Question

Three equal charges $+q$ each, are placed at the three vertices of an equilateral triangle centred at the origin. They are held in equilibrium by a restoring force of magnitude $F\left(r\right)=k{r}^2$ (where $k$ is a constant, $r$ is position of charge from origin) directed towards the origin. The distance of the any charge from the origin is ${\left(\frac{\sqrt{3}{q}^2}{p\pi {ϵ}_{0}k}\right)}^{\frac{1}{4}}$ where $p=$

Answer 1

Here from coloumbs law, $F_q=\frac{q^2}{4\pi {ϵ}_0{a}^2}$
The net force acting on one of the charges is
$F_{net}=2F_{q}cos{30}^{\circ }=\frac{2q^2}{4\pi {ϵ}_0{a}^2}\frac{\sqrt{3}}{2}$

position of charge from origin $r=\frac{a}{\sqrt{3}}\Rightarrow a=r\sqrt{3};$
Hence $F_{net}=\frac{q^2}{4\pi {ϵ}_0\sqrt{3}{r}^2}.$

For equilibrium $F\left(r\right)=F_{net}\Rightarrow k{r}^2=\frac{q^2}{4\pi {ϵ}_0\sqrt{3}{r}^2}\Rightarrow r={\left(\frac{\sqrt{3}{q}^2}{12\pi {ϵ}_{0}k}\right)}^{\frac{1}{4}}$