A long cylindrical solid conductor of cross-sectional radius...

A long cylindrical solid conductor of cross-sectional radius R has linear charge density ?. a) Sketch the cylindrical conductor and a cylindrical Gaussian surface of height h and radius r where r > R. b) Use Gauss's Law to derive an expression for the electric field strength in terms of r and ? at distance r from the center of the conductor.

Transcript

00:01 So in this question, we're going to be using gauss's law, and we're told that we have a cylinder of radius a.

00:08 So i'm drawing this in cross section.

00:12 So that's an a.

00:13 And then outside that, and this is carrying a charge plus q.

00:18 And then outside that, we have a cylinder, a concentric cylinder of radius b.

00:23 So it's meant to be a circle.

00:25 And it's carrying a total charge minus 2q.

00:29 And we're told that these cylinders are conducting.

00:31 And what that means is that the charge is going to be, uh, homogeneously distributed.

00:38 So let's remember that gauss's law is that the integral of e over a surface, so the integral of the flux of the e field through a surface, is equal to the charge enclosed by that surface divided by epsilon.

00:59 So let's think first of a gaussian surface with r less than a.

01:04 This surface contains no charge, and that means that the...

01:09 Well, first of all, let's think, because we have a symmetric situation, this integral, because e is going to be the same at all points of a given radius, because we have circular symmetry, this is going to come down to 2 pi r times the modulus of the e field, equals q enclosed over epsilon naught.

01:38 So here, inside the first cylinder, so for r less than a, q enclosed is zero.

01:48 So we have that e is equal to zero.

01:51 Now let's think about a radius between a and b...

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