An electric dipole is placed at the origin `O` and is directed along the `x`-axis. At a point `P`, far away from the dipole, the electric field is parallel to `y`-axis. `OP` makes an angle `theta` with the `x`-axis then

Options

A. `tan theta = sqrt(3)`
B. `tan theta = sqrt(2)`
C. `theta = 45^(@)`
D. `tan theta = (1)/(sqrt(2))`

Answer

To solve the problem step by step, we will analyze the situation involving the electric dipole and the point P.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- An electric dipole is placed at the origin \( O \) and is directed along the \( x \)-axis.
- A point \( P \) is located far away from the dipole, making an angle \( \theta \) with the \( x \)-axis.
- The electric field at point \( P \) is parallel to the \( y \)-axis.

2. **Identifying Angles**:
- Let \( \alpha \) be the angle between the line \( OP \) and the direction of the electric field.
- Since the electric field is parallel to the \( y \)-axis, the angle \( \alpha + \theta = 90^\circ \).

3. **Relating Angles**:
- From the geometry of the situation, we can express \( \alpha \) in terms of \( \theta \):
\[ \alpha = 90^\circ - \theta \]

4. **Using the Tangent Function**:
- The tangent of angle \( \alpha \) can be expressed as:
\[ \tan \alpha = \frac{\tan \theta}{2} \]
- Substituting \( \alpha = 90^\circ - \theta \) into the tangent function gives:
\[ \tan(90^\circ - \theta) = \cot \theta \]
- Therefore, we have:
\[ \cot \theta = \frac{\tan \theta}{2} \]

5. **Rearranging the Equation**:
- The cotangent can be expressed as:
\[ \cot \theta = \frac{1}{\tan \theta} \]
- Thus, we can rewrite the equation:
\[ \frac{1}{\tan \theta} = \frac{\tan \theta}{2} \]

6. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ 2 = \tan^2 \theta \]

7. **Taking the Square Root**:
- Taking the square root of both sides results in:
\[ \tan \theta = \sqrt{2} \]

8. **Conclusion**:
- Therefore, the solution to the problem is:
\[ \tan \theta = 2 \]
- This means that the angle \( \theta \) can be determined from this relationship.