# Electric field due to infinite line charge distribution

Consider an infinitely long straight line carrying uniform line charge with density $\rho&space;_{L}&space;C/m$ 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 $(0,\rho&space;,0)$ ( a point on y-axis) and assume $dQ$ is a small differential charge confirmed to a point  M $(0,0,Z)$ as co-ordinates.

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

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

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

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

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

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

$\overrightarrow{E}&space;=&space;\int&space;\overrightarrow{dE}$

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

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

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

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

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

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

$\overrightarrow{E}&space;=&space;\frac{\rho&space;_{L}}{4\pi\epsilon&space;_{o}\rho&space;}(\int_{\theta&space;=&space;\frac{-\pi&space;}{2}}^{&space;\frac{\pi&space;}{2}}&space;-\sin&space;\theta&space;\overrightarrow{a_{z}&space;}d\theta&space;+\int_{\theta&space;=&space;\frac{-\pi&space;}{2}}^{&space;\frac{\pi&space;}{2}}&space;\cos\theta&space;d\theta&space;\overrightarrow{a_{\rho&space;}})$

$\overrightarrow{E}=&space;\frac{\rho&space;_{L}}{4\pi\epsilon&space;_{o}\rho&space;}(0+2\overrightarrow{a_{\rho&space;}})$

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

$\overrightarrow{E}$ 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 $\rho&space;_{L}$.

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