Q 12-01-153JEE MainJEE Main 2018 (16 Apr, Shift 1)Easy
Two identical conducting spheres $A$ and $B$ carry an equal charges. They are separated by a distance much larger than their diameters, 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:
Answer: (C) $\dfrac{3F}{8}$
Let each charge be $q$, so $F \propto q^2$.
- $C$ touches $A$: they share equally, so $q_A = q_C = \dfrac q2$.
- $C$ touches $B$: $q_B = q_C = \dfrac{q + q/2}{2} = \dfrac{3q}{4}$.
$$F' \propto \frac q2\cdot\frac{3q}{4} = \frac38q^2 \quad\Rightarrow\quad F' = \frac{3F}{8}$$
Solution by Sreeraj P, M.Sc Physics