Question

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

Answer 1

At P: on 2q, Force due to q is to the left and that due to –3q is to the right.

∴ ${\displaystyle \frac{2q^2}{4\pi {\epsilon }_0{x}^2}=\frac{6q^2}{4\pi {\epsilon }_0{(d+x)}^2}}$

∴ ${\displaystyle {(d+x)}^2=3x^2}$

∴ ${\displaystyle 2x^2-2dx-d^2}$ = 0

${\displaystyle x=\frac{d}{2}+\frac{\sqrt{3}d}{2}}$

(–ve sign would be between q and – 3q and hence is unacceptable.)

${\displaystyle x=\frac{d}{2}+\frac{\sqrt{3}d}{2}=\frac{d}{2}(1+\sqrt{3})}$ to the left of q.