1. Electric Charges and Fields
hard

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

A

$\frac{{3F}}{4}$

B

$\frac{{F}}{2}$

C

$F$

D

$\frac{{3F}}{8}$

(JEE MAIN-2018)

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 $q_{B}$

$\mathrm{F}^{\prime}=\frac{\mathrm{k} \mathrm{q}_{\mathrm{A}} \mathrm{q}_{\mathrm{B}}}{\mathrm{r}^{2}}=\frac{\mathrm{k} \times \frac{\mathrm{q}}{2} \times \frac{3 \mathrm{q}}{4}}{\mathrm{r}^{2}}=\frac{3 \mathrm{kq}^{2}}{8} \mathrm{r}^{2}=\frac{3}{8}\, \mathrm{F}$

Standard 12
Physics

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