Two pith balls carrying equal charges are sus | vedclass

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Question

(A) Let $m$ be the mass of each ball and $q$ be the charge on each ball. The force of electrostatic repulsion is $F = \frac{1}{4 \pi \varepsilon_{0}}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $m$ be the mass of each ball and $q$ be the charge on each ball. The force of electrostatic repulsion is $F = \frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r^{2}}$.In equilibrium,the forces acting on each ball are tension $T$,weight $mg$,and electrostatic force $F$.$T \cos 	heta = mg$  $(i)$$T \sin 	heta = F$  $(ii)$Dividing $(ii)$ by $(i)$,we get $	an 	heta = \frac{F}{mg} = \frac{q^{2}}{4 \pi \varepsilon_{0} r^{2} mg}$.From the geometry of the first case,$	an 	heta = \frac{r/2}{y} = \frac{r}{2y}$.Thus,$\frac{r}{2y} = \frac{q^{2}}{4 \pi \varepsilon_{0} r^{2} mg} \Rightarrow y \propto r^{3}$.In the second case,the strings are clamped at half the height,so the new vertical height is $y' = y/2$. Let the new separation be $r'$.Since the charge $q$ and mass $m$ remain the same,the relationship $y \propto r^{3}$ still holds.Therefore,$\frac{y'}{y} = \left( \frac{r'}{r} ight)^{3}$.Substituting $y' = y/2$,we get $\frac{1}{2} = \left( \frac{r'}{r} ight)^{3}$.$r'^{3} = \frac{r^{3}}{2} \Rightarrow r' = \frac{r}{\sqrt[3]{2}}$.

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 becomes:

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