d^{'}$ and potential difference $^{'} V^{'}$ is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is:

The correct option is C $\frac{C V^{2}}{2 d}$
Force of attraction between the plates is $F = \frac{Q^{2}}{2 A ϵ_{0}}$

or $F = \frac{C^{2} V^{2}}{2 A ϵ_{0}}$ where $Q = C V$

also $C = \frac{A ϵ_{0}}{d} \Rightarrow A ϵ_{0} = C d$

now, $F = \frac{C^{2} V^{2}}{2 C d} = \frac{C V^{2}}{2 d}$