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Two identical conducting spheres $A$ and $B,$ carry equal charge. They are separated by a distance much larger than their diameter, and the force between them is $F$ . A third identical conducting sphere, $C,$ is uncharged. Sphere $C$ is first touched to $A,$ then to $B,$ and then removed. As a result, the force between $A$ and $B$ would be equal to
$\frac{{3F}}{4}$
$\frac{{F}}{2}$
$F$
$\frac{{3F}}{8}$
Solution
Spheres $A$ and $B$ carry equal charge say $'q$
$\therefore $ Force between them, $\mathrm{F}=\frac{\mathrm{k} \mathrm{qq}}{\mathrm{r}^{2}}$
When $A$ and $C$ are touched, charge on both
$\mathrm{q}_{\mathrm{A}}=\mathrm{q}_{\mathrm{C}}=\frac{\mathrm{q}}{2}$
Then when $\mathrm{B}$ and $\mathrm{C}$ are touched, charge or $B$
$\mathrm{q}_{\mathrm{B}}=\frac{\frac{\mathrm{q}}{2}+\mathrm{q}}{2}=\frac{3 \mathrm{q}}{4}$
Now, the force between charge $q_{A}$ and $q_{B}$
$\mathrm{F}^{\prime}=\frac{\mathrm{k} \mathrm{q}_{\mathrm{A}} \mathrm{q}_{\mathrm{B}}}{\mathrm{r}^{2}}=\frac{\mathrm{k} \times \frac{\mathrm{q}}{2} \times \frac{3 \mathrm{q}}{4}}{\mathrm{r}^{2}}=\frac{3 \mathrm{kq}^{2}}{8} \mathrm{r}^{2}=\frac{3}{8}\, \mathrm{F}$