Question 103 of 311

A simple pendulum of mass ' $m$ ', length ' $l$ ' 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

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

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Answer: (C) ${\tan }^{-1}\left[\dfrac{q}{mg}\times \dfrac{{C}_{2}\left({V}_{1}+{V}_{2}\right)}{\left({C}_{1}+{C}_{2}\right)(d-t)}\right]$

Let $E$ be electric field in air
$Tsin\theta =qE$
$Tcos\theta =mg$
$tan\theta =\dfrac{qE}{mg}$

$Q=\left[\dfrac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}\right]\left[{V}_{1}+{V}_{2}\right]$
$E=\dfrac{Q}{A{ϵ}_{o}}=\left[\dfrac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}\right]\dfrac{\left[{V}_{1}+{V}_{2}\right]}{A{ϵ}_{o}}$
${C}_{1}=\dfrac{{ϵ}_{0}A}{d-t}\Rightarrow E=\dfrac{{C}_{2}\left[{V}_{1}+{V}_{2}\right]}{\left({C}_{1}+{C}_{2}\right)(d-t)}$
Now $\theta ={\tan }^{-1}\left[\dfrac{q\cdot E}{mg}\right]$
$\theta ={\tan }^{-1}\left[\dfrac{q}{mg}\times \dfrac{{C}_{2}\left({V}_{1}+{V}_{2}\right)}{\left({C}_{1}+{C}_{2}\right)(d-t)}\right]$