eacn1

Oh, okay. I understand where I was going wrong. Thanks for the correction.

I'm not sure what I am doing wrong but I keep getting 10^-11.

Okay, this is what I have gotten so far..

J=I/A and rho = E/J

So E=rho *(I/4*pi*r^2)

To get the resistance, I'll take the integral of E

R = rho * integral of (1/4*pi*r^2)dr

= rho * 1/4pi * ((10^-6)-(10^-2))

= 1.35*10^-11

Okay this was just a random thought one day.. but think about this. If you were able to build a perfect ring around the earth, supported by beams, and then you took all the beams out at the same time. Would the ring float around the earth because of gravity? Or would it fall over, or spin around the earth??

Well we are given the equation, R = rho*(L/A). Resistance = Resistivity*(Length/Area). In the examples from our book, the length represents the length of a cylindrical object. This is where I am getting stuck. How would I relate the length to a spherical object?

I honestly have no idea where to start one this one. A couple other students in our class have been trying for hours. We have never seen anything like this, and there are no examples in our book.

A hollow sphere with an inner radius r=1 um, and an outer radius R=1 cm is made from copper. A current flows radially; for example, from inside toward outside. Find the resistance of the sphere. The resistivity of copper is p=1.7 * 10^-8