Question

A simple pendulum of mass ${}^{\prime }{m}^{\prime }$, length ${}^{\prime }{l}^{\prime }$ and charge ${}^{\prime }+q^{\prime }$ suspended in the electric field produced by two conducting parallel plates as shown in the figure. The value of deflection of pendulum in equilibrium position will be
$(C_1$ and $C_2$ are the capacitance of capacitors formed by parallel plates, without medium in between and with medium in between, respectively.$)$

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

Answer 1

The correct option is B ${\tan }^{-1}[\frac{q}{mg}\times \frac{C_2(V_1+V_2)}{(C_1+C_2)(d-t)}]$

The equilibrium position of pendulum is shown in the figure.

Let $E$ be electric field in air
$T\sin \theta =qE$
$T\cos \theta =mg$
$\therefore \tan \theta =\frac{qE}{mg}..\left(1\right)$
Now, the given capacitive circuit can be represented as shown below.

Both are in series, so equivalent capacitance can be written as
$C_{eq}=\frac{C_1{C}_2}{C_1+C_2}$
Net charge on equivalent cpacitor is given as
$Q=C_{eq}\Delta V=\left[\frac{C_1{C}_2}{C_1+C_2}\right][V_2+V_1]$
Also, we know that
$E=\frac{Q}{A{ϵ}_0}=\left[\frac{C_1{C}_2}{C_1+C_2}\right]\left[\frac{V_2+V_1}{A{ϵ}_0}\right]$
Also,
$C_1=\frac{{ϵ}_{0}A}{d-t}\Rightarrow E=\frac{C_2[V_2+V_1]}{(C_1+C_2)(d-t)}$
From eq $\left(1\right)$, we have
$\theta ={\tan }^{-1}\left[\frac{q.E}{mg}\right]$
$\Rightarrow \theta ={\tan }^{-1}[\frac{q}{mg}\times \frac{C_2(V_1+V_2)}{(C_1+C_2)(d-t)}]$
Hence, option (c) is correct.