Question

Two point charges $+4e$ and $+e$ are $x$ distance apart. The distance of charge $q$ from charge $+e$ so that it is in equilibrium is :

• A. $\frac{x}{2}$
• B. $\frac{2x}{3}$
• C. $\frac{x}{3}$
• D. $\frac{x}{6}$

Answer 1

The correct option is C $\frac{x}{3}$

The force on $q$ acting must be zero. Let $q$ is kept at distance y in between $4e$ and $e$ for $q$ to be in stable

$\frac{K\left(4e\right)q}{(x-y{)}^2}=\frac{K\left(e\right)\left(q\right)}{y^2}$

$\frac{2}{x-y}=\frac{1}{y}$

$2y=x-y$

$x=3y$

$y=\frac{x}{3}$