Gauss's Law and the Electric Field Due to a Slab of Charge
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5.06 hahahha gay
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what about a point just outside of the slab? I have a homework problem and the first equation you made worked, but when I had a point above the slab, it failed. I tried then making my guassian volume to the surface, but that was also wrong. I.E. My slab exists between -5 and 5 cm. At 1.5 cm, the equation works. at 6.5, it doesnt. I assume that was because there was no charge enclosed above the plated, so I modified guassian surface to go just to 5. still wrong. I have no idea what Im doing wrong
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Why is their electric field inside the long and wide slab?
Since it is a conductor, it should not be having any electric field in it.
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Have your Gaussian Surface be a cylinder that has its bottom at the center of the slab and a top at 6.5 cm. But the total charge enclosed is the volume charge density times the volume of the sphere that is in the slab. That would be pi*r^2*(.05m). The only flux is out the top of the cylinder.
lasseviren1 4 months ago