Question

Two charge $q$ and $-3 q$ are placed fixed on $x$ -axis separated by a distance 'd'. Where should a third charge $2 q$ 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 $2 q$, force due to $q$ is to the left and that due to $-3 q$ is to the right $\therefore \frac{2 q^{2}}{4 \pi \varepsilon_{0} x^{2}}=\frac{6 q^{2}}{4 \pi \varepsilon_{0}(d+x)^{2}}$
$\therefore (d+x)^{2}=3 x^{2}$
$\therefore 2 x^{2}-2 d x-d^{2}=0$
$\Rightarrow x=\frac{d}{2} \pm \frac{\sqrt{3} d}{2}$
(-ve sign would be between $q +-3 q$ and hence is unacceptable)
$x=\frac{d}{2}+\frac{\sqrt{3} d}{2}=\frac{d}{2}(1+\sqrt{3})$ to the left of $q$