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

Hint:

For capacitors in series, the total potential difference divides in the inverse ratio of their capacitances. The electric field inside a parallel plate capacitor is uniform and given by $E = V/d$.

Answer 1

The Correct Option is A

The system consists of two capacitors in series. One with a dielectric medium of thickness $t$ ($C_1$) and another with air of thickness $(d-t)$ ($C_2$).
The total potential difference across the plates is $V = V_2 - (-V_1) = V_1 + V_2$.
Since the capacitors are in series, the potential difference across the air capacitor ($V_{air}$) where the pendulum is located is given by the potential divider rule.
$V_{air} = V \times \frac{C_1}{C_1 + C_2} = (V_1 + V_2) \frac{C_1}{C_1 + C_2}$.
The electric field in the air gap is $E = \frac{V_{air}}{\text{thickness}} = \frac{V_{air}}{d-t}$.
Substituting the expression for $V_{air}$:
$E = \frac{(V_1 + V_2)C_1}{(C_1 + C_2)(d-t)}$.
The electric force ($F_e$) on the charge $+q$ is $F_e = qE$.
$F_e = \frac{qC_1(V_1 + V_2)}{(C_1 + C_2)(d-t)}$.
In equilibrium, let $\theta$ be the angle of deflection. The forces on the pendulum bob are tension (T), gravitational force (mg), and electric force ($F_e$).
Balancing forces in the horizontal and vertical directions:
$T \sin\theta = F_e$
$T \cos\theta = mg$
Dividing the two equations gives $\tan\theta = \frac{F_e}{mg}$.
$\tan\theta = \frac{1}{mg} \left[ \frac{qC_1(V_1 + V_2)}{(C_1 + C_2)(d-t)} \right]$.
Therefore, the deflection angle is $\theta = \tan^{-1}\left[ \frac{q}{mg} \times \frac{C_1(V_1 + V_2)}{(C_1 + C_2)(d-t)} \right]$.