Q.(a) Define electrostatic potential.

Potential is work done per unit charge to bring a test charge from infinity; for a dipole, adding the potentials of +q and −q and using the far-field approximation gives $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}$.

1. Electrostatic potential at a point in an electric field is defined as the work done in bringing a unit positive test charge from infinity to that point, without any acceleration (i.e. quasi-statically, against the electric force): $V = \frac{W_{\infty\to P}}{q_0}$
2. Potential due to a dipole: Let a dipole consist of charges $+q$ at A and $-q$ at B, separated by $2a$, with dipole moment $p = q(2a)$ directed from $-q$ to $+q$. Let P be a point at distance $r$ from the centre O of the dipole, with OP making angle θ with the dipole axis. Let $r_1$ = distance from $+q$ to P, and $r_2$ = distance from $-q$ to P. The potential at P is the sum of the potentials due to the two point charges: $V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{r_1} - \frac{q}{r_2}\right) = \frac{q}{4\pi\varepsilon_0}\cdot\frac{r_2-r_1}{r_1r_2}$ For a point far from the dipole ($r \gg a$), using the geometry (dropping perpendiculars from A and B onto OP): $r_2 - r_1 \approx 2a\cos\theta, \qquad r_1r_2 \approx r^2$ Substituting: …