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Question

Let $m$ be mass of each ball and $q$ be charge on each ball. Force of repulsion

$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r^{2}}$

In equil

Vedclass · Accepted Answer

$\left( {\frac{r}{{\sqrt[3]{2}}}} \right)$

Vedclass · Answer

Let $m$ be mass of each ball and $q$ be charge on each ball. Force of repulsion

$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r^{2}}$

In equilibrium

$T \cos 	heta=m g$    $...(i)$

$T \sin 	heta=F$     $...(ii)$

Divide $(ii)$ by $(i),$ we get,

$	an 	heta=\frac{F}{m g}=\frac{\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}}{m g}$

From figure $(a),$

$\frac{r / 2}{y}=\frac{\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}}{m g}$    $....(iii)$

$	an 	heta^{\prime}=\frac{\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r^{\prime 2}}}{m g}$

From figure $(b)$

$\frac{r^{\prime} / 2}{y / 2}=\frac{\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r^{\prime 2}}}{m g} \ldots(\mathrm{iv})$

Divide $(iv)$ by $(iii),$ we get

$\frac{2 r^{\prime}}{r}=\frac{r^{2}}{r^{\prime 2}} \Rightarrow r^{\prime 3}=\frac{r^{3}}{2}$

$\Rightarrow r^{\prime}=\frac{r}{\sqrt[3]{2}}$

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