A simple pendulum of length L is placed between the plates of a parallel plate capacitor having electric field E, as shown in figure. Its bob has mass m and charge q. The time period of the pendulum is given by:

To find the time period of a simple pendulum when subjected to an external electric field, we can use the concept of effective acceleration ($$a_{\text{eff}}$$).

Here is the structured solution breakdown:

1. Understanding the Time Period Formula

The standard time period of a simple pendulum is given by:

$$T = 2\pi \sqrt{\frac{L}{g}}$$

When multiple constant forces act on the bob, acceleration due to gravity $$g$$ is replaced by the effective net acceleration $$a_{\text{eff}}$$ acting on the bob at its equilibrium position:

$$T = 2\pi \sqrt{\frac{L}{a_{\text{eff}}}}$$

2. Analyzing Forces Acting on the Bob

The bob has a mass $$m$$ and a positive charge $$q$$. Two mutually perpendicular constant forces act on it:

• Downward Gravitational Force:
• Horizontal Electrostatic Force: Since the electric field $$E$$ points horizontally from the positive plate to the grounded negative plate, the force on the positive charge is:
• Correct Option: A

$$F_g = mg$$

This produces a downward acceleration $$a_y = g$$.

$$F_e = qE$$

This produces a horizontal acceleration $$a_x = \frac{qE}{m}$$.

3. Calculating Effective Acceleration ($$a_{\text{eff}}$$)

Because the horizontal and vertical accelerations are perpendicular to each other ($$90^\circ$$ angle), we compute their vector sum using the Pythagorean theorem:

$$a_{\text{eff}} = \sqrt{a_y^2 + a_x^2}$$

$$a_{\text{eff}} = \sqrt{g^2 + \left(\frac{qE}{m}\right)^2} = \sqrt{g^2 + \frac{q^2E^2}{m^2}}$$

4. Final Equation for Time Period

Substituting $$a_{\text{eff}}$$ back into the generalized time period formula:

$$T = 2\pi \sqrt{\frac{L}{\sqrt{g^2 + \frac{q^2E^2}{m^2}}}}$$