The magnitude of electric field intensity at a point on the axis of short dipole is 30V/m. The distance of the point from the centre of the dipole is 2m. Then potential at that point is

Options

A. 30V
B. 25V
C. 20V
D. 15V

Explanation

To find the electric potential at a point on the axis of a short dipole, we can use the relationship between electric field intensity (E) and electric potential (V). The electric field intensity at a point on the axis of a dipole is given by the formula:

\[ E = \frac{1}{4\pi \epsilon_0} \cdot \frac{2p}{r^3} \]

Where:
- \( E \) is the electric field intensity,
- \( p \) is the dipole moment,
- \( r \) is the distance from the center of the dipole.

The electric potential \( V \) at a distance \( r \) from the dipole is given by:

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

Given:
- \( E = 30 \, \text{V/m} \)
- \( r = 2 \, \text{m} \)

### Step 1: Calculate the dipole moment \( p \)

From the electric field formula, we can rearrange it to find \( p \):

\[ E = \frac{1}{4\pi \epsilon_0} \cdot \frac{2p}{r^3} \]

Rearranging gives:

\[ p = \frac{E \cdot 4\pi \epsilon_0 \cdot r^3}{2} \]

Substituting the values:

- \( E = 30 \, \text{V/m} \)
- \( r = 2 \, \text{m} \)
- \( \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)

Calculating \( p \):

\[ p = \frac{30 \cdot 4\pi \cdot (8.85 \times 10^{-12}) \cdot (2^3)}{2} \]

### Step 2: Calculate the potential \( V \)

Now, we can use the dipole moment \( p \) to find the potential \( V \):

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

Substituting \( p \) and \( r \):

\[ V = \frac{1}{4\pi \epsilon_0} \cdot \frac{p}{(2)^2} \]

### Step 3: Substitute the value of \( p \) into the potential formula

Since we already have \( p \) from Step 1, we can substitute it back into the potential formula to find \( V \).

### Final Calculation

After performing the calculations, we find that:

\[ V = 30 \cdot 2 = 60 \, \text{V} \]

### Conclusion

Thus, the potential at that point is \( 60 \, \text{V} \).

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