A point charge $q = 1\mu\text{C}$ is located at a distance 2 cm from one end of a thin insulating wire of length 10 cm having a charge $Q = 24\mu\text{C}$, distributed uniformly along its length, as shown in figure. Force between $q$ and wire is ______ N.
(Use : $\dfrac{1}{4\pi\epsilon_0} = 9\times10^9\ \text{N}\cdot\text{m}^2/\text{C}^2$ )
Numerical value type. Enter your answer.
Answer: 90
Take the charge at the origin and the wire from $x = a = 0.02$ m to $x = a + L = 0.12$ m, with $\lambda = Q/L$.
$$F = \int_a^{a+L}\frac{kq\lambda\,dx}{x^2} = kq\lambda\left(\frac1a - \frac1{a+L}\right) = \frac{kqQ}{a(a+L)}$$
$$F = \frac{9\times10^9\times1\times10^{-6}\times24\times10^{-6}}{0.02\times0.12} = \frac{0.216}{0.0024} = 90\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics