Two equal charges `q` of opposite sign separated by a distance 2a constitute an electric dipole of dipole moment `p`. If P is a point at a distance r from the centre of the dipole and the line joining the centre of the dipole to this point makes an angle `theta` with the axis of the dipole, then then potential at P is given by `(r gt gt 2a)` where (p = 2 qa)

Options

A. `V=(p cos theta)/(4 piepsilon_(0)r^(2))`
B. `V=(p cos theta)/(4 piepsilon_(0)r)`
C. `V=(p sin theta)/(2 piepsilon_(0)r)`
D. `V=(p cos theta)/(2 piepsilon_(0)r^(2))`

Answer

To solve the problem, we need to find the electric potential \( V \) at a point \( P \) due to an electric dipole consisting of two equal and opposite charges \( +q \) and \( -q \) separated by a distance \( 2a \). The dipole moment \( p \) is given by \( p = 2qa \).

### Step-by-Step Solution:

1. **Understanding the Dipole Configuration**:
- We have two charges: \( +q \) and \( -q \) separated by a distance \( 2a \).
- The dipole moment \( p \) is defined as \( p = q \cdot d \), where \( d \) is the separation between the charges. Here, \( d = 2a \), so \( p = 2qa \).

2. **Positioning the Dipole**:
- Place the dipole along the x-axis with the positive charge at \( (-a, 0) \) and the negative charge at \( (a, 0) \).
- The center of the dipole is at the origin \( (0, 0) \).

3. **Identifying the Point \( P \)**:
- The point \( P \) is at a distance \( r \) from the center of the dipole.
- The line joining the center of the dipole to point \( P \) makes an angle \( \theta \) with the dipole axis (x-axis).

4. **Calculating the Electric Potential**:
- The electric potential \( V \) due to a dipole at a point in space is given by the formula:
\[ V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{p \cos \theta}{r^2} \]
- Here, \( \epsilon_0 \) is the permittivity of free space.

5. **Substituting the Dipole Moment**:
- Substitute \( p = 2qa \) into the potential formula:
\[ V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2qa \cos \theta}{r^2} \]

6. **Final Expression**:
- Thus, the electric potential \( V \) at point \( P \) is:
\[ V = \frac{2qa \cos \theta}{4 \pi \epsilon_0 r^2} \]

### Conclusion:
The potential at point \( P \) due to the dipole is given by:
\[ V = \frac{p \cos \theta}{4 \pi \epsilon_0 r^2} \]
where \( p = 2qa \).