Question

A charge $Q$ is fixed at each of two opposite corners of a square. A charge $q$ is placed at each of the other two corners. If the resultant electric force on $Q$ is zero, then $Q$ and $q$ are related as

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

Answer 1

Answer: (B)

Let $a$ be the sides of the square.
$F_I=F_2=\frac{Qq}{4\pi {\epsilon }_0{a}^2}$
$F_3=\frac{QQ}{4\pi {\epsilon }_0(a\sqrt{2}{)}^2}$
$=\frac{Q^2}{4\pi {\epsilon }_0(a\sqrt{2}{)}^2}$
As the resultant force on $Q$ is zero,
therefore $F_3\cos 45^{\circ }=-F_1$
$\frac{Q^2}{4\pi {\epsilon }_0(a\sqrt{2}{)}^2}\frac{1}{\sqrt{2}}=-\frac{Qq}{4\pi {\epsilon }_0{a}^2}$
or $Q=-2\sqrt{2}q$