Question

Find the electric field intensity at a point $P$ which is at a distance $R$ (point lying on the perpendicular drawn to the wire at one of its end) from a semi-infinite uniformly charged wire. (Linear charge density $=\lambda$.)

Answer 1

Field at point $P$ is due to an elemental charge $dq(=\lambda d\ell )$
$dE=k\frac{dq}{(R^2+{\ell }^2)}$
The component along x-axis
$d{E}_x=\frac{kdq\cos \theta }{(R^2+{\ell }^2)}=\frac{k\left(dq\right)R}{(R^2+{\ell }^2{)}^{3/2}}$
$\therefore E_x=\underset{0}{\stackrel{\infty }{\int }}\frac{1}{4\pi {\in }_0}\frac{\lambda d\ell R}{(R^2+{\ell }^2{)}^{3/2}}⟹E_x=\frac{\lambda }{4\pi {\in }_{0}R}$
Similarly, $E_y=\frac{1}{4\pi {\in }_0}\frac{\lambda \left(d\ell \right)\ell }{(R^2+{\ell }^2{)}^{3/2}}$
$\therefore E_y=\frac{\lambda }{4\pi {\in }_{0}R}$
$\therefore E=\sqrt{E_x^2+E_y^2}=\frac{\lambda }{2\sqrt{2}\pi {\in }_{0}R},\tan \theta =\frac{E_y}{E_x}$
$⟹\theta ={45}^o$