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 can follow these steps:

### Step 1: Understand the Electric Field of a Dipole
The electric field \( E \) due to an electric dipole at a point along the dipole axis is given by:
\[ E_{\text{dipole}} = \frac{2Kp \cos \theta}{r^3} \]
where \( K \) is a constant, \( p \) is the dipole moment, \( \theta \) is the angle between the dipole moment and the line connecting the dipole to the point, and \( r \) is the distance from the dipole.

### Step 2: Electric Field Components
At an angle \( \theta \) to the dipole axis, we can resolve the electric field into two components:
- The component along the dipole axis:
\[ E_{\text{radial}} = \frac{2Kp \cos \theta}{r^3} \]
- The component perpendicular to the dipole axis:
\[ E_{\text{perpendicular}} = \frac{Kp \sin \theta}{r^3} \]

### Step 3: Condition for Perpendicularity
For the net electric field to be perpendicular to the dipole axis, the tangent of the angle \( \alpha \) formed by the resultant electric field with respect to the dipole axis must satisfy:
\[ \tan \alpha = \frac{E_{\text{perpendicular}}}{E_{\text{radial}}} \]

### Step 4: Substitute the Components
Substituting the expressions for \( E_{\text{perpendicular}} \) and \( E_{\text{radial}} \):
\[ \tan \alpha = \frac{\frac{Kp \sin \theta}{r^3}}{\frac{2Kp \cos \theta}{r^3}} = \frac{\sin \theta}{2 \cos \theta} \]
Thus, we have:
\[ \tan \alpha = \frac{1}{2} \tan \theta \]

### Step 5: Relate Angles
Since the resultant electric field must be perpendicular to the dipole axis, we have:
\[ \alpha + \theta = 90^\circ \]
This implies:
\[ \alpha = 90^\circ - \theta \]

### Step 6: Substitute for \( \alpha \)
Now substituting \( \alpha \) into the tangent equation:
\[ \tan(90^\circ - \theta) = \cot \theta \]
Thus:
\[ \cot \theta = \frac{1}{2} \tan \theta \]

### Step 7: Solve for \( \tan \theta \)
From the cotangent identity, we have:
\[ \cot \theta = \frac{1}{\tan \theta} \]
Substituting gives:
\[ \frac{1}{\tan \theta} = \frac{1}{2} \tan \theta \]
Multiplying both sides by \( 2 \tan \theta \) yields:
\[ 2 = \tan^2 \theta \]
Thus:
\[ \tan \theta = \sqrt{2} \]

### Step 8: Find \( \theta \)
Finally, we find \( \theta \):
\[ \theta = \tan^{-1}(\sqrt{2}) \]

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