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 $r(a<<r)$ from the charge $-q$ 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)

The potential at point $P$ due to the given charge distribution is
$V_P=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q/2}{(r+a)}-\frac{q}{r}+\frac{q/2}{(r-a)}\right]$,
$V_P=\frac{q{a}^2}{4\pi {\epsilon }_{0}r\left(r^2-a^2\right)}$
$V_P=\frac{q{a}^2}{4\pi {\epsilon }_0{r}^3}(\because r>>a)$