Question

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
• B. Greater than that before the balls touched
• C. Same as that before the balls touched
• 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.
$∴ \, \, q_{1}^{′}=\frac{q_{1} + q_{2}}{2}$
Similarly $q_{2 \, }^{′}=\frac{q_{1} + q_{2}}{2}$
Therefore, new force of repulsion between spheres $A \, \, and \, B$ is
$F^{′}=\frac{1}{4 \pi ε_{0}}\frac{\left[\frac{q_{1} + q_{2}}{2}\right] \left[\frac{q_{1} + q_{2}}{2}\right]}{r^{2}}$
$ \, =\frac{\left[\frac{q_{1} + q_{2}}{2}\right]^{2}}{4 \pi ε_{0} r^{2}}$
As $\left[\frac{q_{1 + q_{2}}}{2}\right]^{2}>q_{1}q_{2}$
$∴ \, \, F^{′}>F$