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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 $x=a+r$ (if $a< < r)$ is : ( $\varepsilon_0$ is the permittivity of free space)
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The correct answer is:
$\frac{q a^2}{4 \pi \varepsilon_0 r^3}$
We have to find the electric potential at point $P$.
$V_P=\left[\frac{1}{4 \pi \varepsilon_0} \frac{q / 2}{(r+a)}-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q / 2}{(r-a)}\right]$
$=\frac{q}{4 \pi \varepsilon_0}\left[\frac{1}{2(r+a)}-\frac{1}{r}+\frac{1}{2(r-a)}\right]$
$=\frac{q}{4 \pi \varepsilon_0}\left[\frac{r(r-a)-2\left(r^2-a^2\right)+(r+a) r}{2 r\left(r^2-a^2\right)}\right]$
$=\frac{q}{4 \pi \varepsilon_0}\left[\frac{r^2-a r-2 r^2+2 a^2+r^2+a r}{2 r\left(r^2-a^2\right)}\right]$
$=\frac{q a^2}{4 \pi \varepsilon_0 r\left(r^2-a^2\right)}$
Since, $r>>a$, so we have $r^2-a^2 \approx r^2$
$\therefore \quad V_P=\frac{q a^2}{4 \pi \varepsilon_0 r^3}$
$V_P=\left[\frac{1}{4 \pi \varepsilon_0} \frac{q / 2}{(r+a)}-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q / 2}{(r-a)}\right]$
$=\frac{q}{4 \pi \varepsilon_0}\left[\frac{1}{2(r+a)}-\frac{1}{r}+\frac{1}{2(r-a)}\right]$
$=\frac{q}{4 \pi \varepsilon_0}\left[\frac{r(r-a)-2\left(r^2-a^2\right)+(r+a) r}{2 r\left(r^2-a^2\right)}\right]$
$=\frac{q}{4 \pi \varepsilon_0}\left[\frac{r^2-a r-2 r^2+2 a^2+r^2+a r}{2 r\left(r^2-a^2\right)}\right]$
$=\frac{q a^2}{4 \pi \varepsilon_0 r\left(r^2-a^2\right)}$
Since, $r>>a$, so we have $r^2-a^2 \approx r^2$
$\therefore \quad V_P=\frac{q a^2}{4 \pi \varepsilon_0 r^3}$
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