Thank you for reporting, we will resolve it shortly
Q.
Two point charges $+4 e$ and $+ e$ are kept at
distance $x$ apart. At what distance a charge $q$
must be placed from charge $+ e$, so that $q$ is in
equilibrium :-
Solution:
If $q$ is in equilibrium at position y from $+e$ then
$\frac{+ e }{4 \pi \in_{0} y ^{2}}=\frac{+4 e }{4 \pi \in_{0}( x - y )^{2}}$
$\Rightarrow x-y=2 y $
$\Rightarrow y=\frac{x}{3}$
