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. (a)`tan theta = sqrt3`
B. (b)`tan theta = sqrt2`
C. (c)`theta= 45^@`
D. (d)None the these

Answer

To solve the problem step by step, we will analyze the given information about the electric dipole and the point P.

### Step 1: Understand the Configuration
We have an electric dipole placed at the origin O and directed along the x-axis. The point P is located far away from the dipole, and the electric field at point P is parallel to the y-axis. The line OP makes an angle θ with the x-axis.

**Hint**: Visualize the setup with a diagram showing the dipole at the origin and point P at an angle θ from the x-axis.

### Step 2: Analyze the Electric Field Direction
Since the electric field at point P is parallel to the y-axis, this means that the x-component of the electric field must be zero at point P.

**Hint**: Recall that the electric field due to a dipole decreases with distance and has a specific directional dependence based on the angle from the dipole axis.

### Step 3: Relationship Between Angles
The angle θ is the angle between the line OP and the x-axis. Since the electric field is parallel to the y-axis, we can establish a relationship between θ and the angle α (the angle between the dipole moment and the line OP).

**Hint**: Use the property that the sum of angles in a right triangle or the relationship of complementary angles can help establish the relationship between θ and α.

### Step 4: Establish the Tangent Relationship
From the geometry, we can say that:
- The angle α is complementary to θ, which means α = 90° - θ.
- The tangent of angle α can be expressed in terms of θ.

Using the tangent function:
\[ \tan(α) = \frac{1}{2} \tan(θ) \]

**Hint**: Remember the definition of tangent in a right triangle and how it relates to the opposite and adjacent sides.

### Step 5: Substitute and Solve
Substituting α = 90° - θ into the tangent relationship gives us:
\[ \tan(90° - θ) = \cot(θ) = \frac{1}{2} \tan(θ) \]

This leads to:
\[ \cot(θ) = \frac{1}{2} \tan(θ) \]

Now, using the identity that \( \cot(θ) = \frac{1}{\tan(θ)} \), we can rewrite the equation:
\[ \frac{1}{\tan(θ)} = \frac{1}{2} \tan(θ) \]

Multiplying both sides by \( \tan(θ) \):
\[ 1 = \frac{1}{2} \tan^2(θ) \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ \tan^2(θ) = 2 \]

Taking the square root of both sides results in:
\[ \tan(θ) = \sqrt{2} \]

### Conclusion
Thus, the value of \( \tan(θ) \) is \( \sqrt{2} \).

**Final Answer**: \( \tan(θ) = \sqrt{2} \)