00:01
Okay, so we're doing chapter 22, problem 33 here.
00:04
So it says a long cylindrical shell of radius big r0 and length little l, with r0 much smaller than l, so it's very long rod.
00:16
And it possesses a uniform surface charge density, sigma.
00:20
So the charge is uniform across the surface of the cylindrical shell.
00:25
So for part a, we want to determine the electric field at points r greater than r not, so outside of the cylinder.
00:38
So if we follow similarly how example 226 goes, we can draw a gaussian cylinder around our wire here, around the cylindrical shell of radius big r, and we should see that the electric field is going to be distributed perpendicular, to this surface at all points.
01:08
And that's what we need for gauss's law to apply.
01:12
So let's write out gauss's law here.
01:13
So we need the electric field if it's perpendicular to the surface at all points.
01:18
We then can pull it out and evaluate just the surface area.
01:27
And this then becomes e, the surface area of our cylindrical shell.
01:33
Then at this point is given by 2 pi r times the length l.
01:40
And this equals the charge enclosed over epsilon not.
01:46
From that we can rearrange and solve for what e is, and this is given as the surface density times the area enclosed, the surface area enclosed here, over 2 pi, epsilon, big r .l...