Two identical conducting balls A and B have positive charges q₁ and q₂ respectively. But q₁ ≠ q₂. The balls are brought together so that they touch each other and then kept in their original positions. The force between them is

less than that before the balls touched

greater than that before the balls touched

same as that before the balls touched

zero

Solution

greater than that before the balls touched

According to Coulomb's law, the force of repulsion between them is given by $F = \frac{q_{1} q_{2}}{4 {πε}_{0} r^{2}}$

when the charged spheres A and B are brought in contact, each sphere will attain equal charge q'.

$q' = \frac{q_{1} + q_{2}}{2}$

Now, the force of repulsion between them at the same distance r is

$F' = \frac{q' x q'}{4 {πε}_{0} r^{2}} = \frac{1}{4 {πε}_{0}} \frac{(\frac{q_{1} + q_{2}}{2}) (\frac{q_{1} + q_{2}}{2})}{r^{2}} = \frac{{(\frac{q_{1} + q_{2}}{2})}^{2}}{4 {πε}_{0} r^{2}} As {(\frac{q_{1} + q_{2}}{2})}^{2} > q_{1} q_{2} ∴ F' > F$

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