Question

A 10-cm long rod carries a charge of +50 μC distributed uniformly along its length. Find the magnitude of the electric field at a point 10 cm from both ends of the rod.

Answer 1

Net electric field = ∫dE cos θ
$$\begin{gathered} \int \frac{1}{4\mathrm{\pi }{\in }_0}\frac{{\displaystyle \frac{q}{l}}.dx}{\left(75-x^2\right)}.\frac{x}{\sqrt{75-x^2}} \\ =\underset{-5}{\overset{+5}{\int }}\frac{1}{4\mathrm{\pi }{\in }_0}.\frac{q}{l}\frac{xdx}{\sqrt{75-x^2.}\left(75-x^2\right)} \\ =5.2\times {10}^7\mathrm{N}/\mathrm{C} \end{gathered}$$