Two identical conducting balls \(A\) and \(B\) have positive charges \(\mathrm{q}_{1}\) and \(\mathrm{q}_{2}\) respectively but \(\mathrm{q}_{1} \neq \mathrm{q}_{2}\). The balls are brought together so that they touch each other and then kept in their original positions. The force between them is

Original charges on spheres \(A\) and \(B\) be \(\mathrm{q}_{1}\) and \(\mathrm{q}_{2}\) respectively.
Distance between the two spheres \(=\mathrm{r}\)
Since, both the spheres are of same size, they will possess equal charges on being brought in contact.
\(\therefore \mathrm{q}_{1}^{\prime}=\frac{\mathrm{q}_{1}+\mathrm{q}_{2}}{2}\)
Similarly, \(\mathrm{q}_{2}^{\prime}=\frac{\mathrm{q}_{1}+\mathrm{q}_{2}}{2}\)
Therefore, new force of repulsion between spheres A and B is
\(\mathrm{F}^{\prime} =\frac{1}{4 \pi \varepsilon_{0}} \frac{\left[\frac{\mathrm{q}_{1}+\mathrm{q}_{2}}{2}\right]\left[\frac{\mathrm{q}_{1}+\mathrm{q}_{2}}{2}\right]}{\mathrm{r}^{2}} \)
\(=\frac{\left[\frac{\mathrm{q}_{1}+\mathrm{q}_{2}}{2}\right]^{2}}{4 \pi \varepsilon_{0} \mathrm{r}^{2}} \)
\(\therefore \text {As} \quad\left[\frac{\mathrm{q}_{1}+\mathrm{q}_{2}}{2}\right]^{2}>\mathrm{q}_{1} \mathrm{q}_{2} \)
\(\therefore \mathrm{F}^{\prime}>\mathrm{F}\)