Question

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 $r(a < < r)$ from the charge $-q$ is $\left(\varepsilon_{0}\right.$ is the permittivity of free space)

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

Answer 1

Answer: (B)

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