Question

Two particles of charges $+e$ and $+2e$ are at $16cm$ away from each other. Where should another charge $q$ be placed between them, so that the system remain in equilibrium?

• A. $24cm$ from $+e$
• B. $12.23cm$ from $+e$
• C. $80cm$ from $+e$
• D. $6.63cm$ from $+e$

Answer 1

The correct option is D $6.63cm$ from $+e$

For the system to be in equilibrium, we have force on charge $q$ due to $+e=$ force on charge $q$ due to $+2e$.
i.e. $\frac{1}{4\pi {ϵ}_0}\cdot \frac{e.q}{x^2}=\frac{1}{4\pi {ϵ}_0}\cdot \frac{2e\cdot q}{(16-x{)}^2}$
$\Rightarrow 2x^2=(16-x{)}^2$
$\Rightarrow 2x^2=256+x^2-32x$
$[\because (a-b{)}^2=a^2+b^2-2ab]$
$\Rightarrow x^2+32x-256=0....\left(i\right)$
On solving Eq. (i), we get
$x=6.63cm$.