Question

I was able to figure it out one way. I used a Cylindrical coordinate system and first integrated from 0 to length S.Then from 0->2pi. Took s to a point where it was much, much bigger than R, and found it all worked out. I'm not sure what the other two ways were.

We were told to just set up the integral and that using Wolfram Alpha was okay. I'm just not sure where to go from here. We've also discussed fundamental theorem of gradient and curl but I don't know if that will help you here. Any help would be appreciated.

Prove that the electric field just above a plate with an evenly distributed charge “o” is... E = zî YO using three different methods. (Hint: If the point is close enough, the plate will appear infinite)

Answer 1

direction will be perpendicular to the plate as you can see in the figure itself.

+ + + + + + + + + + + + + + + + + + + I can suggest one method wing quass law (i) Assume infinite charged plate Pod → È will be perpendicular to the plate Assume a Cylinder in such a way that is located symmetrically to plate and P is located on one of its face No flux will pass through the cylendrical surface as Ě is parallel to it È passing through both Surface of faces will be equal and opposite using Guass Law f Edi = Renclosed offe, d3 + 1 (28), (d) = (x+2) Es face- I face-2 > 2 (S732 + 2) - O Le