A simple pendulum of mass \('m',\) length \('I'\) and charge \('+q'\) suspended in the electric field produced by two conducting parallel plates as shown. The value of deflection of pendulum in equilibrium position will be -

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Correct Answer

(D) \(\tan ^{-1}\left[\frac{{q}}{{mg}} \times \frac{{C}_{2}\left({V}_{1}+{V}_{2}\right)}{\left({C}_{1}+{C}_{2}\right)({d}-{t})}\right]\)

Detailed explanation

Let \(E\) be electric field in air

\(T \sin \theta=q E\)

\(T \cos \theta=m g\)

\(\tan \theta=\frac{q {E}}{m g}\)

\({Q}=\left[\frac{{C}_{1} {C}_{2}}{{C}_{1}+{C}_{2}}\right]\left[{V}_{1}+{V}_{2}\right]\)

\({E}=\frac{{Q}}{{A} \in_{0}}=\left[\frac{{C}_{1} {C}_{2}}{{C}_{1}+{C}_{2}}\right] \frac{\left[{V}_{1}+{V}_{2}\right]}{{A} \in_{0}}\)

\({C}_{1}=\frac{\epsilon_{0} {A}}{{d}-{t}} \Rightarrow {E}=\frac{{C}_{2}\left[{V}_{1}+{V}_{2}\right]}{\left({C}_{1}+{C}_{2}\right)({d}-{t})}\)

Now \(\theta=\tan ^{-1}\left[\frac{{q} \cdot {E}}{{mg}}\right]\)

\(\theta=\tan ^{-1}\left[\frac{{q}}{{mg}} \times \frac{{C}_{2}\left({V}_{1}+{V}_{2}\right)}{\left({C}_{1}+{C}_{2}\right)({d}-{t})}\right]\)

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