The potential at a point P' on the axial line of the short dipole on the side of positive charge is 10V. Then the potential at the same point when dipole was reversed will be

Options

A. `10V`
B. `-10V`
C. 5V
D. `5V`

Explanation

To solve the problem step by step, let's analyze the situation involving the electric dipole and the potential at point P' on the axial line.

### Step 1: Understand the Initial Condition
We are given that the potential at point P' on the axial line of the dipole (on the side of the positive charge) is 10V. This means that when the dipole is oriented with the positive charge on the side of point P', the potential is positive.

### Step 2: Define the Potential Due to a Dipole
The potential \( V \) at a point on the axial line of a 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,
- \( r \) is the distance from the center of the dipole to the point.

For a point on the axial line of the dipole, \( \theta = 0^\circ \) (for the positive charge) and \( \cos(0) = 1 \).

### Step 3: Analyze the Reversed Dipole
When the dipole is reversed, the positive charge becomes negative and the negative charge becomes positive. The potential at point P' when the dipole is reversed can be calculated using the same formula, but now the dipole moment \( p \) will effectively be negative because the charges are reversed.

### Step 4: Calculate the New Potential
Since the potential due to the dipole when it was in the original orientation was \( +10V \), when the dipole is reversed, the potential becomes:
\[ V' = -V = -10V \]

### Conclusion
Thus, the potential at point P' when the dipole is reversed will be \( -10V \).

### Summary of the Solution
1. The initial potential at point P' is \( 10V \).
2. When the dipole is reversed, the potential becomes \( -10V \).