The electric potential at a distance of 3 m on the axis of a short dipole of dipole moment `4 xx 10^(-12)` coulomb-metre is

Options

A. 1.33 `xx 10^(-3)` V
B. 4 mV
C. 12 mV
D. 27 mV

Answer

To find the electric potential at a distance of 3 m on the axis of a short dipole with a dipole moment of \(4 \times 10^{-12}\) coulomb-metre, we can use the formula for the electric potential \(V\) on the axis of a dipole:

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

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

### Step-by-Step Solution:

1. **Identify the given values**:
- Dipole moment \(p = 4 \times 10^{-12} \, \text{C m}\)
- Distance \(r = 3 \, \text{m}\)
- Coulomb's constant \(k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\)

2. **Substitute the values into the formula**:
\[ V = \frac{(9 \times 10^9) \cdot (4 \times 10^{-12})}{(3)^2} \]

3. **Calculate \(r^2\)**:
\[ r^2 = 3^2 = 9 \]

4. **Substitute \(r^2\) back into the equation**:
\[ V = \frac{(9 \times 10^9) \cdot (4 \times 10^{-12})}{9} \]

5. **Simplify the equation**:
\[ V = (9 \times 10^9) \cdot (4 \times 10^{-12}) \cdot \frac{1}{9} \]
The \(9\) in the numerator and denominator cancels out:
\[ V = 4 \times 10^{-12} \times 10^9 \]

6. **Calculate the final value**:
\[ V = 4 \times 10^{-3} \, \text{V} \]

7. **Convert to millivolts**:
\[ V = 4 \, \text{mV} \]

### Final Answer:
The electric potential at a distance of 3 m on the axis of the dipole is \(4 \, \text{mV}\).