Electric field due to infinite line charge distribution

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 }[\int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} -\sin \theta \overrightarrow{a_{z} }d\theta +\int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} \cos\theta d\theta \overrightarrow{a_{\rho }}]

\overrightarrow{E}= \frac{\rho _{L}}{4\pi\epsilon _{o}\rho }[0+2\overrightarrow{a_{\rho }}]

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

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

 

 

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Author: Lakshmi Prasanna

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.