Question

A charge $Q$ is placed at each of two opposite corners of a square. A charge $q$ is laced at each of the two opposite corners of the square. If the resultant electric field on $Q$ is zero, then

• A. $Q=-\frac{q}{2\sqrt{2}}$
• B. $Q=-2\sqrt{2}q$
• C. $Q=-2q$
• D. $Q=2\sqrt{2}q$

Answer 1

The correct option is A $Q=-\frac{q}{2\sqrt{2}}$

$\begin{array}{c}{F}_{net}onq=\overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_3}\\ 0=F\hat{i}+F\hat{j}+\overrightarrow{F_3}\\ 0=\frac{kq}{a^2}\left(\hat{i}+\hat{j}\right)+\frac{k{q}^2}{{\left(\sqrt{2}a\right)}^2}\frac{\hat{i}+\hat{j}}{\sqrt{2}}\\ 0=\left(\frac{kqQ}{a^2}+\frac{kqQ}{2\sqrt{2}{a}^2}\right)\left(\hat{i}+\hat{j}\right)\\ qQ+\frac{q^2}{2\sqrt{2}}=0\\ q=-2\sqrt{2}Q\\ Q=-\frac{q}{2\sqrt{2}}\end{array}$

Hence,

option $\left(A\right)$ is correct answer.