Q.(a) Draw equipotential surfaces for a positive point charge.

(a) A point charge's equipotentials are concentric spheres. (b) The dipole potential formula gives zero on the equatorial plane, since every point there is equidistant from both charges.

(a) Equipotential surfaces of a positive point charge:

Since the potential of a point charge $q$ at distance $r$ is $V = \dfrac{q}{4\pi\varepsilon_0 r}$, which depends only on $r$, every point at the same distance from the charge has the same potential. The equipotential surfaces are therefore a family of concentric spheres centred on the charge, with potential decreasing as the radius increases (surfaces farther from the charge have lower potential, spaced progressively farther apart for equal potential steps).

(b) Electric potential — definition and dipole derivation:

Definition: Electric potential at a point in an electrostatic field is the work done per unit positive test charge in bringing it from infinity to that point, without acceleration:

$V = \frac{W}{q_0}$

Derivation for a dipole: consider a dipole with charges $+q$ at A and $-q$ at B, separated by $2a$, dipole moment $p = q(2a)$. Let P be a point at distance $r$ from the centre O of the dipole, at angle $\theta$ from the dipole axis, with $r \gg a$.

Let $r_1$ = distance from $+q$ to P, $r_2$ = distance from $-q$ to P. The total potential at P (superposition of the two point-charge potentials):

$V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{r_1} - \frac{q}{r_2}\right) = \frac{q}{4\pi\varepsilon_0}\cdot\frac{r_2 - r_1}{r_1r_2}$

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