Problem 1 The figure below shows a section of a long...

Problem 1 The figure below shows a section of a long thin-walled metal tube of radius R = 3.0 cm, with a positive charge per unit length of ? = 2.0 × 10^-8 C/m on its surface: Problem 1 - Cont. Obtain equations for the electric field E as a function of the distance r from the tube axis, for both: (a) Positions outside the tube (r > R) (b) Positions inside the tube (r < R) (c) Draw E(r) for the range r = 0 to r = 5.0 cm

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Transcript

00:01 Okay, so here we're looking at a section of tube like this.

00:06 It's like a cylinder.

00:07 And its radius r is 0 .03 meters.

00:13 So that's r.

00:15 And we're told that its charged density is 2 times 10 to the minus 8 kouloms per meter.

00:21 So for each meter in length, you have 2 times 10 to the minus 8 koulogs splying on it.

00:28 So we're asked to figure out the electric field for, firstly, the electric field for our bigger, for our gaussian, we're going to take our gauss surface to be a cylinder again.

00:43 And to start off with, they want the electric field for outside of this tube.

00:49 So i .e.

00:51 For little r greater than big r, where little r is the radius of our gaussian cylinder and big r is the radius of our actual tube.

01:00 So we use gauss's law.

01:08 Now, for a little or greater than bigger than the enclosed charge q is going to be just lambda times the length l of our gaussian cylinder.

01:31 And over here we're going to have the electric field in the direction of the normal to the area.

01:40 This is going to be the surface area of the cylinder, of our gals cylinder, which is 2 pi r times l, times the electric field.

01:54 They're going in the same direction, right? the electric field's radiating outwards like this, and so is the normal to the area of the curved surface of the cylinder...

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