Question

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. ${\tan }^{-1}\left[\frac{q}{mg}\times \frac{C_1\left(V_2-V_1\right)}{\left(C_1+C_2\right)(d-t)}\right]$
• B. ${\tan }^{-1}\left[\frac{q}{mg}\times \frac{C_2\left(V_2-V_1\right)}{\left(C_1+C_2\right)(d-t)}\right]$
• C. ${\tan }^{-1}\left[\frac{q}{mg}\times \frac{C_2\left(V_2+V_1\right)}{\left(C_1+C_2\right)(d-t)}\right]$
• D. ${\tan }^{-1}\left[\frac{q}{mg}\times \frac{C_1\left(V_2+V_1\right)}{\left(C_1+C_2\right)(d-t)}\right]$

Answer 1

Answer: (C)

$Q=\left[\frac{C_1{C}_2}{C_1+C_2}\right]\left[V_1+V_2\right]$
$E=\frac{Q}{A{\in }_0}=\left[\frac{C_1{C}_2}{C_1+C_2}\right]\frac{\left[V_1+V_2\right]}{A{\in }_0}$
$C_1=\frac{{ϵ}_{o}A}{d-t}\Rightarrow E=\frac{C_2\left[V_1+V_2\right]}{\left(C_1+C_2\right)(d-t)}$
Now $\theta ={\tan }^{-1}\left[\frac{q\cdot E}{mg}\right]$
$\theta ={\tan }^{-1}\left[\frac{q}{mg}\times \frac{C_2\left(V_2+V_1\right)}{\left(C_1+C_2\right)(d-t)}\right]$