Question

Three point charges $q=2q$ and q are located along the x-axis as y shown in figure. The electric field at P $(y>>A)$ along the y-axis is:

• A. $\overrightarrow{E}=-\frac{1}{8\pi {\epsilon }_0}\frac{3q{a}^2}{y^4}\hat{j}$
• B. $\overrightarrow{E}=-\frac{1}{4\pi {\epsilon }_0}\frac{6q{a}^2}{y^4}\hat{j}$
• C. $\overrightarrow{E}=-\frac{1}{4\pi {\epsilon }_0}\frac{3q{a}^2}{y^4}\hat{j}$
• D. $0$

Answer 1

The correct option is C $\overrightarrow{E}=-\frac{1}{4\pi {\epsilon }_0}\frac{3q{a}^2}{y^4}\hat{j}$

$\overrightarrow{E}=2\times \frac{1}{4\pi ϵ}\frac{q}{y^2+a^2}cos\left(ta{n}^{-1}\frac{a}{y}\right)-\frac{1}{4\pi ϵ}\frac{2q}{y^2}\hat{j}$

$=\frac{2}{4\pi ϵ}q[\frac{y}{(y^2+a^2{)}^{3/2}}-\frac{1}{y^2}]\hat{j}$

$=\frac{2}{4\pi ϵ}q\left[\frac{y^3-(y^2+a^2{)}^{3/2}}{y^2(y^2+a^2{)}^{3/2}}\right]\hat{j}$

using Taylor expansion,

$=\frac{2}{4\pi ϵ}q\left[\frac{(y^3-y^3-\frac{3}{2}y{a}^2+..)}{y^2(y^2+a^2{)}^{3/2}}\right]\hat{j}$

As a<<y ,

$=\frac{2}{4\pi ϵ}q\left[\frac{-\frac{3}{2}y{a}^2+}{y^2(y^2{)}^{3/2}}\right]\hat{j}$

$=\frac{2}{4\pi ϵ}q\left[\frac{-\frac{3}{2}y{a}^2}{y^5}\right]\hat{j}$

$=-\frac{1}{4\pi ϵ}q\frac{3a^2+}{y^4}]\hat{j}$