Question

Two point charges +9q and +q are kept 16 cm apart. Where should a third charge Q be placed between them so that the system remains in equilibrium ?

Answer 1

Two point charges $+9q$ and $+9$.

Kept apart $=16cm=16\times {10}^{-2}m$

Two cases will be needed.

Given charges, $+q$, $+9q$. let, the third charge $q$ be placed at a distance $x$ from $q$ and hence its distance from $9q$ and $(16-x)cm$ for the system to remain in equilibrium $F_1=F_2$

Therefore,

$F_1=\frac{K(q\times Q)}{x^2}$

$F_2=\frac{K(9q\times Q)}{{(16-x)}^2}$

Equating $F_1$ and $F_2$ we get,

$\frac{9}{x^2}=\frac{1}{{(16-x)}^2}$

$F_1=F_2$

or, $\frac{K(q\times Q)}{x^2}=\frac{K(9q\times Q)}{{(16-x)}^2}$

or, $\frac{1}{x^2}=\frac{9}{{(16-x)}^2}$

or, $\frac{1}{x}=\frac{9}{(16-x)}$

or, $16-x=3x$

or, $4x=16$

or, $x=4$

$\therefore$ $(16-4)cm=12cm$ from $+9q$.