The potential due to a short electric field dipole moment`2xx10^-6` C-m along its axis point 4 m from dipole is

Options

A. `(9/8)xx10^3` V
B. `(9/8)xx10^-3 ` V
C. `(8/9)xx10^2 ` V
D. Zero

Answer

To find the electric potential due to a short electric dipole moment along its axis at a point 4 m from the dipole, we can use the formula for the potential \( V \) due to a dipole:

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

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 connecting the dipole to the point where the potential is being calculated,
- \( r \) is the distance from the dipole to the point.

### Step 1: Identify the given values
- Dipole moment \( p = 2 \times 10^{-6} \, \text{C m} \)
- Distance \( r = 4 \, \text{m} \)
- Since we are calculating the potential along the axis of the dipole, \( \theta = 0^\circ \).

### Step 2: Substitute the values into the formula
Since \( \cos(0^\circ) = 1 \), the formula simplifies to:

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

### Step 3: Calculate \( r^2 \)
\[ r^2 = (4 \, \text{m})^2 = 16 \, \text{m}^2 \]

### Step 4: Substitute \( k \), \( p \), and \( r^2 \) into the equation
\[ V = \frac{9 \times 10^9 \, \text{N m}^2/\text{C}^2 \cdot 2 \times 10^{-6} \, \text{C m}}{16 \, \text{m}^2} \]

### Step 5: Perform the multiplication in the numerator
\[ 9 \times 10^9 \cdot 2 \times 10^{-6} = 18 \times 10^3 = 1.8 \times 10^4 \]

### Step 6: Divide by \( r^2 \)
\[ V = \frac{1.8 \times 10^4}{16} \]

### Step 7: Calculate the final value
\[ V = 1.125 \times 10^3 \, \text{V} = 1125 \, \text{V} \]

Thus, the electric potential due to the dipole at a point 4 m along its axis is **1125 V**.