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Along the $x$-axis, three charges $\frac{q}{2},-q$ and $\frac{q}{2}$ are placed at $x=0, x=a$ and $x=2 a$ respectively. The resultant electric potential at a point $P$ located at a distance $r$ from the charge $-q(a < r)$ is $\left(\varepsilon_0\right.$ is the permittivity of free space)
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The correct answer is:
$\frac{q a^2}{4 \pi \varepsilon_0 r^3}$
As $r \gg a$, so $r>2 a$ So, potential at point $P$
$\begin{aligned} V & =\frac{1}{4 \pi \varepsilon_0}\left[\frac{q / 2}{r+a}-\frac{q}{r}+\frac{q / 2}{r-a}\right] \\ & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{2}\left[\frac{1}{r+a}-\frac{2}{r}+\frac{1}{r-a}\right] \\ & =\frac{q}{8 \pi \varepsilon_0}\left[\frac{r(r-a)-2\left(r^2-a^2\right)+r(r+a)}{r\left(r^2-a^2\right)}\right] \\ & =\frac{q}{8 \pi \varepsilon_0} \cdot \frac{2 a^2}{r\left(r^2-a^2\right)} \\ & =\frac{q a^2}{4 \pi \varepsilon_0 r^3}(\text { as } r>a)\end{aligned}$
$\begin{aligned} V & =\frac{1}{4 \pi \varepsilon_0}\left[\frac{q / 2}{r+a}-\frac{q}{r}+\frac{q / 2}{r-a}\right] \\ & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{2}\left[\frac{1}{r+a}-\frac{2}{r}+\frac{1}{r-a}\right] \\ & =\frac{q}{8 \pi \varepsilon_0}\left[\frac{r(r-a)-2\left(r^2-a^2\right)+r(r+a)}{r\left(r^2-a^2\right)}\right] \\ & =\frac{q}{8 \pi \varepsilon_0} \cdot \frac{2 a^2}{r\left(r^2-a^2\right)} \\ & =\frac{q a^2}{4 \pi \varepsilon_0 r^3}(\text { as } r>a)\end{aligned}$
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