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

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. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now becomes:

Accepted Answer

At equilibrium, the electrostatic repulsion force ($F_e$) is balanced by the component of tension and gravity ($mg$). For a small angle $	heta$, $	an 	heta \approx \frac{r/2}{y}$, where $y$ is the vertical height from the suspension point to the line joining the balls.
The equilibrium condition is $F_e = mg 	an 	heta$.
Substituting Coulomb's law: $\frac{kq^2}{r^2} = mg \frac{r}{2y}$.
Rearranging gives $r^3 = \frac{2kq^2 y}{mg}$, which implies $r^3 \propto y$.
When the strings are clamped at half the height, the new height $y' = y/2$.
Let the new separation be $r'$. Then $(r')^3 \propto y'$.
Taking the ratio: $\frac{(r')^3}{r^3} = \frac{y'}{y} = \frac{y/2}{y} = \frac{1}{2}$.
Therefore, $r' = \frac{r}{\sqrt{2}}$.

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