Question

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

Answer 1

Net electric field $=\int dE\cos \theta$

$=\int \frac{1}{4\pi {ϵ}_0}\frac{\frac{q}{1}.dx}{(x^2+125)}.\frac{x}{\sqrt{x^2+125}}={\int }_{-10}^{+10}\frac{1}{4\pi {ϵ}_0}.\frac{q}{1}\frac{xdx}{\sqrt{x^2+125.}(x^2+125)}=5.2\times {10}^{7}N/C$