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

Answer: (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.
$\therefore q_1^{\prime }=\frac{q_1+q_2}{2}$
Similarly, $q_2^{\prime }=\frac{q_1+q_2}{2}$
Therefore, new force of repulsion between spheres $A$ and $B$ is
$F^{\prime }=\frac{1}{4\pi {\epsilon }_0}\frac{\left[\frac{q_1+q_2}{2}\right]\left[\frac{q_1+q_2}{2}\right]}{r^2}$
$=\frac{{\left[\frac{q_1+q_2}{2}\right]}^2}{4\pi {\epsilon }_0{r}^2}$
As ${\left[\frac{q_1+q_2}{2}\right]}^2>q_1{q}_2$
$\therefore F^{\prime }>F$