Question

Three charged particles having charges $q,-2 q$ and $q$ are placed in a line at points $(-a, 0),(0,0)$ and $(a, 0)$ respectively. The expression for electric potential at $P(r, 0)$ for $r>>a$ is

• A. $\frac{1}{4 \pi \varepsilon_{0}} \frac{q a^{2}}{r^{4}}$
• B. $\frac{1}{4 \pi \varepsilon_{0}} \frac{2 q a^{2}}{r^{3}}$
• C. $\frac{1}{4 \pi \varepsilon_{0}} \frac{4 q a^{2}}{r^{2}}$
• D. $\frac{1}{4 \pi \varepsilon_{0}} \frac{8 q a^{2}}{r}$

Answer 1

Answer: (B)

$E_{1}=\frac{2 k p}{\left(r+\frac{a}{2}\right)^{3}} $
$E_{2}=\frac{2 k p}{\left(r-\frac{a}{2}\right)^{3}}$
$E_{2}-E_{1}=\frac{2 k p}{\left(r-\frac{a}{2}\right)^{3}}-\frac{2 k p}{\left(r+\frac{a}{2}\right)^{3}}$
$\frac{2 k p}{r^{3}}\left[\frac{1}{\left(1-\frac{a}{2 r}\right)^{3}}-\frac{1}{\left(1+\frac{a}{2 r}\right)^{3}}\right]$
$\frac{2 k p}{r^{3}}\left[1+\frac{3 a}{2 r}-\left(1-\frac{3 a}{2 r}\right)\right]$
$\frac{2 k p}{r^{3}}\left[\frac{3 a}{r}\right]$
$E=\frac{6 k p a}{r^{4}}=\frac{3\, \not{6}}{2\,\not{4} \pi \varepsilon_{0}} \frac{p a}{r^{4}}$
$=\frac{3\, p a}{2 \pi \varepsilon_{0} r^{4}}$
$V=\frac{-k p}{\left(r+\frac{a}{2}\right)^{2}}+\frac{k p}{\left(r-\frac{a}{2}\right)^{2}}$
$=\frac{k p}{r^{2}}\left[1+\frac{2 a}{2 r}-1+\frac{2 a}{2 r}\right]=\frac{2 k p a}{r^{3}}$