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 -
JEE MAIN 2021, Diffcult
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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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