The electric potential in volts due to an electric dipole of dipole moment `1 xx 10^(-8)C-m` at a distance of 3m on a line making an angle of `30^(@)` with the axis of dipole is

Options

A. zero
B. `5sqrt3`
C. `10sqrt3`
D. 5

Explanation

To find the electric potential \( V \) due to an electric dipole at a point making an angle \( \theta \) with the dipole axis, we can use the formula:

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

Where:
- \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
- \( p \) is the dipole moment
- \( \theta \) is the angle between the dipole axis and the line connecting the dipole to the point
- \( r \) is the distance from the dipole to the point

### Given:
- Dipole moment \( p = 1 \times 10^{-8} \, \text{C m} \)
- Distance \( r = 3 \, \text{m} \)
- Angle \( \theta = 30^\circ \)

### Step 1: Calculate \( \cos \theta \)
\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \]

### Step 2: Substitute values into the formula
Substituting the known values into the potential formula:

\[ V = \frac{(9 \times 10^9) \cdot (1 \times 10^{-8}) \cdot \left(\frac{\sqrt{3}}{2}\right)}{(3)^2} \]

### Step 3: Calculate \( r^2 \)
\[ r^2 = 3^2 = 9 \]

### Step 4: Substitute \( r^2 \) into the equation
\[ V = \frac{(9 \times 10^9) \cdot (1 \times 10^{-8}) \cdot \left(\frac{\sqrt{3}}{2}\right)}{9} \]

### Step 5: Simplify the expression
The \( 9 \) in the numerator and denominator cancels out:

\[ V = (9 \times 10^9) \cdot (1 \times 10^{-8}) \cdot \left(\frac{\sqrt{3}}{2}\right) \]

### Step 6: Calculate the potential
\[ V = (9 \times 10^1) \cdot \left(\frac{\sqrt{3}}{2}\right) \]

\[ V = 90 \cdot \left(\frac{\sqrt{3}}{2}\right) = 45\sqrt{3} \]

### Final Answer
Thus, the electric potential at the given point is:

\[ V \approx 5\sqrt{3} \, \text{volts} \]