Question

Two identical conducting balls $A$ and $B$ have positive charges $q_1$ and $q_2$ respectively but $q_1 \ne q_2$ The balls are brought together so that they touch each other and then kept in their original positions. the force between them is

• A. less than that before the balls touched
• B. greater than that before the balls touched
• C. same as that before the balls toucher
• D. zero

Answer 1

The Correct Option is B

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