In th electric potential on the axis of an electric dipole at a distance 'r' from it is V, then the potential at a point on its equatorial line at the same distance away frornt it will be

Options

A. 2V
B. `(2V)/(2)`
C. 0
D. `- V`

Answer

To solve the problem, we need to understand the electric potential due to an electric dipole at different points in space.

### Step-by-Step Solution:

1. **Understanding the Electric Dipole**:
An electric dipole consists of two equal and opposite charges, +Q and -Q, separated by a small distance 'd'. The dipole moment \( p \) is given by \( p = Q \cdot d \).

2. **Electric Potential on the Axis of the Dipole**:
The electric potential \( V \) at a point on the axis of the dipole at a distance \( r \) from the center of the dipole is given by the formula:
\[ V = \frac{k \cdot p \cdot \cos(\theta)}{r^2} \]
where \( k \) is Coulomb's constant, \( p \) is the dipole moment, and \( \theta \) is the angle between the dipole moment and the line joining the dipole to the point where the potential is being calculated.

3. **Calculating Potential on the Axis**:
For a point on the axis of the dipole, \( \theta = 0^\circ \). Therefore, \( \cos(0) = 1 \). The potential at this point becomes:
\[ V = \frac{k \cdot p \cdot 1}{r^2} = \frac{k \cdot p}{r^2} \]

4. **Electric Potential on the Equatorial Line**:
Now, we need to find the potential at a point on the equatorial line of the dipole at the same distance \( r \). For a point on the equatorial line, \( \theta = 90^\circ \). Thus, the potential at this point is given by:
\[ V' = \frac{k \cdot p \cdot \cos(90^\circ)}{r^2} \]
Since \( \cos(90^\circ) = 0 \), we have:
\[ V' = \frac{k \cdot p \cdot 0}{r^2} = 0 \]

5. **Conclusion**:
Therefore, the electric potential at a point on the equatorial line at the same distance \( r \) from the dipole is \( 0 \).

### Final Answer:
The potential at a point on the equatorial line at the same distance \( r \) from the dipole is \( 0 \).

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