An elastic dipole is formed by two equal and opposite charge with separation d. The charges have same mass m. It is Kept in a uniform electric field E. If it is slightly rotated from its equilibrium frequency `omega` is

Options

A. `sqrt((qm)/(md))`
B. `sqrt((2qe)/(md))`
C. `2sqrt((qE)/(md))`
D. `sqrt((qe)/(2md))`

Answer

To find the frequency of an elastic dipole formed by two equal and opposite charges in a uniform electric field when slightly rotated from its equilibrium position, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the System**:
We have an elastic dipole consisting of two charges, +q and -q, separated by a distance d. The dipole is placed in a uniform electric field E.

2. **Forces Acting on the Charges**:
Each charge experiences a force due to the electric field:
- The positive charge (+q) experiences a force \( F_+ = qE \) in the direction of the electric field.
- The negative charge (-q) experiences a force \( F_- = -qE \) opposite to the direction of the electric field.

3. **Torque Calculation**:
When the dipole is slightly rotated by an angle \( \theta \), the torque \( \tau \) acting on the dipole can be calculated. The torque is given by:
\[ \tau = r \times F \]
where \( r \) is the distance from the pivot point to the line of action of the force. For our dipole, the distance from the center to each charge is \( \frac{d}{2} \). Thus, the total torque is:
\[ \tau = \left( \frac{d}{2} \right) (qE) \sin(\theta) - \left( \frac{d}{2} \right) (-qE) \sin(\theta) = d \cdot qE \sin(\theta) \]

4. **Moment of Inertia**:
The moment of inertia \( I \) of the dipole about the center is:
\[ I = 2 \cdot m \left( \frac{d}{2} \right)^2 = \frac{md^2}{2} \]

5. **Relating Torque to Angular Acceleration**:
The torque is also related to angular acceleration \( \alpha \) by:
\[ \tau = I \alpha \]
Substituting the expressions for torque and moment of inertia, we get:
\[ d \cdot qE \sin(\theta) = \frac{md^2}{2} \alpha \]

6. **Small Angle Approximation**:
For small angles, we can use the approximation \( \sin(\theta) \approx \theta \):
\[ d \cdot qE \theta = \frac{md^2}{2} \alpha \]

7. **Expressing Angular Acceleration**:
Rearranging gives:
\[ \alpha = \frac{2qE}{md} \theta \]

8. **Relating Angular Acceleration to Frequency**:
The equation \( \alpha = \omega^2 \theta \) relates angular acceleration to angular frequency \( \omega \):
\[ \frac{2qE}{md} \theta = \omega^2 \theta \]

9. **Solving for Angular Frequency**:
Dividing both sides by \( \theta \) (assuming \( \theta \neq 0 \)):
\[ \omega^2 = \frac{2qE}{md} \]
Taking the square root gives:
\[ \omega = \sqrt{\frac{2qE}{md}} \]

### Final Result:
The frequency \( \omega \) of the elastic dipole when slightly rotated from its equilibrium position is:
\[ \omega = \sqrt{\frac{2qE}{md}} \]