Question

Two particle $A$ and $B$ having charges $q$ and $2q$ respectively are placed on a smooth table with a separation $d$. A third particle is $C$ is to be clamped on the table in such a way that the particle $A$ and $B$ remain at rest on the table under electrical forces. What should be the charge on $C$ and where should it be claimped?

Answer 1

Let the charge on $C=q$

So, net force on c is equal to zero

So ${\overrightarrow{F}}_{AC}+{\overrightarrow{F}}_{BA}=0$, But $F_{AC}=F_{BC}\Rightarrow \frac{kqQ}{x^2}=\frac{k2qQ}{(d-x{)}^2}$

$\Rightarrow 2x^2=(d-x{)}^2\Rightarrow \sqrt{2}x=d-x$

$\Rightarrow x=\frac{d}{\sqrt{2}+1}=\frac{d}{(\sqrt{2}+1)}\times \frac{(\sqrt{2}-1)}{(\sqrt{2}-1)}$

$=d(\sqrt{2}-1)$