JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Electrostatics

The wire carries uniform linear charge density $\lambda =\frac{Q}{L}=\frac{24\times 10^{-6}}{0.1}\text{C/m}$. An element at distance $x$ from the point charge exerts $dF=\frac{kq\lambda dx}{x^2}$, and integrating across the wire, whose near and far ends are $0.02$ m and $0.12$ m from $q$, $F=kq\lambda {\int }_{0.02}^{0.12}\frac{dx}{x^2}=kq\lambda \left(\frac{1}{0.02}-\frac{1}{0.12}\right)$. Substituting, $F=9\times 10^9\times 10^{-6}\times \frac{24\times 10^{-6}}{0.1}\times \frac{125}{3}=90$ N. The force between the charge and the wire is $90$ N.