Question

Three point charges $-q,+q$ and $+q$ are placed at the points $(0,a),(0,0)$ and $(0,a)$ respectively in the $x-y$plane. Prove that the potential $V$ at the distance r on the line making an angle 9 with the axis would be
$V=\frac{1}{4\pi {\epsilon }_0}(\frac{q}{r}+\frac{2qa\cos \theta }{r^2}),r\gg a$

Answer 1

Here $(0,a)$ and $(0,-a)$ are position of charges $-q$ and $+q.$ They form an electric dipole. Electric potential is
$V_2$ at point P due to this electric dipole.
$V_2=\frac{1}{4\pi {\epsilon }_0}\frac{q\times 2a\cos \theta }{r^2-a^2{\cos }^2\theta }$
Here $r^2⋙r^2\ge a^2{\cos }^2\theta$
$r^2-a^2{\cos }^2\theta \simeq r^2$
$V_2=\frac{1}{4\pi {\epsilon }_0}\frac{q\times 2a\cos \theta }{r^2}$
Electric potential at point $P$ due to charge at (0,0)
$V_1=\frac{1}{4\pi {\epsilon }_0}-\frac{q}{r}$
Net potential at $P$
$V_P=V_1+V_2$
$\begin{array}{c}{V}_P=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}+\frac{1}{4\pi {\epsilon }_0}\frac{q\times 2a\cos \theta }{r^2}\\ V_P=\frac{1}{4\pi {\epsilon }_0}[\frac{q}{r}+\frac{q\times 2a\cos \theta }{r^2}]\text{where}r\gg a\end{array}$