Question

Three charges $+q,+q$ and $-q$ are situated in x-y plane at the point $(0,a),(0,0),(0,-a)$. Then, the electric potential at a point in first quadrant whose position vector makes an angle $\theta$ with the y-axis and at a distance $r(r>>a)$ from the origin is :

• A. $\frac{qacos\theta }{2\pi {\epsilon }_0{r}^2}$
• B. $\frac{q}{4\pi {\epsilon }_{0}r}(1+\frac{2acos\theta }{r})$
• C. $\frac{q}{4\pi {\epsilon }_{0}r}$
• D. $\frac{q}{4\pi {\epsilon }_{0}r}(1-\frac{2acos\theta }{r})$

Answer 1

Answer: (B)

$\frac{q}{4\pi {\epsilon }_{0}r}(1+\frac{2acos\theta }{r})$

Given that the point $A$ makes an angle $\theta$ with the $y$ axis and lies in the first quadrant.

The coordinates of $A$ are $(r\sin \theta ,r\cos \theta )$

Given charge distribution is $+q$ at $(0,a)$, $+q$ at $(0,0)$ and $-q$ at $(0,-a)$

Potential at $A$ due to this distribution is

$V=\frac{q}{4\pi {ϵ}_0}(\frac{1}{\sqrt{(r\sin \theta {)}^2+(r\cos \theta -a{)}^2}}+\frac{1}{r}-\frac{1}{\sqrt{(r\sin \theta {)}^2+(r\cos \theta +a{)}^2}})=\frac{q}{4\pi {ϵ}_{0}r}(1+\frac{1}{\sqrt{1-2\frac{a}{r}\cos \theta +\frac{a^2}{r^2}}}-\frac{1}{\sqrt{1+2\frac{a}{r}\cos \theta +\frac{a^2}{r^2}}})$

As $r>>a$, $\sqrt{1+k\frac{a}{r}+\frac{a^2}{r^2}}\approx 1+\frac{ka}{2r}\approx (1-\frac{ka}{2r}{)}^{-1}$

Using the above approximations, $V=\frac{q}{4\pi {ϵ}_{0}r}(1+(1+\frac{a}{r}\cos \theta )-(1-\frac{a}{r}\cos \theta ))=\frac{q}{4\pi {ϵ}_{0}r}(1+2\frac{a}{r}\cos \theta )$