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

Options

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

Answer

To find the electric potential \( V \) due to a short electric dipole at a distance \( r \) from the dipole, we can use the formula:

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

where:
- \( p \) is the dipole moment,
- \( \theta \) is the angle made with the dipole axis,
- \( r \) is the distance from the dipole,
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

### Step-by-Step Solution

1. **Identify the 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. **Calculate the constant \( k = \frac{1}{4\pi \epsilon_0} \)**:
\[ k \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \]

4. **Substitute the values into the potential formula**:
\[ V = k \cdot \frac{p \cdot \cos \theta}{r^2} \]
\[ V = 9 \times 10^9 \cdot \frac{(2 \times 10^{-8}) \cdot \left(\frac{1}{2}\right)}{(3)^2} \]

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

6. **Substitute \( r^2 \) into the equation**:
\[ V = 9 \times 10^9 \cdot \frac{(2 \times 10^{-8}) \cdot \frac{1}{2}}{9} \]

7. **Simplify the expression**:
\[ V = 9 \times 10^9 \cdot \frac{(2 \times 10^{-8})}{2 \cdot 9} \]
\[ V = 9 \times 10^9 \cdot \frac{10^{-8}}{9} \]
\[ V = 10 \, \text{volts} \]

### Final Answer
The electric potential at the point is \( 10 \, \text{volts} \).

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