The electric potential in volt due to an electric dipole of dipole moment `2 xx 10^(-8)` coulomb-metre at a distance of 3m on a line making an angle of `60^(@)` with the axis of the dipole is

Options

A. zero
B. 10
C. 20
D. 40

Answer

To find the electric potential \( V \) due to an electric dipole at a given distance and angle, we can use the formula:

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

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,
- \( \theta \) is the angle between the dipole moment and the line from the dipole to the point where the potential is being calculated,
- \( r \) is the distance from the dipole.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Dipole moment \( p = 2 \times 10^{-8} \, \text{C m} \)
- Distance \( r = 3 \, \text{m} \)
- Angle \( \theta = 60^\circ \)

2. **Calculate \( \cos \theta \):**
- \( \cos 60^\circ = \frac{1}{2} \)

3. **Substitute Values into the Formula:**
\[ V = \frac{(9 \times 10^9) \cdot (2 \times 10^{-8}) \cdot \left(\frac{1}{2}\right)}{(3)^2} \]

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

5. **Simplify the Expression:**
\[ V = \frac{(9 \times 10^9) \cdot (2 \times 10^{-8}) \cdot \left(\frac{1}{2}\right)}{9} \]

6. **Cancel Out Terms:**
- The \( 9 \) in the numerator and denominator cancels out:
\[ V = (9 \times 10^9) \cdot (2 \times 10^{-8}) \cdot \left(\frac{1}{2}\right) \]

7. **Calculate Further:**
\[ V = (9 \times 10^9) \cdot (1 \times 10^{-8}) = 9 \times 10^1 = 90 \, \text{V} \]

8. **Final Result:**
\[ V = 10 \, \text{V} \]

### Conclusion:
The electric potential at a distance of 3m from the dipole making an angle of \( 60^\circ \) with the dipole axis is \( 10 \, \text{V} \).