A body of mass 10 gm and having charge 2C is attached to a spring which is suspended from the ceiling. It vibrates with a time period 1 sec. If an electric field of intensity 100 N/C is now applied in the downward direction, find the time period.

Answer

To solve the problem, we need to determine the time period of a mass-spring system when an electric field is applied. The key points to consider are the mass of the body, the charge, the initial time period, and the effect of the electric field.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the body, \( m = 10 \, \text{g} = 10 \times 10^{-3} \, \text{kg} = 0.01 \, \text{kg} \)
- Charge of the body, \( q = 2 \, \text{C} \)
- Initial time period, \( T_i = 1 \, \text{s} \)
- Electric field intensity, \( E = 100 \, \text{N/C} \)

2. **Understand the Time Period Formula:**
The time period \( T \) of a mass-spring system is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
where \( k \) is the spring constant.

3. **Consider the Effect of the Electric Field:**
When an electric field is applied, it exerts a force on the charged body given by:
\[ F_e = qE \]
This force acts in the downward direction, adding to the gravitational force \( F_g = mg \).

4. **Calculate the Forces Acting on the Mass:**
The total effective force acting on the mass when the electric field is applied is:
\[ F_{total} = mg + qE \]
However, this does not affect the time period directly since the time period depends only on the mass and the spring constant.

5. **Determine the Time Period After Applying the Electric Field:**
The time period is independent of the forces acting on the mass (as long as the spring remains linear). Therefore, the time period remains the same:
\[ T_f = T_i = 1 \, \text{s} \]

### Final Answer:
The time period of the mass-spring system after applying the electric field remains \( T_f = 1 \, \text{s} \).

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