Electrostatics Part 3

Concept:The wire is a continuous distribution of charge, so the net force on the point charge $q$ is obtained by integrating the Coulomb force due to each infinitesimal element of the wire.Both $q$ and $Q$ are positive, so the force is repulsive (directed away from the wire).Step 1 — Linear charge density of the wire:$\lambda = \dfrac{Q}{L} = \dfrac{24 \times 10^{-6}}{0.1} = 2.4 \times 10^{-4}\ \mathrm{C/m}$.Step 2 — Force due to one element:Take an element of length $dx$ at a distance $x$ from $q$, so its charge is $dQ = \lambda\,dx$.$$dF = \frac{1}{4\pi\varepsilon_0}\frac{q\,dQ}{x^2} = \frac{k\,q\,\lambda}{x^2}\,dx, \qquad k = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \mathrm{N\,m^2/C^2}$$Step 3 — Integrate along the wire:The near end of the wire is at $x_1 = 2\ \mathrm{cm} = 0.02\ \mathrm{m}$ and the far end at $x_2 = 2 + 10 = 12\ \mathrm{cm} = 0.12\ \mathrm{m}$.$$F = \int_{0.02}^{0.12} \frac{k\,q\,\lambda}{x^2}\,dx = k\,q\,\lambda\left[-\frac{1}{x}\right]_{0.02}^{0.12} = k\,q\,\lambda\left(\frac{1}{0.02} - \frac{1}{0.12}\right)$$Since $\lambda = \dfrac{Q}{L}$ and $L = x_2 - x_1$, this reduces to the compact result$$F = \frac{k\,q\,Q}{x_1 x_2} = \frac{k\,q\,Q}{0.02 \times 0.12}$$Step 4 — Substitution:$k = 9 \times 10^9\ \mathrm{N\,m^2/C^2}$, $q = 1\,\mu\mathrm{C} = 10^{-6}\ \mathrm{C}$, $Q = 24\,\mu\mathrm{C} = 24 \times 10^{-6}\ \mathrm{C}$.$$F = \frac{9 \times 10^9 \times 10^{-6} \times 24 \times 10^{-6}}{0.02 \times 0.12} = \frac{0.216}{0.0024} = 90\ \mathrm{N}$$Answer:The force between $q$ and the wire is $90\ \mathrm{N}$ (repulsive, since both charges are positive), so the net force is $90$ N.