Question

Derive an expression for electric potential at a point due to an electric dipole. Explain the special cases.

Answer 1

Two charges $-q$ at $A$ and $+q$ at $B$ separated by a small distance $2d$ constitute an electric dipole and its dipole moment is $p$ (figure).
Let $r_1$ and $r_2$ be the distances of the point $P$ from $+q$ and $-q$ charges respectively. Let $P$ be the point at a distance $r$ from the midpoint of the dipole $O$ and $\theta$ be the angle between $PO$ and the axis of the dipole $OB$.
Potential at $P$ due to charge $(+q)=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r_1}$
Potential at $P$ due to charge $(-q)=\frac{1}{4\pi {\epsilon }_0}(-\frac{q}{r_2})$
Total potential at $P$ due to dipole is,
$V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r_1}-\frac{1}{4\pi {\epsilon }_0}\frac{q}{r_2}$
$V=\frac{q}{4\pi {\epsilon }_0}(\frac{1}{r_1}-\frac{1}{r_2})$ ....(1)
Applying cosine law,
$r_1^2=r^2+d^2-2rd\cos \theta$
$r_1^2=r^2(1-2d\frac{\cos \theta }{r}+\frac{d^2}{r^2})$
Since $d$ is very much smaller than $r$, $\frac{d^2}{r^2}$ can be neglected.
$\therefore r_1=r{(1-\frac{2d}{r}\cos \theta )}^{1/2}$
(or) $\frac{1}{r_1}=\frac{1}{r}{(1-\frac{2d}{r}\cos \theta )}^{-1/2}$
Using the Binomial theorem and neglecting higher powers,
$\therefore \frac{1}{r_1}=\frac{1}{r}(1+\frac{d}{r}\cos \theta )$ ....(2)
Similarly,
$r_2^2=r^2+d^2-2rd\cos (180-\theta )$
or $r_2^2=r^2+d^2+2rd\cos \theta$
$r_2=r{(1+\frac{2d}{r}\cos \theta )}^{1/2}$ ($\because \frac{d^2}{r^2}$ is negligible)
(or) $\frac{1}{r_2}=\frac{1}{r}{(1+\frac{2d}{r}\cos \theta )}^{-1/2}$
Using the Binomial theorem and neglecting higher powers,
$\frac{1}{r_2}=\frac{1}{r}(1-\frac{d}{r}\cos \theta )$ ...(3)
Substituting equation (2) and (3) in equation (1) and simplifying
$V=\frac{q}{4\pi {\epsilon }_0}\frac{1}{r}(1+\frac{d}{r}\cos \theta -1+\frac{d}{r}\cos \theta )$
$\therefore V=\frac{q2d\cos \theta }{4\pi {\epsilon }_0\cdot r^2}=\frac{1}{4\pi {\epsilon }_0}\frac{p\cdot \cos \theta }{r^2}$ ....(4)
Special cases :
(i) When the point $P$ lies on the axial line of the dipole on the side of $+q$, then $\theta =0$.
$\therefore V=\frac{p}{4\pi {\epsilon }_0{r}^2}$
(ii) When the point $P$ lies on the axial line of the dipole on the side of $-q$, then $\theta ={180}^o$.
$V=-\frac{p}{4\pi {\epsilon }_0{r}^2}$
(iii) When the point $P$ lies on the equatorial line of the dipole, then, $\theta ={90}^o$,
$V=0$