I have tried to demonstrate the classic result that afirms that the electric field in a charged surface is proportional to the surface densitity in each point. In the books I read they assume, when using Gauss theorem with a cylinder, that the field in the lateral is orthogonal to the normal vector. It is clear to me that the field is normal to the surface, but I think that is false in the case of the cylinder, so I dont get to solve the problem.
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Thats right. The fact is that the electic field its orthogonal in the surface, but not in the rest of points of the cylinder above the surface, so the flux is not zero in the lateral. – Carlos Oct 30 '17 at 23:22
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The flux is zero on the lateral, for an infinitely large plane. The electric field is perpendicular to the plane, so the flux through any vertical surface is 0. – Andrei Oct 31 '17 at 02:14
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I think the assumption has to be that this is an electrostatic field and that the surface is the surface of a conductor. The electric potential is constant over the conductor, otherwise the charges would move. The electric field, which is the gradient of the potential, is orthogonal to surfaces of constant potential.
Robert Israel
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