E due to infinite line charge distribution

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Consider an infinitely long straight line carrying uniform line charge with density $\rho _{L} C/m$ and lies on Z-axis from $-\infty$ to $+\infty$ .

Consider a point P at which Electric field intensity has to be determined which is produced by the line charge distribution.

from the figure let the co-ordinates of P are $(0,\rho ,0)$ ( a point on y-axis) and assume $dQ$ is a small differential charge confirmed to a point M $(0,0,Z)$ as co-ordinates.

$\therefore dQ$ produces a differential field $\overrightarrow{dE}$

$\overrightarrow{dE}=\frac{dQ}{4\pi \epsilon _{o}R^{2}}\widehat{a_{r}}$

the position vector $\overrightarrow{R}=-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}$ and the corresponding unit vector $\widehat{a_{r}} =\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{\sqrt{\rho ^{2}+Z^{2}}}$

$\therefore \overrightarrow{dE} =\frac{dQ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})$

$therefore \overrightarrow{dE} =\frac{\rho _{L}dZ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})$

then the Electric field strength $\overrightarrow{E}$ produced by the infinite line charge distribution $\rho _{L}$ is

$\overrightarrow{E} = \int \overrightarrow{dE}$

$\overrightarrow{E} = \int_{z=-\infty }^{\infty }\frac{\rho _{L}dZ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})$

to solve this integral let $Z= \rho \tan \theta \Rightarrow dZ=\rho \sec ^{2}\theta d\theta$

as $Z\rightarrow -\infty \Rightarrow \theta \rightarrow \frac{-\pi }{2}$

$Z\rightarrow \infty \Rightarrow \theta \rightarrow \frac{\pi }{2}$

$\therefore \overrightarrow{E} = \int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} \frac{\rho _{L}}{4\pi\epsilon _{o}}(\frac{-\rho ^{2}\\sec ^{2}\theta \tan \theta d\theta \overrightarrow{a_{z}}+\rho ^{2}\sec ^{2}\theta d\theta \overrightarrow{a_{\rho }}}{(\rho ^{2}+\rho ^{2}\tan ^{2}\theta )^{\frac{3}{2}}})$

$\overrightarrow{E} = \int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} \frac{\rho _{L}}{4\pi\epsilon _{o}}(\frac{-\rho ^{2}\\sec ^{2}\theta \tan \theta d\theta \overrightarrow{a_{z}}+\rho ^{2}\sec ^{2}\theta d\theta \overrightarrow{a_{\rho }}}{\rho ^{3}\sec ^{3}\theta })$

$\overrightarrow{E} = \frac{\rho _{L}}{4\pi\epsilon _{o}\rho }$

$\overrightarrow{E}= \frac{\rho _{L}}{4\pi\epsilon _{o}\rho }$

$\therefore \overrightarrow{E}= \frac{\rho _{L}}{2\pi\epsilon _{o}\rho }\overrightarrow{a_{\rho }} Newtons/Coulomb$ .

$\overrightarrow{E}$ is a function of $\rho$ only, there is no $\overrightarrow{a_{z}}$ component and $\rho$ is the perpendicular distance from the point P to line charge distribution $\rho _{L}$ .