An electric dipole is placed along the x-axis at the origin `O.A` point `P` is at a distance of `20 cm` from this origin such that `OP` makes an angle `(pi)/(3)` with the `x`-axis. If the electric field at `P` makes an angle `theta` with x-axis, the value of `theta` would be

Options

A. `(pi)/(3)`
B. `(pi)/(3)+tan^(-1)((sqrt(3))/(2))`
C. `(2pi)/(3)`
D. `tan^(-1)((sqrt(3))/(2))`

Answer

To solve the problem, we need to find the angle \( \theta \) that the electric field at point \( P \) makes with the x-axis. Given the information, we can break it down step by step.

### Step 1: Understand the Geometry
- The electric dipole is placed along the x-axis at the origin \( O \).
- Point \( P \) is located at a distance of \( 20 \, \text{cm} \) from \( O \) and makes an angle of \( \frac{\pi}{3} \) (or \( 60^\circ \)) with the x-axis.

### Step 2: Define the Angles
- Let \( \alpha \) be the angle that the electric field \( E \) at point \( P \) makes with the line \( OP \).
- The angle \( \theta \) that we need to find is given by:
\[ \theta = \frac{\pi}{3} + \alpha \]

### Step 3: Calculate the Electric Field Components
- The electric field \( E \) due to a dipole at a point in space can be resolved into two components:
- The component along the direction of \( OP \): \( E \cos \alpha \)
- The component perpendicular to \( OP \): \( E \sin \alpha \)

### Step 4: Use the Dipole Electric Field Formula
- The electric field \( E \) due to a dipole at a distance \( r \) is given by:
\[ E = \frac{1}{4\pi \epsilon_0} \cdot \frac{2p \cos \theta}{r^3} \]
where \( p \) is the dipole moment and \( \theta \) is the angle between the dipole moment and the line joining the dipole to the point.

### Step 5: Set Up the Equations
- For the component along \( OP \):
\[ E \cos \alpha = \frac{2p \cos(\frac{\pi}{3})}{4\pi \epsilon_0 r^3} \]
- For the component perpendicular to \( OP \):
\[ E \sin \alpha = \frac{p \sin(\frac{\pi}{3})}{4\pi \epsilon_0 r^3} \]

### Step 6: Divide the Two Equations
- Dividing the second equation by the first gives:
\[ \frac{E \sin \alpha}{E \cos \alpha} = \frac{\frac{p \sin(\frac{\pi}{3})}{4\pi \epsilon_0 r^3}}{\frac{2p \cos(\frac{\pi}{3})}{4\pi \epsilon_0 r^3}} \]
- This simplifies to:
\[ \tan \alpha = \frac{\sin(\frac{\pi}{3})}{2 \cos(\frac{\pi}{3})} \]

### Step 7: Substitute Known Values
- We know that:
- \( \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \)
- \( \cos(\frac{\pi}{3}) = \frac{1}{2} \)
- Thus:
\[ \tan \alpha = \frac{\frac{\sqrt{3}}{2}}{2 \cdot \frac{1}{2}} = \frac{\sqrt{3}}{2} \]

### Step 8: Find \( \alpha \)
- Therefore, \( \alpha = \tan^{-1}(\frac{\sqrt{3}}{2}) \).

### Step 9: Find \( \theta \)
- Now substituting back into the equation for \( \theta \):
\[ \theta = \frac{\pi}{3} + \tan^{-1}(\frac{\sqrt{3}}{2}) \]

### Final Answer
- The value of \( \theta \) is:
\[ \theta = \frac{\pi}{3} + \tan^{-1}(\frac{\sqrt{3}}{2}) \]