An electric point dipole is placed at the origin O with its dipolemoment along the X-axis . A point A is at a distance r from the origin such that OA makes an angle `pi//3` with the X -axis if the electric field `vec( E)` due to the dipole at A makes an angle `theta` with the positive X-axis, the value of `theta` is

Options

A. `pi//3`
B. `(pi//3)+tan^(-1)(sqrt(3)//2)`
C. `(pi//3) -tan^(-1) (sqrt(3)//2)`
D. `tan^(-1) (sqrt(3)//2)`

Answer

To solve the problem, we need to find the angle \( \theta \) that the electric field \( \vec{E} \) due to a dipole makes with the positive X-axis when the dipole is placed at the origin and the point A is at a distance \( r \) from the origin, making an angle \( \frac{\pi}{3} \) with the X-axis.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- The dipole is placed at the origin \( O \) with its dipole moment \( \vec{p} \) along the X-axis.
- Point \( A \) is at a distance \( r \) from the origin and makes an angle \( \frac{\pi}{3} \) with the X-axis.

2. **Components of the Electric Field**:
- The electric field \( \vec{E} \) due to a dipole at a point in space can be resolved into two components:
- Radial component \( E_r \) along the line OA.
- Perpendicular component \( E_\theta \) perpendicular to OA.

3. **Calculating the Radial Component \( E_r \)**:
- The formula for the radial component of the electric field due to a dipole is given by:
\[ E_r = \frac{2kp \cos \theta}{r^3} \]
- Here, \( \theta \) is the angle between the dipole moment and the line connecting the dipole to point A. Since point A makes an angle \( \frac{\pi}{3} \) with the X-axis, we have:
\[ E_r = \frac{2kp \cos \left(\frac{\pi}{3}\right)}{r^3} \]
- Knowing that \( \cos \left(\frac{\pi}{3}\right) = \frac{1}{2} \):
\[ E_r = \frac{2kp \cdot \frac{1}{2}}{r^3} = \frac{kp}{r^3} \]

4. **Calculating the Perpendicular Component \( E_\theta \)**:
- The formula for the perpendicular component of the electric field is given by:
\[ E_\theta = \frac{kp \sin \theta}{r^3} \]
- Since \( \theta = \frac{\pi}{3} \):
\[ E_\theta = \frac{kp \sin \left(\frac{\pi}{3}\right)}{r^3} \]
- Knowing that \( \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \):
\[ E_\theta = \frac{kp \cdot \frac{\sqrt{3}}{2}}{r^3} = \frac{kp \sqrt{3}}{2r^3} \]

5. **Finding the Angle \( \theta \)**:
- The angle \( \theta \) that the electric field \( \vec{E} \) makes with the positive X-axis can be found using the tangent of the angle:
\[ \tan \theta = \frac{E_\theta}{E_r} \]
- Substituting the values of \( E_\theta \) and \( E_r \):
\[ \tan \theta = \frac{\frac{kp \sqrt{3}}{2r^3}}{\frac{kp}{r^3}} = \frac{\sqrt{3}}{2} \]
- Therefore, we have:
\[ \theta = \tan^{-1} \left(\frac{\sqrt{3}}{2}\right) \]

6. **Final Angle Calculation**:
- The angle \( \theta \) can be calculated as:
\[ \theta = \frac{\pi}{3} + \tan^{-1} \left(\frac{\sqrt{3}}{2}\right) \]
- Since \( \tan^{-1} \left(\frac{\sqrt{3}}{2}\right) \) corresponds to an angle that is less than \( \frac{\pi}{3} \), the final value of \( \theta \) can be approximated or calculated based on the context.

### Final Answer:
The value of \( \theta \) is:
\[ \theta = \frac{\pi}{3} + \tan^{-1} \left(\frac{\sqrt{3}}{2}\right) \]