Question

Two charge $q$ and $-3q$ are placed fixed on $x$ -axis separated by a distance 'd'. Where should a third charge $2q$ be placed such that it will not experience any force?

• A. $\frac{d}{2}(1+\sqrt{3})$ to the left of $q$
• B. $\frac{d}{2}(1+\sqrt{3})$ to the right of $q$
• C. $\frac{d}{2}(1+\sqrt{5})$ to the left of $q$
• D. $\frac{d}{2}(1+\sqrt{5})$ to the right of $q$

Answer 1

Answer: (A)

If on $2q$, force due to $q$ is to the left and that due to $-3q$ is to the right $\therefore \frac{2q^2}{4\pi {\epsilon }_0{x}^2}=\frac{6q^2}{4\pi {\epsilon }_0(d+x{)}^2}$
$\therefore (d+x{)}^2=3x^2$
$\therefore 2x^2-2dx-d^2=0$
$\Rightarrow x=\frac{d}{2}\pm \frac{\sqrt{3}d}{2}$
(-ve sign would be between $q+-3q$ and hence is unacceptable)
$x=\frac{d}{2}+\frac{\sqrt{3}d}{2}=\frac{d}{2}(1+\sqrt{3})$ to the left of $q$