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 solve the problem, we need to determine the time period of a pendulum bob that carries a negative charge (-q) while a positive charge (+q) is held at the point of support. The presence of the electric field due to the positive charge will affect the effective gravitational acceleration experienced by the pendulum bob.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Bob:**
The forces acting on the pendulum bob are:
- The gravitational force (weight) acting downward: \( F_g = mg \)
- The electric force due to the positive charge at the support acting upward: \( F_e = \frac{kq^2}{L^2} \) (where \( k \) is Coulomb's constant)

2. **Net Force on the Bob:**
The net force acting on the bob can be expressed as:
\[ F_{net} = mg - F_e \]
Substituting the expression for the electric force:
\[ F_{net} = mg - \frac{kq^2}{L^2} \]

3. **Effective Acceleration:**
The net force can also be expressed in terms of mass and acceleration:
\[ F_{net} = ma \]
Therefore, we can equate the two expressions for net force:
\[ ma = mg - \frac{kq^2}{L^2} \]
Dividing through by \( m \):
\[ a = g - \frac{kq^2}{mL^2} \]
This shows that the effective acceleration \( g_{effective} \) is:
\[ g_{effective} = g - \frac{kq^2}{mL^2} \]

4. **Time Period of the Pendulum:**
The time period \( T \) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
When considering the effective gravitational acceleration, the new time period \( T' \) becomes:
\[ T' = 2\pi \sqrt{\frac{L}{g_{effective}}} \]

5. **Comparing Time Periods:**
Since \( g_{effective} < g \) (because of the upward electric force), it follows that:
\[ T' > T \]
Thus, the new time period \( T' \) is greater than the original time period \( T \).

6. **Conclusion:**
Therefore, the time period of the pendulum bob with the given charges is greater than \( 2\pi \).

### Final Answer:
The time period of the bob is greater than \( 2\pi \).