Finding E-field of 2 concentric clyinders

[mr_coffee]

Hello everyone. I'm having troubles figuring out the E-field of two long, charged, thin-walled, concentric clyinders. Its question #6 on the picutre. The image has my work, it seems my apporach was too simplistic, because the problem gives you 2 radi, but then gives u another 2 radi to find the E field at, so I totally didn't even use the first 2 radi they gave me. I also was wondering if someone could check #4 and #5 to see if i did them right. Thanks! http://img138.imageshack.us/img138/743/jkl9pg.jpg

 

[whozum]

The field is created by the cylinders with the first two radii, and they want you to find the E field strength at those two points a) and b) 's radii.

For a) remember what Gauss' Law says about E fields inside conductors.

 

[mr_coffee]

Thanks for the reply. For a, when you said remember what Gauss's Law says about E-fields inside conductors...I know a charge inside a conductor is 0. Also I know if the charge is enclosed in a conductor it will be radially outward, but i don't see anywhere in the book that says about E-fields in a conductor. I found the magnitude and direction of the e-fields of the inner and outter clyinder. You said now they want me to find the E field strength at those two points. I'm confused on what equation I would use. I already used the E = \gamma/(2PIEoR). Any hints on which one i should use? Thanks.

 

[whozum]

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gausur.html#c2


The E field inside any closed conductor is zero.

I know a charge inside a conductor is 0.

This isn't correct and is probably what your mistaking with the correct statement above. The conducting surface will arrange its charges so that the net field inside the conductor is zero.

If you have an expression for the E field of each cylinder, then by superposition, the E field at any point is the vector sum of all the individual E fields.