Along the $x$-axis, three charges $\frac{q}{2},-q$ and $\frac{q}{2}$ are placed at $x=0, x=a$ and $x=2 a$…
- $\frac{q a}{4 \pi \varepsilon_0 r^2}$
- $\frac{q a^2}{4 \pi \varepsilon_0 r^3}$
- $\frac{q\left(\frac{a^2}{4}\right)}{4 \pi \varepsilon_0 r^3}$
- $\frac{q}{4 \pi \varepsilon_0 r}$
Solution

$\begin{aligned} V & =\frac{1}{4 \pi \varepsilon_0}\left[\frac{q / 2}{r+a}-\frac{q}{r}+\frac{q / 2}{r-a}\right] \\ & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{2}\left[\frac{1}{r+a}-\frac{2}{r}+\frac{1}{r-a}\right] \\ & =\frac{q}{8 \pi \varepsilon_0}\left[\frac{r(r-a)-2\left(r^2-a^2\right)+r(r+a)}{r\left(r^2-a^2\right)}\right] \\ & =\frac{q}{8 \pi \varepsilon_0} \cdot \frac{2 a^2}{r\left(r^2-a^2\right)} \\ & =\frac{q a^2}{4 \pi \varepsilon_0 r^3}(\text { as } r>a)\end{aligned}$
Asked in: AP EAMCET 2007