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 (as shown in Fig. I). Now, as shown in Fig. II, the strings are rigidly clamped at half the height. The equilibrium separation between the balls now becomes:

Accepted Answer

At equilibrium, the electrostatic repulsive force ($F_e$) is balanced by the component of gravity acting along the tangent of the arc. For small angles $	heta$, the condition for equilibrium is $	an 	heta = \frac{F_e}{mg}$.

1.  **Geometry:** From the figure, $	an 	heta \approx \sin 	heta = \frac{r/2}{y} = \frac{r}{2y}$, where $y$ is the vertical distance from the point of suspension to the line joining the balls.
2.  **Force Equation:** Substituting Coulomb's law ($F_e = \frac{kq^2}{r^2}$), we get $\frac{kq^2}{r^2 mg} = \frac{r}{2y}$.
3.  **Proportionality:** Rearranging the terms, $r^3 \propto y$.
4.  **New Condition:** When clamped at half height, the new vertical distance is $y' = y/2$. Let the new separation be $r'$.
5.  **Calculation:** $\frac{(r')^3}{r^3} = \frac{y'}{y} = \frac{1}{2} \implies r' = \frac{r}{\sqrt{2}}$.

Step-by-Step Solution

Practice All PHYSICS Questions