Question

Two identical small equally charged conducting balls are suspended from long threads secured at one point. The charges and masses of the balls are such that they are in equilibrium when the distance between them is $a$ (the length of the thread $L >> a$). One of the balls is then discharged. Again for the certain value of distance $b$ ($b << L$) between the balls, the equilibrium is restored, the value of $(\frac{a^3}{b^3}) = k$, where k is:

Answer 1

4

Answer 2

Initially, two identical balls have charge $q$ each. They are in equilibrium at a distance $a$. The electrostatic force is $F_{e1} = k_e q^2/a^2$. For small angles, the equilibrium condition is $\frac{a}{2L} = \frac{F_{e1}}{mg}$, which gives $a^3 = \frac{2L k_e q^2}{mg}$.

When one ball is discharged, we assume the process results in the total charge being halved and distributed equally on the two identical conducting balls. The initial total charge was $2q$. If the total charge becomes $q$ and is distributed equally, each ball has charge $q/2$. The balls are in equilibrium at a distance $b$. The electrostatic force is $F_{e2} = \frac{k_e (q/2)^2}{b^2} = \frac{k_e q^2}{4b^2}$. For small angles, the equilibrium condition is $\frac{b}{2L} = \frac{F_{e2}}{mg}$, which gives $b^3 = \frac{2L k_e q^2}{4mg}$.

Taking the ratio: $\frac{a^3}{b^3} = \frac{2L k_e q^2 / mg}{2L k_e q^2 / (4mg)} = 4$.

Thus, $k=4$.