Question

Obtain the formula for the electric field due to a long thin wire of uniform linear change density $X$ without using Gauss's law.

Answer 1

Consider a long thin wire of uniform linear charge density $\lambda$. Let us find electric field due to the wire at the point $P$ at a normal distance. $PN=r$ from the wire. Let $AB=dx$be a small elementary portion of the wire at a distance $x$ from point $N$.

Then charge on the elementary portion $AB,dq=\lambda dn$

Therefore, electric field due to the small portion $AB$ at the point $P$.

$dE=\frac{1}{4\pi {\in }_0}\times \frac{dp}{(op{)}^2}=\frac{1}{4\pi {\in }_0}\times \frac{\lambda dx}{(r^2+x^2)}$

The component $dE\sin \theta$ due to the different small portions, such as $AB$ will cancel rach other. Therefore, effective field due to the small portion $AB$,

$d{E}^1=sE=\cos \theta =\frac{1}{4\pi {\in }_0}\times \frac{\lambda dx}{(r^2+x^2)}\cos \theta$

Here,

$x=r\tan \theta$

$\therefore dx=r{\sec }^2\theta d\theta$

$r^2+x^2=r^2+r^2{\tan }^2\theta =r^2{\sec }^2\theta$

$\therefore d{E}^1=\frac{1}{4\pi {\in }_0}\times \frac{\lambda {\sec }^2\theta d\theta }{r^2{\sec }^2\theta }\cos \theta$

$=\frac{1}{4\pi {\in }_0}\times \frac{\lambda }{r}\cos \theta d\theta$

The electric field due to the whole long wire can be found by integrating the above between proper limits i.e., $\theta =-\pi /2$ to $\theta =-\pi /2$

$\therefore E={\int }_{-\pi /2}^{\pi /2}\frac{1}{4\pi {\in }_0}\times \frac{\lambda }{r}\cos \theta d\theta$

$=\frac{1}{4\pi {\in }_0}\times \frac{\lambda }{r}{\int }_{-\pi /2}^{\pi /2}\cos \theta d\theta =\frac{1}{4\pi {\in }_0}\times \frac{\lambda }{r}|\sin \theta {|}_{-\pi /2}^{\pi /2}$

$=\frac{1}{4\pi {\in }_0}\times \frac{\lambda }{r}|\sin \frac{\pi }{2}-\sin \left(\frac{\pi }{2}\right)|$

$=\frac{1}{4\pi {\in }_0}\times \frac{\lambda }{r}|1-(-1)|=\frac{1}{4\pi {\in }_0}\times \frac{\lambda }{r}\times 2$

$=\frac{\lambda }{2\pi {\in }_{0}r}$