Question

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

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

Answer 1

The correct option is D greater than that before the balls touched

Given two identical conducting balls A and B have positive charges $q_1$ and $q_2$ respectively. The balls are brought together so that they touch each other and then kept in their original positions. We have to find the force between them.

So, initially let the charges are at a distance $r$ apart. So, force between them is $F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$.

So, when they are touched, charges on each ball is $\frac{q_1+q_2}{2}$

[Since two balls are identical]

So, now force between them is: $F=\frac{1}{4\pi {\epsilon }_0}{\left(\frac{q_1+q_2}{2}\right)}^2\frac{1}{r^2}$

We can clearly see that $F^{\prime }>F$.

So, the force between the balls is greater than that before the balls touched.