Question

Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is $r$. Fig. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls, now becomes.

Answer 1

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, $\tan \theta =\frac{F}{mg}$ ...(i)

and also, $\tan \theta =\frac{r/2}{y}$ ...(ii)

Equating (i) and (ii), we get

$\frac{F}{mg}=\frac{r}{2y}$

also, Force, $F=\frac{k{q}^2}{r^2}$

where $k=$ constant $=9\times {10}^9$

So, we get $r^3=\left(\frac{k{q}^2}{mg}\right)y$ ...(iii)

Let $r^{\prime }$ be the equilibrium separation when $y\to \frac{y}{2}$

So, equation (iii) then reduces to:

$r^{\prime 3}=\left(\frac{k{q}^2}{mg}\right)\frac{y}{2}$ ...(iv)

Dividing (iv) by (iii) we get,

So, $\frac{r^{\prime 3}}{r^3}=\frac{1}{2}$

$r^{\prime }=r{\left(\frac{1}{2}\right)}^{1/3}$