The locus of the points (in the `xy`-plane) where the electric field due to a dipole (dipole axis is along `x`-axis and its equatorial is along `y`-axis) is perpendicular to its axis, is

Options

A. straight line perpendicular to the axis
B. circle
C. parabola
D. straigth line having inclination `theta=tan^(-1)sqrt(2)` with the axis

Explanation

To solve the problem of finding the locus of points in the `xy`-plane where the electric field due to a dipole is perpendicular to its axis, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Dipole Configuration**:
- A dipole consists of two equal and opposite charges separated by a small distance. In this case, the dipole axis is aligned along the x-axis, and its equatorial plane is along the y-axis.

2. **Defining the Electric Field**:
- The electric field \( \vec{E} \) due to a dipole at a point in space can be derived from the dipole moment \( \vec{p} \). The dipole moment is defined as \( \vec{p} = q \cdot \vec{d} \), where \( q \) is the charge and \( \vec{d} \) is the vector pointing from the negative charge to the positive charge.

3. **Finding the Condition for Perpendicularity**:
- We need to find the points where the electric field \( \vec{E} \) is perpendicular to the dipole axis. The dipole axis is along the x-axis, so we want \( \vec{E} \) to be in the y-z plane.

4. **Using Geometry**:
- Let’s denote a point \( P(x, y) \) in the xy-plane. The angle \( \theta \) is the angle between the dipole moment direction (x-axis) and the line connecting the origin to point \( P \). The angle \( \alpha \) is the angle between the electric field direction and the x-axis.

5. **Applying Trigonometric Relationships**:
- From the geometry, we can establish that:
\[ \tan \theta = 2 \tan \alpha \]
- Since \( \alpha + \theta = 90^\circ \), we can express \( \alpha \) as:
\[ \alpha = 90^\circ - \theta \]

6. **Substituting into the Equation**:
- Substitute \( \alpha \) into the first equation:
\[ \tan \theta = 2 \tan(90^\circ - \theta) \]
- Using the identity \( \tan(90^\circ - \theta) = \cot \theta \), we get:
\[ \tan \theta = 2 \cot \theta \]
- This can be rewritten as:
\[ \tan^2 \theta = 2 \]

7. **Solving for \( \theta \)**:
- Taking the square root gives:
\[ \tan \theta = \sqrt{2} \]
- Thus, \( \theta = \tan^{-1}(\sqrt{2}) \).

8. **Finding the Locus**:
- The angle \( \theta \) represents a constant angle in the xy-plane. The locus of points where the electric field is perpendicular to the dipole axis will be a straight line making an angle \( \theta = \tan^{-1}(\sqrt{2}) \) with the x-axis.

### Conclusion:
The locus of points in the xy-plane where the electric field due to the dipole is perpendicular to its axis is a straight line inclined at an angle \( \tan^{-1}(\sqrt{2}) \) to the x-axis.