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. $\tan ^{-1}\left[\frac{q}{m g} \times \frac{C_{1}\left(V_{2}-V_{1}\right)}{\left(C_{1}+C_{2}\right)(d-t)}\right]$
• B. $\tan ^{-1}\left[\frac{q}{m g} \times \frac{C_{2}\left(V_{2}-V_{1}\right)}{\left(C_{1}+C_{2}\right)(d-t)}\right]$
• C. $\tan ^{-1}\left[\frac{q}{m g} \times \frac{C_{2}\left(V_{2}+V_{1}\right)}{\left(C_{1}+C_{2}\right)(d-t)}\right]$
• D. $\tan ^{-1}\left[\frac{q}{m g} \times \frac{C_{1}\left(V_{2}+V_{1}\right)}{\left(C_{1}+C_{2}\right)(d-t)}\right]$

Answer 1

Answer: (C)

$Q=\left[\frac{C_{1} C_{2}}{C_{1}+C_{2}}\right]\left[V_{1}+V_{2}\right] $
$E=\frac{Q}{A \in_{0}}=\left[\frac{C_{1} C_{2}}{C_{1}+C_{2}}\right] \frac{\left[V_{1}+V_{2}\right]}{A \in_{0}}$
$C_{1}=\frac{\epsilon_{o} A}{d-t} \Rightarrow E=\frac{C_{2}\left[V_{1}+V_{2}\right]}{\left(C_{1}+C_{2}\right)(d-t)}$
Now $\theta=\tan ^{-1}\left[\frac{q \cdot E}{m g}\right]$
$\theta=\tan ^{-1}\left[\frac{q}{m g} \times \frac{C_{2}\left(V_{2}+V_{1}\right)}{\left(C_{1}+C_{2}\right)(d-t)}\right]$