A and B are two points on the axis and the perpendicular bisector of an electric dipole. A and B are far away from the dipole and at equal distances from it. The potentials at A and B are `V_(A) and V_(B)` respectively. Then

Options

A. `V_(A) = V_(B) = 0`
B. `V_(A) = 2V_(B)`
C. `V_(A) ne 0, V_(B) = 0`
D. `V_(A) ne 0, V_(B) ne 0`

Answer

To solve the problem, we need to analyze the electric potential at points A and B, which are located on the axis and the perpendicular bisector of an electric dipole.

### Step-by-Step Solution:

1. **Understanding the Electric Dipole:**
An electric dipole consists of two equal and opposite charges, +q and -q, separated by a distance of 2L. The dipole moment \( P \) is defined as:
\[ P = q \cdot 2L \]

2. **Identifying Points A and B:**
Points A and B are located at equal distances from the dipole along the perpendicular bisector. Let's denote the distance from the center of the dipole to points A and B as \( R \).

3. **Electric Potential Due to a Dipole:**
The electric potential \( V \) at a point due to an electric dipole is given by the formula:
\[ V = \frac{1}{4\pi \epsilon_0} \cdot \frac{P \cdot \cos \theta}{r^2} \]
where \( P \) 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 to the point.

4. **Calculating Potential at Point A:**
At point A, which is on the axis of the dipole, the angle \( \theta = 0 \) degrees. Thus, \( \cos \theta = 1 \). The potential at point A can be calculated as:
\[ V_A = \frac{1}{4\pi \epsilon_0} \cdot \frac{P}{R^2} \]

5. **Calculating Potential at Point B:**
At point B, which is on the perpendicular bisector of the dipole, the angle \( \theta = 90 \) degrees. Thus, \( \cos \theta = 0 \). The potential at point B can be calculated as:
\[ V_B = \frac{1}{4\pi \epsilon_0} \cdot \frac{P \cdot 0}{R^2} = 0 \]

6. **Comparing Potentials at A and B:**
From the calculations, we find:
\[ V_A \neq 0 \quad \text{and} \quad V_B = 0 \]
Therefore, the relationship between the potentials at points A and B is:
\[ V_A > V_B \]

### Conclusion:
The potential at point A is non-zero, while the potential at point B is zero. Thus, we conclude:
\[ V_A \neq V_B \]