Solved: Gauss's Law Problem w/Infinite Slab of Charge

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The discussion focuses on solving a Gauss's Law problem involving an infinite slab of charge with density ρ for -b < x < b. The user initially calculated the electric field inside the slab as E = ρx / (2ε₀), which was incorrect; the correct solution is E = ρx / ε₀. The error stemmed from incorrectly applying the net flux as 2EA for the Gaussian surface. The user clarified that the Gaussian surface must be centered on the origin to maintain symmetry, which is crucial for applying Gauss's Law correctly.

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[SOLVED] Gauss's Law problem

Homework Statement


An infinite slab of charge parallel to the yz-plane has density p for -b < x < b, 0 otherwise. Find the electric field at all points.

The Attempt at a Solution


I am able to do the electric field outside the slab. But I am off by a factor of 1/2 for the electric field inside. I made a Gaussian cylinder that starts at the axis to x. I'll call the area of the circular faces A = \pi r^2. Since there are two of these faces, net flux is 2EA. The charge enclosed in the surface is \rho A x.

2 E A = \frac{\rho A x}{\epsilon_0} =&gt; E = \frac{\rho x}{2 \epsilon_0}

My book says the solution is just \rho x / \epsilon_0. So, I know the error probably came from saying the flux was 2EA. But, for the electric field outside the slab, I used 2EA for the flux, and since the charge enclosed in a cylinder running from -x to x, for |x| > b, was 2b, the factors of 2's cancel out. Here they don't. I can't figure out what I'm doing wrong.

I suppose I could just say the flux was one EA because the field at x = 0 is 0. But, for the next problem, I have to find the electric field for a slab whose density is \rho(x) = \rho_0 e^{-|x/b|}, and I still have that extra 1/2 factor in there, without the field at x = 0 being 0.
 
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By having your Gaussian surface extend from the origin to a distance x from the origin you lose the symmetry needed to apply Gauss's law. You must have your Gaussian volume centered on the origin--that way you know that the electric field is the same at each end. Have it go from -x to +x.
 
Ah, that makes sense. Thanks.
 

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