Question

Along the $x$ - axis, three charges $\frac{q}{2},-q$ and $\frac{q}{2}$ are placed at $x=0,x=a$ and $x=2a$ respectively. The resultant electric potential at a point $P$ located at a distance $r$ from the charge $-q(a<<r)$ is $({\epsilon }_0$ is the permittivity of free space)

• A. $\frac{qa}{4\pi {\epsilon }_0{r}^2}$
• B. $\frac{q{a}^2}{4\pi {\epsilon }_0{r}^3}$
• C. $\frac{q\left(\frac{a^2}{4}\right)}{4\pi {\epsilon }_0{r}^3}$
• D. $\frac{q}{4\pi {\epsilon }_{0}r}$

Answer 1

Answer: (B)

As $r>>a$, so $r>2a$
So, potential at point $P$

$V=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q/2}{r+a}-\frac{q}{r}+\frac{q/2}{r-a}\right]$
$=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q}{2}\left[\frac{1}{r+a}-\frac{2}{r}+\frac{1}{r-a}\right]$
$=\frac{q}{8\pi {\epsilon }_0}\left[\frac{r(r-a)-2\left(r^2-a^2\right)+r(r+a)}{r\left(r^2-a^2\right)}\right]$
$=\frac{q}{8\pi {\epsilon }_0}\cdot \frac{2a^2}{r\left(r^2-a^2\right)}$
$=\frac{q{a}^2}{4\pi {\epsilon }_0{r}^3}(\text{ as }r>>a)$