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Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is $r.$ Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now become
$\left( {\frac{r}{{\sqrt[3]{2}}}} \right)$
$\left( {\frac{{2r}}{{\sqrt 3 }}} \right)$
$\left( {\frac{{2r}}{3}} \right)$
${\left( {\frac{1}{{\sqrt 2 }}} \right)^2}$
Solution
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
$T \cos \theta=m g$ $…(i)$
$T \sin \theta=F$ $…(ii)$
Divide $(ii)$ by $(i),$ we get,
$\tan \theta=\frac{F}{m g}=\frac{\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}}{m g}$
From figure $(a),$
$\frac{r / 2}{y}=\frac{\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}}{m g}$ $….(iii)$
$\tan \theta^{\prime}=\frac{\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r^{\prime 2}}}{m g}$
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
$\frac{2 r^{\prime}}{r}=\frac{r^{2}}{r^{\prime 2}} \Rightarrow r^{\prime 3}=\frac{r^{3}}{2}$
$\Rightarrow r^{\prime}=\frac{r}{\sqrt[3]{2}}$
