Two identical conducting balls $A$ and $B$ have positive charges $q_1$ and $q_2$ respectively. But $q_1 \neq q_2$. 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
• Correct answer
  B

  greater than that before the balls touched
• C

  same as that before the balls touched
• D

  zero

Solution

According to coulomb's law, the force of repulsion between them is given by $F=\frac{q_1 q_2}{4 \pi \varepsilon_0 r^2}$ 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 distance $r$ is $F^{\prime}=\frac{q^{\prime} \times q^{\prime}}{4 \pi \varepsilon_0 r^2}=\frac{1}{4 \pi \varepsilon_0} \frac{\left(\frac{q_1+q_2}{2}\right)\left(\frac{q_1+q_2}{2}\right)}{r^2}$ $=\frac{\left(\frac{q_1+q_2}{2}\right)^2}{4 \pi \varepsilon_0 r^2}$ $\operatorname{As}\left(\frac{q_1+q_2}{2}\right)^2 \gt q_1 q_2$ $\therefore F^{\prime} \gt F$

Hence, the correct answer is option: 'B'.