Q.A charge q is moved from a point A above a dipole of dipole moment p to a point B below the dipole in equatorial plane without acceleration. Find the work done in this process.

The electric potential of a dipole at a general point depends on $\vec p\cdot\hat r$, not just on $p$ and $r$.

For a short dipole of moment $\vec p$, the potential at a point whose position vector from the centre of the dipole is $\vec r$ (with $r \gg$ the dipole's charge separation) is

$V = \frac{1}{4\pi\epsilon_0}\,\frac{\vec p \cdot \hat r}{r^2}$

This general formula correctly reduces to $V=\dfrac{1}{4\pi\epsilon_0}\dfrac{p}{r^2}$ on the axial line (where $\hat r$ is parallel to $\vec p$) and to $V=0$ on the equatorial line (where $\hat r \perp \vec p$), so it is the only one of the four options valid at a general point.

On the equatorial (broad-side-on) line the two charges are equidistant, so their potentials cancel → V = 0.

The potential due to a dipole at a point making angle θ with the axis is

$V = \frac{1}{4\pi\varepsilon_0}\,\frac{p\cos\theta}{r^2}$

At the broad-side-on (equatorial) position, θ = 90°, so cos θ = 0 and V = 0. Physically, this point is equidistant from +q and −q, so their potentials (equal and opposite) cancel.

On the axial line of an electric dipole, at a distance r from the centre (with r much larger than the charge separation), the potential falls off as p/r², unlike a single point charge's 1/r fall-off.

For a dipole of moment $p = q(2a)$, with charges $+q$ and $-q$ separated by $2a$, the potential at an axial point at distance $r$ from the centre O is obtained by superposing the potentials of the two point charges:

$V = \frac{1}{4\pi\varepsilon_0}\left[\frac{q}{r-a} - \frac{q}{r+a}\right] = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q\,(2a)}{r^2-a^2} = \frac{1}{4\pi\varepsilon_0}\cdot\frac{p}{r^2-a^2}$

For a short dipole (the usual approximation, $r \gg a$), this simplifies to the standard result:

$V = \frac{1}{4\pi\varepsilon_0}\cdot\frac{p}{r^2}$