(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Non uniform slab of charge density [tex]\rhov = \rho_{0}cos(pi*x/(2d))[/tex]

extends infinitely in the y and z planes is present between -d < x < d.

Find the Electric Field everywhere.

2. Relevant equations

[tex]\int\epsilon_{0}{E^\rightarrow}\bullet{ds^\rightarrow}= \int\rho_{v}dv[/tex]

3. The attempt at a solution

So I have an understanding of how Gauss' Integral Law is supposed to function but this semi-infinite slab is confusing me.

First I will attempt to solve the rhs of Gauss' integral law

[tex]\int\rho_{v}dv = \int_{0}^{L}\int_{0}^{W}\int_{-d}^{d}\rho_{0}cos(\pi x/(2d))dxdydz = \int_{0}^{L}\int_{0}^{W}\frac{2d\rho_{0}sin(\pi x/(2d))}{\pi} |^{d}_{-d}dydz = \frac{4d\rho_{0}WL}{\pi} [/tex]

Now I believe I did that correct the lhs has me much more confused.

Since it is infinite about the y and z axis wouldn't it observe some of the same principles as a infinite sheet charge?

I dont believe we can define ds as the surface area is infinite. So that forces the use of the integral. Which is where I believe I am going wrong. There is only charge at the +x and -x axis outside of d and -d. How do I set up the Lhs?

[tex]\int_{0}^{L}\int_{0}^{W}\epsilon_{0}E a_{x}dydz = \epsilon_{0}EWL[/tex]

This cannot be correct because when applying Gauss' Differential Law there is no way to get the surface charge density without a x variable in which to differentiate.

Thats where I am stuck how is the LHS of Gauss' Integral Law to be implemented such that when applying Gauss' Differential Law I receive the surface charge density [tex]\rho_{0}cos(pi*x/(2d))[/tex]

Thanks for the help

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# Homework Help: Finding the Electric Field of a non-uniform Semi-Infinite Slab of Charge

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