Question

Three charged particles having charges $q,-2q$ 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. $\frac{1}{4\pi {\epsilon }_0}\frac{q{a}^2}{r^4}$
• B. $\frac{1}{4\pi {\epsilon }_0}\frac{2q{a}^2}{r^3}$
• C. $\frac{1}{4\pi {\epsilon }_0}\frac{4q{a}^2}{r^2}$
• D. $\frac{1}{4\pi {\epsilon }_0}\frac{8q{a}^2}{r}$

Answer 1

Answer: (B)

$E_1=\frac{2kp}{{\left(r+\frac{a}{2}\right)}^3}$
$E_2=\frac{2kp}{{\left(r-\frac{a}{2}\right)}^3}$
$E_2-E_1=\frac{2kp}{{\left(r-\frac{a}{2}\right)}^3}-\frac{2kp}{{\left(r+\frac{a}{2}\right)}^3}$
$\frac{2kp}{r^3}\left[\frac{1}{{\left(1-\frac{a}{2r}\right)}^3}-\frac{1}{{\left(1+\frac{a}{2r}\right)}^3}\right]$
$\frac{2kp}{r^3}\left[1+\frac{3a}{2r}-\left(1-\frac{3a}{2r}\right)\right]$
$\frac{2kp}{r^3}\left[\frac{3a}{r}\right]$
$E=\frac{6kpa}{r^4}=\frac{3/6}{2/4\pi {\epsilon }_0}\frac{pa}{r^4}$
$=\frac{3pa}{2\pi {\epsilon }_0{r}^4}$
$V=\frac{-kp}{{\left(r+\frac{a}{2}\right)}^2}+\frac{kp}{{\left(r-\frac{a}{2}\right)}^2}$
$=\frac{kp}{r^2}\left[1+\frac{2a}{2r}-1+\frac{2a}{2r}\right]=\frac{2kpa}{r^3}$