Question

Four positive point charges $(+q)$ are kept at the four corners of a square of side ${}^{\prime }{l}^{\prime }$. The net electric field at the midpoint of any one side of the square is(Take $\frac{1}{4\pi {\epsilon }_0}=K$):

• A. $\frac{4Kq}{l^2}$
• B. $\frac{16Kq}{5\sqrt{5}.l^2}$
• C. $\frac{8Kq}{\sqrt{5}.l^2}$
• D. $\frac{Kq}{l^2}$

Answer 1

The correct option is B $\frac{16Kq}{5\sqrt{5}.l^2}$

The electric field due to the two charges on that particular edge will be cancelled by one another. The electric field due to one of the charge at other corner will be:

$E=\frac{Kq}{r^2}=\frac{Kq}{l^2+{\left(\frac{l}{2}\right)}^2}=\frac{4}{5}\frac{Kq}{l^2}$.

Each charge will have an electric field of this magnitude with an angle $2\theta$ between them where

$tan\theta =\frac{1}{2}$

$cos\theta =\frac{1}{\sqrt{1+ta{n}^2\theta }}=\frac{2}{\sqrt{5}}$

as $0<\theta <\pi /2$

Hence, resultant electric field at this point is :

$E_R=2Ecos\theta =2\left(\frac{4}{5}\frac{Kq}{l^2}\right)\left(\frac{2}{\sqrt{5}}\right)=\frac{16}{5\sqrt{5}}\frac{Kq}{l^2}$