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.
$F_{ I }=F_{2}=\frac{Q q}{4 \pi \varepsilon_{0} a^{2}}$
$F_{3}=\frac{Q Q}{4 \pi \varepsilon_{0}(a \sqrt{2})^{2}}$
$=\frac{Q^{2}}{4 \pi \varepsilon_{0}(a \sqrt{2})^{2}}$
As the resultant force on $Q$ is zero,
therefore $F_{3} \cos\, 45^{\circ}=-F_{1}$
$\frac{Q^{2}}{4 \pi \varepsilon_{0}(a \sqrt{2})^{2}} \frac{1}{\sqrt{2}} =-\frac{Q q}{4 \pi \varepsilon_{0} a^{2}} $
or $ Q=-2 \sqrt{2} q$