Question

Two equal point charges, $Q=+\sqrt{2}\mu C$ are placed at each of the two opposite corners of a square and equal point charges $q$ at each of the other two corners. The value of $q$, so that the resultant force on $Q$ is zero is

• A. $+0.5\mu C$
• B. $-0.5\mu C$
• C. $1\mu C$
• D. $-1\mu C$

Answer 1

Answer: (B)

From Coulomb's law, $F=\frac{1}{4\pi {\epsilon }_0}\frac{Q^2}{2a^2}$
$F_1$ and $F_2$ will be directed as shown, for this both $q$ should be negative
$F_1=F_2=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Qq}{a^2}$
$F_{12}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\sqrt{2}Qq}{a^2}$

For equilibrium of $Q$, we have
$F=-F_{12}$
$\frac{Q^2}{2a^2}=-\frac{\sqrt{2}Qq}{a^2}$
$\Rightarrow q=\frac{Q}{2\sqrt{2}}$
Given $Q=+\sqrt{2}\mu C$
$\therefore q=-\frac{\sqrt{2}\mu C}{2\sqrt{2}}=-0.5\mu C$