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

As $r >>a$, so $r>2 a$
So, potential at point $P$

$V =\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q / 2}{r+ a}-\frac{q}{r}+\frac{q / 2}{r-a}\right]$
$=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{2}\left[\frac{1}{r +a}-\frac{2}{r}+\frac{1}{r-a}\right] $
$=\frac{q}{8 \pi \varepsilon_{0}}\left[\frac{r(r-a)-2\left(r^{2}-a^{2}\right)+r(r+a)}{r\left(r^{2}-a^{2}\right)}\right]$
$=\frac{q}{8 \pi \varepsilon_{0}} \cdot \frac{2 a^{2}}{r\left(r^{2}-a^{2}\right)}$
$=\frac{q a^{2}}{4 \pi \varepsilon_{0} r^{3}}(\text { as } r>>a)$