Question

Two equal charges $‘q^{\prime }$ of opposite sign are separated by a small distance $‘d^{\prime }$. The electric intensity $‘E^{\prime }$ at a point on the straight line passing through the two charges at a very large distance $‘r^{\prime }$ from the midpoint of two charges is :

• A. $\frac{1}{4\pi {\epsilon }_0}\frac{qd}{r^2}$
• B. $\frac{1}{4\pi {\epsilon }_0}\frac{2qd}{r^2}$
• C. $\frac{1}{4\pi {\epsilon }_0}\frac{qd}{r^3}$
• D. $\frac{1}{4\pi {\epsilon }_0}\frac{2qd}{r^3}$

Answer 1

The correct option is D $\frac{1}{4\pi {\epsilon }_0}\frac{2qd}{r^3}$

As shown in the fig there are two electric fields $E_{+q}$ and $E_{-q}$ due to $+q$ and $-q$ charges respectively.

Resultant of this two oppositely directed electric fields gives the net electric field at that point.

$\therefore$ Electric field at that point is given by,
$E=E_{+q}-E_{-q}=Kq/(r-d/2{)}^2-Kq/(r+d/2{)}^2$

$=Kq[\frac{1}{(r^2-rd+d^4/4)}-\frac{1}{(r^2+rd+d^4/4)}]$

$=Kq[\frac{1}{(r^2-rd)}-\frac{1}{(r^2+rd)}]$ ......($d^2/4\approx 0$)

$=Kq[\frac{(r^2+rd)}{r^4-r^2{d}^2}-\frac{(r^2-rd)}{r^4-r^2{d}^2}]$

$=Kq\left[\frac{\left(2rd\right)}{r^2(r^2-d^2)}\right]$

$=Kq\frac{\left(2rd\right)}{r^4}$ ....$(d^2\approx 0)$

$\therefore E=\frac{1}{4\pi {ϵ}_o}\frac{2qd}{r^3}$