Question: Q. 2. Two charges $q$ and $-3 q$ are placed on $x$-axis separated by distance $d$. Where a third charge $2 q$ should be placed such that it will not experience any force? A [SQP 2017] [NCERT Exemplar]
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Solution:
Ans. At $P$ on $2 q$, force due to $q$ is to the left and that due to $-3 q$ is to the right.
$$ \begin{aligned} & \therefore \quad \frac{2 q^{2}}{4 \pi \varepsilon_{0} x^{2}}=\frac{6 q^{2}}{4 \pi \varepsilon_{0}(d+x)^{2}} \ & \therefore \quad(d+x)^{2}=3 x^{2} \ & \therefore \quad 2 x^{2}-2 d x-d^{2}=0 \ & \therefore \quad x=\frac{d}{2} \pm \frac{\sqrt{3} d}{2} \end{aligned} $$
As the negative charge will be in between $q$ and $-3 q$, hence it is unacceptable.
$x=\frac{d}{2} \pm \frac{\sqrt{3} d}{2} \Rightarrow \frac{d}{2}(1+\sqrt{3})$ to the left of $q$.
[CBSE Marking Scheme, 2017] 2
