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forestmine
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Gauss's Law -- Insulating Material
A slab of insulating material of uniform thickness d, lying between -{d}/{2} to +{d}/{2} along the x axis, extends infinitely in the y and z directions. The slab has a uniform charge density rho. The electric field is zero in the middle of the slab, at x=0.
What is E_out, the magnitude of the electric field outside the slab?
As implied by the fact that E_out is not given as a function of x, this magnitude is constant everywhere outside the slab, not just at the surface.
Flux = E*A
ρ=q/V
I'm finding myself really confused about this problem. If I'm visualizing it correctly, if we use Gaussian surfaces, then the charge enclosed by the surface is a line charge which extends infinitely along x=0. Since the electric field is 0 at that point, the angle the field makes with the surface of the slab is simply 90 degrees. But already right there I'm confused...if the electric field is 0 at that point, can a charge exist at that point?
At that point, when asked, "What is q, the charge enclosed by the Gaussian surface," I get q=ρAd/2 coming from the fact that the charge density in this case equals the charge over a volume.
From here, given a charge, I don't understand how to get to the electric field, presumably through calculating the flux.
I'm really lost, and I'm having a hard time connecting these concepts.
I'm not looking for an answer...I just don't think I'm thinking of any of this correctly. Any help would be greatly appreciated! Thank you!
Homework Statement
A slab of insulating material of uniform thickness d, lying between -{d}/{2} to +{d}/{2} along the x axis, extends infinitely in the y and z directions. The slab has a uniform charge density rho. The electric field is zero in the middle of the slab, at x=0.
What is E_out, the magnitude of the electric field outside the slab?
As implied by the fact that E_out is not given as a function of x, this magnitude is constant everywhere outside the slab, not just at the surface.
Homework Equations
Flux = E*A
ρ=q/V
The Attempt at a Solution
I'm finding myself really confused about this problem. If I'm visualizing it correctly, if we use Gaussian surfaces, then the charge enclosed by the surface is a line charge which extends infinitely along x=0. Since the electric field is 0 at that point, the angle the field makes with the surface of the slab is simply 90 degrees. But already right there I'm confused...if the electric field is 0 at that point, can a charge exist at that point?
At that point, when asked, "What is q, the charge enclosed by the Gaussian surface," I get q=ρAd/2 coming from the fact that the charge density in this case equals the charge over a volume.
From here, given a charge, I don't understand how to get to the electric field, presumably through calculating the flux.
I'm really lost, and I'm having a hard time connecting these concepts.
I'm not looking for an answer...I just don't think I'm thinking of any of this correctly. Any help would be greatly appreciated! Thank you!