On the axis of a short electric dipole at a point potential is V. If the dipole is rotated through `90^(@)` potential at same point is

Options

A. `V`
B. `(V)/(2)`
C. `Zero`
D. `2V`

Explanation

To solve the problem, we need to understand how the electric potential due to a dipole changes when the dipole is rotated.

### 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 small distance 'd'. The dipole moment \( p \) is defined as \( p = q \cdot d \).

2. **Electric Potential on the Axis of a Dipole**:
The electric potential \( V \) at a point on the axis of a dipole (along the line extending from the positive charge to the negative charge) at a distance \( r \) from the center of the dipole is given by the formula:
\[ V = \frac{kp}{r^2} \]
where \( k \) is a constant (Coulomb's constant), \( p \) is the dipole moment, and \( r \) is the distance from the dipole.

3. **Initial Position**:
Initially, we have a dipole oriented such that the point where we measure the potential is on the axis of the dipole. At this point, the potential is given as \( V \).

4. **Rotating the Dipole**:
When the dipole is rotated through \( 90^\circ \), the orientation changes. Now, the point of interest is still the same, but it is now perpendicular to the dipole moment.

5. **Potential After Rotation**:
The potential at a point on the perpendicular bisector of the dipole (which is the new position after rotation) is given by:
\[ V' = \frac{kp \cos(90^\circ)}{r^2} \]
Since \( \cos(90^\circ) = 0 \), we have:
\[ V' = \frac{kp \cdot 0}{r^2} = 0 \]

6. **Final Result**:
Therefore, the potential at the same point after the dipole is rotated through \( 90^\circ \) is \( 0 \).

### Conclusion:
The potential at the same point after rotating the dipole through \( 90^\circ \) is \( 0 \).

---