Q.(a) Derive an expression for electrostatic potential due to an electric dipole. OR

(a) Superposing the potentials of the dipole's two charges and using the far-field approximation gives $V=kp\cos\theta/r^2$; (b) using the path-difference condition for bright fringes in YDSE gives the fringe (band) width $\beta=\lambda D/d$. Both alternatives answered below.

(a) Electrostatic potential due to an electric dipole

Consider a dipole with charges $+q$ at $A$ and $-q$ at $B$, separated by distance $2a$ (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 $\vec{OP}$ making angle $\theta$ with the dipole axis; let $r_1,r_2$ be the distances of $P$ from $+q$ and $-q$ respectively.

By superposition, the net potential at $P$:

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

For a point far from the dipole ($r\gg a$), using the geometric approximations $r_2-r_1\approx2a\cos\theta$ and $r_1r_2\approx r^2$:

$V\approx\dfrac{q}{4\pi\varepsilon_0}\cdot\dfrac{2a\cos\theta}{r^2}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}$

Special cases: on the axial line ($\theta=0°$), $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2}$ (maximum); on the equatorial line ($\theta=90°$), $V=0$.

(b) Bandwidth (fringe width) in Young's Double Slit Experiment

Two coherent slits $S_1$, $S_2$ separated by distance $d$ illuminate a screen placed at distance $D$ ($D\gg d$). Consider a point $P$ on the screen at distance $y$ from the central point $O$ (on the perpendicular bisector of $S_1S_2$).

The path difference between the two waves reaching $P$ is (using the standard small-angle geometry of YDSE): …