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 thread L >> a). Then one of the balls is discharged. What will be the distance b(b << L) between the balls when equilibrium is restored?

Answer 1

As we see

$Tsin\theta =F=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{a^2}$ (i)

$Tcos\theta =mg$ (ii)

From equation (i) and (ii), we get

$tan\theta =\frac{q^2}{4\pi {\epsilon }_0{a}^{2}mg}$

$\frac{a/2}{L}=\frac{q^2}{4\pi {\epsilon }_0{a}^{2}mg}$ ($\because$ for small $\theta ,tan\theta \approx a/2L$)

or $\frac{a^3}{L}=\frac{q^2}{2\pi {\epsilon }_{0}mg}$ (iii)

When one of the balls is discharged, the balls come closer and touch each other and again separate due to repulsion. The charge on each ball after touching each other is q/2. Replacing q with q/2 in equation (iii), we get

$\frac{b^3}{L}=\frac{(q/2{)}^2}{2\pi {\epsilon }_{0}mg}$ (iv)

From equations (iii) and (iv), we get

$\frac{b^3}{a^3}=\frac{1}{4}orb=\frac{a}{2^{2/3}}$