Animated Solution for Physics - Electrostatics: A point dipole $\mathbf{p}=-p_0\hat{\mathbf{x}}$ is kept at the origin. The potential and electric field due to this dipole on the Y-axis at a distance $d$ are, respectively [Take, $V=0$ at infinity]

The problem asks us to find the electric potential and the electric field at a point on the Y-axis due to a point dipole placed at the origin. The dipole moment is given as $\mathbf{p}=-p_0\hat{\mathbf{x}}$.

The dipole is placed exactly at the origin. Its dipole moment vector $\mathbf{p}$ points along the negative X-axis. This means the positive charge $+q$ is on the negative X-axis, and the negative charge $-q$ is on the positive X-axis.

We are interested in a point $P$ located on the Y-axis at a distance $d$ from the origin. Geometrically, the Y-axis is the perpendicular bisector of the dipole. In physics terms, any point on the Y-axis lies on the equatorial plane of this dipole.

The total potential is simply the scalar sum of the potentials created by the individual charges of the dipole.

Because point $P$ is on the perpendicular bisector, its distance $r$ to the positive charge is exactly equal to its distance to the negative charge.

When we add them up, the positive and negative terms perfectly cancel each other out.

This is a fundamental property: The electric potential at any point on the equatorial plane of a dipole is always zero.

Unlike potential, the electric field is a vector. The field from the positive charge points away from it, and the field from the negative charge points towards it. When you resolve these vectors, their vertical components cancel out, but their horizontal components add up.

For a short dipole (where the distance $d$ is much larger than the separation between the charges), the standard formula for the electric field on the equatorial line is:

The negative sign is crucial here. It physically signifies that the electric field on the equatorial line is always anti-parallel to the dipole moment vector.

In our specific problem, the distance $r$ is given as $d$. Substituting this into our formula, we get:

The electric potential is $0$, and the electric field is $\frac{-\mathbf{p}}{4\pi {\epsilon }_0{d}^3}$. Comparing this with the given options, we can confidently select option (b).