The potential due to a short electric field dipole moment`2xx10^-6` C-m along its axis point 4 m from dipole is

Options

A. `(9/8)xx10^3` V
B. `(9/8)xx10^-3 ` V
C. `(8/9)xx10^2 ` V
D. Zero

Answer

To find the electric potential due to a dipole at a certain distance along its axis, we can use the formula for the potential \( V \) due to a dipole:

\[ V = \frac{k \cdot p}{r^2} \]

where:
- \( V \) is the electric potential,
- \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \),
- \( p \) is the dipole moment,
- \( r \) is the distance from the dipole.

### Step-by-Step Solution:

1. **Identify the given values**:
- Dipole moment \( p = 2 \times 10^{-6} \, \text{C m} \)
- Distance \( r = 4 \, \text{m} \)

2. **Substitute the values into the formula**:
\[ V = \frac{9 \times 10^9 \times (2 \times 10^{-6})}{(4)^2} \]

3. **Calculate \( r^2 \)**:
\[ r^2 = 4^2 = 16 \]

4. **Substitute \( r^2 \) into the equation**:
\[ V = \frac{9 \times 10^9 \times (2 \times 10^{-6})}{16} \]

5. **Calculate the numerator**:
\[ 9 \times 10^9 \times 2 \times 10^{-6} = 18 \times 10^3 = 1.8 \times 10^4 \]

6. **Now divide by 16**:
\[ V = \frac{1.8 \times 10^4}{16} \]

7. **Perform the division**:
\[ V = 1.125 \times 10^3 \, \text{V} = 1125 \, \text{V} \]

### Final Answer:
The potential at a distance of 4 m from the dipole is \( 1125 \, \text{V} \).