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 follow these steps:

### Step-by-Step Solution:

1. **Identify the formula for electric potential (V) on the axis of a dipole**:
The electric potential \(V\) at a distance \(r\) from a dipole on its axis is given by the formula:
\[ V = \frac{K \cdot p}{r^2} \]
where:
- \(K\) is the electrostatic 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.

2. **Substitute the known values into the formula**:
Given:
- \(p = 4 \times 10^{-12} \, \text{C m}\),
- \(r = 3 \, \text{m}\),
- \(K = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\).

Substitute these 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. **Calculate the potential**:
Now substitute \(r^2\) back into the equation:
\[ V = \frac{9 \times 10^9 \cdot (4 \times 10^{-12})}{9} \]
Simplifying this gives:
\[ V = 9 \times 10^9 \cdot 4 \times 10^{-12} \cdot \frac{1}{9} \]
The \(9\) in the numerator and denominator cancels out:
\[ V = 4 \times 10^9 \cdot 10^{-12} \]
\[ V = 4 \times 10^{-3} \, \text{V} \]

5. **Convert to millivolts**:
Since \(1 \, \text{V} = 1000 \, \text{mV}\), we can convert:
\[ V = 4 \, \text{mV} \]

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

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