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.

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Aicharya · Accepted Answer

Given 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, we have to find the equilibrium separation when the strings are clamped at half the height.
We can see from the figure that under equilibrium, $\tan \theta=\frac{\mathrm{F}}{\mathrm{mg}}$...(i) and
also, $\tan \theta=\frac{\mathrm{r} / 2}{\mathrm{y}} \ldots$ (ii)
Equating (i) and (ii), we get
$\frac{\mathrm{F}}{\mathrm{mg}}=\frac{\mathrm{r}}{2 \mathrm{y}}$
also, Force, $\mathrm{F}=\frac{\mathrm{kq}^2}{\mathrm{r}^2}$
where $\mathrm{k}=$ constant $=9 \times 10^9$
So, we get $\mathrm{r}^3=\left(\frac{\mathrm{kq}^2}{\mathrm{mg}}\right) \mathrm{y} \ldots$ (iii)
Let $r^{\prime}$ be the equilibrium separation when $y \rightarrow \frac{y}{2}$ So, equation (iii) then reduces to:
$\mathrm{r}^{\prime 3}=\left(\frac{\mathrm{kq}^2}{\mathrm{mg}}\right) \frac{\mathrm{y}}{2} \ldots \text { (iv) }$
Dividing (iv) by (iii) we get,
So, $\frac{\mathrm{r}^{\prime 3}}{\mathrm{r}^3}=\frac{1}{2}$
$r^{\prime}=r\left(\frac{1}{2}\right)^{1 / 3}$

Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the…

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