Two identical charged conducting spheres $A$ and $B$ have their centres separated by a certain dista

Question

Two identical charged conducting spheres $A$ and $B$ have their centres separated by a certain distance. Charge on each sphere is $q$ and the force of repulsion between them is $F$. $A$ third identical uncharged conducting sphere is brought in contact with sphere $A$ first and then with $B$ and finally removed from both. New force of repulsion between spheres $A$ and $B$ (Radii of $A$ and $B$ are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:

Professor PC Thomas Classes · Answer

(b) : According to Coulomb's law, $F=\frac{k q^2}{R^2}$ By touching an uncharged sphere $C$ with charged sphere $A$, charge on each sphere $A$ and $C$ becomes $q / 2$. \Now, when charged sphere $C$ kept in contact with sphere $B$, then total charge is, $\frac{q}{2}+\frac{q}{1}=\frac{3 q}{2}$ Then, Charge on $B=\frac{3 q}{4}$ Charge on C $=\frac{3 q}{4}$ \ Now, charge on $A$ is $\frac{q}{2}$ and charge on $B$ is $\frac{3 q}{4}$.\ So, $F^{\prime}=\frac{k}{R^2}\left(\frac{q}{2}ight)\left(\frac{3 q}{4}ight)=k \frac{3}{8} \frac{q^2}{R^2} \quad$ or $\quad F^{\prime}=\frac{3}{8} F$