Assume an infinite charge distribution that lies on the xy plane. If the charge density is σ, then what is the electric field of the distribution ?
First of all we should notice that since the distribution is infinite, any rotation shouldn't change anything; that is, a symmetry exists. This means that we can wisely use Gauss's Law in the right gaussian surface we select. If we take a gaussian surface to be a cylinder whose axis is perpendicular to xy plane, then things are going to be very easy.
The upper and lower surface of the cylinder have equal electric fluxes. That is, Φ1=2ΕΑ, where E is the electric field and A is the upper (or bottom) surface. In addition, the flux of the remaining surface of the cylinder is zero - the electric field is perpendicular to xy plane (due to symmetry). Thus, the total flux is equal to Φ=Φ1.
Using Gauss's Law, we get that Φ= q/ε , where q the total charge inside the cylinder. But q=σΑ. So, 2EA=σΑ/ε. Thus, E = σ/2ε.
We see that the electric field is independent of the height above the distribution - the field is homogenous and no matter where we are in space it has the same value.
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