JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Electrostatics

Let the initial charges on A and B be $q_A=q_B=q$. The initial force is $F=\frac{K{q}^2}{r^2}$.

When C (uncharged) is placed in contact with A, charge distributes equally: each gets $\frac{q}{2}$.

When C (now carrying $\frac{q}{2}$) is placed in contact with B (carrying $q$), the total charge $q+\frac{q}{2}=\frac{3q}{2}$ distributes equally: each gets $\frac{3q}{4}$.

Now C (charge $\frac{3q}{4}$) is placed at the midpoint between A (charge $\frac{q}{2}$) and B (charge $\frac{3q}{4}$), each at distance $\frac{r}{2}$ from C.

Force on C from A: $F_1=\frac{K\cdot \frac{q}{2}\cdot \frac{3q}{4}}{(r/2{)}^2}=\frac{3K{q}^2}{2r^2}$ (repulsive, toward B)

Force on C from B: $F_2=\frac{K\cdot \frac{3q}{4}\cdot \frac{3q}{4}}{(r/2{)}^2}=\frac{9K{q}^2}{4r^2}$ (repulsive, toward A)

Net force: $F^{\prime }=F_2-F_1=\frac{9K{q}^2}{4r^2}-\frac{3K{q}^2}{2r^2}=\frac{9K{q}^2-6K{q}^2}{4r^2}=\frac{3K{q}^2}{4r^2}=\frac{3F}{4}$