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 :

• A

  $\frac{3 F}{5}$
• B

  $\frac{2 F}{3}$
• C

  $\frac{F}{2}$
• Correct answer
  D

  $\frac{3 F}{8}$

Solution

$\mathrm{F}=\frac{\mathrm{kQ}^2}{\mathrm{r}^2}$ on touching; $F^{\prime}=\frac{k(Q / 2)(3 Q / 4)}{r^2}$ $F^{\prime}=\frac{3 F}{8}$
After touching $\rightarrow$ $(\mathrm{A}) \&(\mathrm{C}) \rightarrow$ Total charge $\rightarrow \mathrm{q}+0=\mathrm{q}$ $\mathrm{Q}_{\mathrm{A}}^{\prime}=\frac{\mathrm{q}}{2}, \quad \mathrm{Q}_{\mathrm{C}}^{\prime}=\frac{\mathrm{q}}{2}$ (B) & (C) $\begin{aligned} & \text { Total charge }=q+\frac{q}{2}=\frac{3}{2} q \\ & Q_B^{\prime}=\frac{3}{4} q \quad Q_C^{\prime}=\frac{3}{4} q \end{aligned}$

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