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

We need to determine the angle of deflection ($$\theta$$) for a simple pendulum of mass $$m$$, length $$l$$, and charge $$+q$$ suspended in equilibrium inside a complex parallel-plate system.

1. Understand the Equilibrium Condition

When the pendulum is in its deflected equilibrium position, it experiences three primary forces:

• The downward gravitational force: $$F_g = mg$$
• The horizontal electric force: $$F_e = qE$$ (where $$E$$ is the effective net electric field acting on the charge)
• The tension in the string: $$T$$

Balancing the horizontal and vertical components of these forces gives:

$$T \sin\theta = qE$$

$$T \cos\theta = mg$$

Dividing the two equations yields the standard relationship for the deflection angle:

$$\tan\theta = \frac{qE}{mg} \implies \theta = \tan^{-1}\left[ \frac{q}{mg} \times E \right]$$

2. Determine the Effective Electric Field ($$E$$)

From the problem geometry , the system consists of a dielectric or plate combination setup between conducting parallel boundaries split across thicknesses $$t$$ and $$d-t$$. This arrangement acts equivalent to two capacitors, $$C_1$$ and $$C_2$$, connected in series across a total potential difference.

The total net potential difference driving the system is the sum of the magnitudes of the boundary voltages:

$$V_{\text{total}} = V_1 + V_2$$

In a series combination of capacitors, the potential difference across a single component (let's find the voltage $$V$$ across the active region hosting the pendulum) is inversely proportional to its capacitance:

$$V = \left( \frac{C_2}{C_1 + C_2} \right) V_{\text{total}} = \frac{C_2(V_1 + V_2)}{C_1 + C_2}$$

The uniform electric field $$E$$ in a region of space with a remaining gap thickness of $$(d - t)$$ is defined as:

$$E = \frac{V}{\text{thickness}} = \frac{C_2(V_1 + V_2)}{(C_1 + C_2)(d - t)}$$

3. Substitute the Field into the Deflection Angle Formula

Now, substitute this expression for the electric field $$E$$ back into our equilibrium equation:

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

Conclusion

The exact value of the deflection of the pendulum at its equilibrium position matches Option C:

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