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
At $P:$ on $2 \mathrm{q}$, Force due to $\mathrm{q}$ is to the left and that due to $-3 \mathrm{q}$ is to the right.
$\therefore \frac{2 q^{2}}{4 \pi \varepsilon_{0} x^{2}}=\frac{6 q^{2}}{4 \pi \varepsilon_{0}(d+x)^{2}}$
$\therefore(d+x)^{2}=3 x^{2}$
$\therefore 2 x^{2}-2 d x-d^{2}=0$
$x=\frac{d}{2} \pm \frac{\sqrt{3} d}{2}$
($-ve$ sign would be between $\mathrm{q}$ and $-3 \mathrm{q}$ and hence is unacceptable.)
$x=\frac{d}{2}+\frac{\sqrt{3} d}{2}=\frac{d}{2}(1+\sqrt{3})$ to the left of $q$
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