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 touched
• D. Zero

Answer 1

Answer: (B)

greater than that before the balls touched

According to coulomb's law, the force of repulsion between them is given by $F=\frac{q_1{q}_2}{4\pi {\epsilon }_0{r}^2}$
When the charged spheres A and B are brought in contact, each sphere will attain equal charge $q^{\prime }$
$q^{\prime }=\frac{q_1+q_2}{2}$
Now, the force of repulsion between them at the same direction r is
$F^{\prime }=\frac{q^{\prime }\times q^{\prime }}{4\pi {\epsilon }_0{r}^2}=\frac{1}{4\pi {\epsilon }_0}\frac{\left(\frac{q_1+q_2}{2}\right)\left(\frac{q_1+q_2}{2}\right)}{r^2}$
$=\frac{{\left(\frac{q_1+q_2}{2}\right)}^2}{4\pi {\epsilon }_0{r}^2}$
As ${\left(\frac{q_1+q_2}{2}\right)}^2>q_1{q}_2$
$\therefore F^{\prime }>F$.