The electric potential in volts due to an electric dipole of dipole moment at a `2 xx 10^(-5)C-m` distance of 2m on perpendicular bisector is

Options

A. zero
B. 10
C. 20
D. 15

Explanation

To find the electric potential \( V \) due to an electric dipole at a point on its perpendicular bisector, we can use the formula for the potential due to a dipole:

\[ V = \frac{1}{4\pi \epsilon_0} \cdot \frac{\mathbf{p} \cdot \hat{r}}{r^2} \]

Where:
- \( \mathbf{p} \) is the dipole moment,
- \( \hat{r} \) is the unit vector in the direction from the dipole to the point where the potential is being calculated,
- \( r \) is the distance from the dipole to the point.

### Step 1: Identify the given values
- Dipole moment \( p = 2 \times 10^{-5} \, \text{C-m} \)
- Distance \( r = 2 \, \text{m} \)

### Step 2: Calculate the electric potential
Since we are on the perpendicular bisector, the angle between the dipole moment and the line connecting the dipole to the point is \( 90^\circ \). Therefore, \( \cos(90^\circ) = 0 \), and the formula simplifies to:

\[ V = 0 \]

This is because the contributions to the potential from the positive and negative charges of the dipole cancel each other out at this point.

### Final Answer:
The electric potential \( V \) at a distance of 2 m on the perpendicular bisector of the dipole is:

\[ V = 0 \, \text{volts} \]

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