MHTCET 25 Apr 2025 Shift 2 Paper

Concept:When a charged ball touches an identical uncharged ball, the charge is equally shared between them.The net force is found by the vector sum of the forces from both balls.Explanation:Initially, the force between two equal charges $q$ separated by distance $r$ is:$$F=K\frac{q^2}{r^2}, \quad K=\frac{1}{4\pi\varepsilon_0}$$When the uncharged ball touches one charged ball, the charge $q$ is shared equally.So, the touched ball now has charge $\frac{q}{2}$, and the middle ball also gets charge $\frac{q}{2}$.The untouched ball still has charge $q$.The middle ball is placed midway, so its distance from each ball is $\frac{r}{2}$.Force on the middle ball due to the untouched ball:$$F_1=K\frac{q\cdot \frac{q}{2}}{\left(\frac{r}{2}\right)^2}=K\frac{2q^2}{r^2}=2F$$Force on the middle ball due to the touched ball:$$F_2=K\frac{\frac{q}{2}\cdot \frac{q}{2}}{\left(\frac{r}{2}\right)^2}=K\frac{q^2}{r^2}=F$$These two forces act in opposite directions.Therefore, the net force is:$$F'=F_1-F_2=2F-F=F$$Answer:The force experienced by the middle ball is $F$.Correct option: B.