The electric field at a point due to an electric dipole, on an axis inclined at an angle `theta ( lt 90^(@))` to the dipole axis, is perpendicular to the dipole axis, if the angle `theta` is

Options

A. `tan^(-1) (2)`
B. `tan^(-1) ((1)/(2))`
C. `tan^(-1) (sqrt2)`
D. `tan^(-1) ((1)/(sqrt2))`

Answer

To solve the problem of finding the angle \( \theta \) at which the electric field due to an electric dipole is perpendicular to the dipole axis, we will follow these steps:

### Step 1: Understand the Configuration
We have an electric dipole, which consists of two equal and opposite charges separated by a distance. The dipole moment \( \mathbf{P} \) is defined as \( \mathbf{P} = q \cdot \mathbf{d} \), where \( q \) is the charge and \( \mathbf{d} \) is the separation vector pointing from the negative to the positive charge.

### Step 2: Identify the Electric Field Components
The electric field \( \mathbf{E} \) at a point due to a dipole can be resolved into two components:
- The axial component \( E_{\text{axial}} \) along the dipole axis.
- The equatorial component \( E_{\text{equatorial}} \) perpendicular to the dipole axis.

The expressions for these components are:
- \( E_{\text{axial}} = \frac{2kP}{r^3} \cos \theta \)
- \( E_{\text{equatorial}} = \frac{kP}{r^3} \sin \theta \)

Where \( k \) is a constant, \( P \) is the dipole moment, \( r \) is the distance from the dipole, and \( \theta \) is the angle between the dipole axis and the line joining the dipole to the point where the field is being calculated.

### Step 3: Set Up the Condition for Perpendicularity
For the electric field to be perpendicular to the dipole axis, we need to find the angle \( \theta \) such that:
\[ \tan \alpha = \frac{E_{\text{equatorial}}}{E_{\text{axial}}} \]
Where \( \alpha \) is the angle between the resultant electric field and the dipole axis.

### Step 4: Substitute the Electric Field Expressions
Substituting the expressions for the electric fields into the equation gives:
\[ \tan \alpha = \frac{\frac{kP}{r^3} \sin \theta}{\frac{2kP}{r^3} \cos \theta} \]
This simplifies to:
\[ \tan \alpha = \frac{\sin \theta}{2 \cos \theta} = \frac{1}{2} \tan \theta \]

### Step 5: Relate \( \theta \) and \( \alpha \)
From the geometry of the situation, we have:
\[ \theta + \alpha = 90^\circ \implies \alpha = 90^\circ - \theta \]
Taking the tangent of both sides gives:
\[ \tan \theta = \cot \alpha = \frac{1}{\tan \alpha} \]

### Step 6: Substitute for \( \tan \alpha \)
Substituting \( \tan \alpha \) from the previous step into this equation gives:
\[ \tan \theta = \frac{1}{\frac{1}{2} \tan \theta} \]
Cross-multiplying leads to:
\[ \tan^2 \theta = 2 \]
Thus:
\[ \tan \theta = \sqrt{2} \]

### Step 7: Find the Angle \( \theta \)
Finally, we find:
\[ \theta = \tan^{-1}(\sqrt{2}) \]

### Conclusion
The angle \( \theta \) at which the electric field due to an electric dipole is perpendicular to the dipole axis is:
\[ \theta = \tan^{-1}(\sqrt{2}) \]