Q 12-01-104JEE MainJEE Main 2022 (27 Jul, Shift 2)Easy
A charge of $4\ \mu$C is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be:
Answer: (B) $2\ \mu$C and $2\ \mu$C
With parts $q$ and $Q - q$, $F \propto q(Q - q)$.
$$\frac{d}{dq}\left[q(Q - q)\right] = Q - 2q = 0 \Rightarrow q = \frac Q2 = 2\ \mu\text{C}$$
So the charges are $2\ \mu$C and $2\ \mu$C.
Solution by Sreeraj P, M.Sc Physics