1. Find the electric field between the plates

The pendulum bob has charge $+q$, so in equilibrium it is acted upon by:

• weight $mg$ downward,
• electric force $qE$ horizontally,
• tension along the string.

Hence, if the deflection is $\theta$, $\tan\theta = \frac{qE}{mg}.$

So the main task is to determine $E$ between the plates.

2. Equivalent capacitor arrangement

From the given options, the setup clearly corresponds to two capacitors $C_1$ and $C_2$ in series with applied potentials $V_1$ and $V_2$ arranged so that the effective potential difference across the gap containing the pendulum is determined by the combination.

For two capacitors in series, the common charge is $Q = C_{\text{eq}}(V_1+V_2),$ where $C_{\text{eq}} = \frac{C_1C_2}{C_1+C_2}.$ Thus, $Q = \frac{C_1C_2}{C_1+C_2}(V_1+V_2).$

The potential difference across capacitor $C_1$ is $V_{C_1} = \frac{Q}{C_1} = \frac{C_2(V_1+V_2)}{C_1+C_2}.$

The electric field in the plate separation $(d-t)$ is therefore $E = \frac{V_{C_1}}{d-t} = \frac{C_2(V_1+V_2)}{(C_1+C_2)(d-t)}.$

3. Deflection of the pendulum

Now, $\tan\theta = \frac{qE}{mg}$ so $\tan\theta = \frac{q}{mg}\cdot \frac{C_2(V_1+V_2)}{(C_1+C_2)(d-t)}.$

Therefore, $\theta = \tan^{-1}\left[\frac{q}{mg}\times \frac{C_2(V_1+V_2)}{(C_1+C_2)(d-t)}\right].$

4. Match with options

This matches Option C: $\tan ^{ - 1}\left[ \frac{q}{mg} \times \frac{C_2(V_1+V_2)}{(C_1+C_2)(d-t)} \right].$

5. Comparison with stored answer

Stored correct answer: C

My derived answer: C

So they agree.