A pendulum of length L carries a negative charge - q on the bob. A positive charge +q is held at the point of support . Then, the time period of the bob is

Options

A. Greater than `2pisqrt(L/g)`
B. Less than `2pisqrt(L/g)`
C. equal to `2pisqrt(L/g)`
D. Equal to `2pisqrt((2L)/g)`

Answer

To find the time period of a pendulum bob carrying a negative charge -q, with a positive charge +q held at the point of support, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Forces Acting on the Bob**:
- The bob of the pendulum has a negative charge (-q) and is influenced by the gravitational force acting downwards (mg) and the electrostatic force due to the positive charge (+q) at the point of support.
- The electrostatic force (F) between the charges can be calculated using Coulomb's law:
\[ F = k \frac{|q \cdot q|}{r^2} = k \frac{q^2}{L^2} \]
where \( k \) is Coulomb's constant, \( q \) is the magnitude of the charge, and \( L \) is the length of the pendulum.

2. **Torque Calculation**:
- When the bob is displaced by a small angle \( \theta \), the torque (\( \tau \)) due to the weight of the bob is given by:
\[ \tau = -mgL \sin(\theta) \approx -mgL\theta \quad (\text{for small } \theta) \]
- The torque due to the electrostatic force also acts to restore the bob to its equilibrium position. The torque due to the electrostatic force is:
\[ \tau_{elec} = F \cdot L \sin(\theta) \approx k \frac{q^2}{L^2} \cdot L \theta = k \frac{q^2}{L} \theta \]

3. **Net Torque**:
- The net torque acting on the bob when displaced is:
\[ \tau_{net} = -mgL\theta + k \frac{q^2}{L} \theta \]
- This can be simplified to:
\[ \tau_{net} = \left(-mgL + k \frac{q^2}{L}\right) \theta \]

4. **Equation of Motion**:
- The equation of motion for the pendulum can be expressed as:
\[ I \alpha = \tau_{net} \]
- The moment of inertia \( I \) of the pendulum bob is \( mL^2 \), and the angular acceleration \( \alpha \) can be expressed as \( \alpha = \frac{d^2\theta}{dt^2} \). Thus:
\[ mL^2 \frac{d^2\theta}{dt^2} = \left(-mgL + k \frac{q^2}{L}\right) \theta \]

5. **Finding the Time Period**:
- Rearranging the equation gives:
\[ \frac{d^2\theta}{dt^2} + \left(\frac{g}{L} - \frac{k q^2}{mL^3}\right) \theta = 0 \]
- This is a simple harmonic motion equation of the form \( \frac{d^2\theta}{dt^2} + \omega^2 \theta = 0 \), where:
\[ \omega^2 = \frac{g}{L} - \frac{k q^2}{mL^3} \]
- The time period \( T \) is given by:
\[ T = 2\pi \sqrt{\frac{L}{g - \frac{k q^2}{mL^2}}} \]

6. **Conclusion**:
- However, since the attractive electrostatic force does not change the torque due to the weight of the pendulum significantly, the time period remains effectively unchanged from that of a simple pendulum:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]