An infinitely long cylindrical conductor has radius r and uniform surface charge density σ. (a) In terms of σ and R, what is the charge per unit length λ for the cylinder?

Video transcript

Hey everyone. This problem is dealing with linear charge density. So we are asked to determine the linear density of a very long cylindrical rod. In terms of the charge per unit area given by this term sigma and the radius of the rod, which is given by this uppercase R. So this is a pretty straightforward question. The first thing that we need to recall is that the linear charge density equation is given by lambda equals Q. Over L. Where Q is the net charge. And so we're given a charge per unit area of sigma and Q equals the net charge. So that's gonna be the charge per unit area times the area. So sigma. And then let's recall that the area of the cylinder is two pi R times H. Or in this case. And so um by definition we can then equate the first part. This first equation will rewrite that as Q equals lambda L. And then we can set these two equations equal to each other. So we've got lambda L equals sigma times two pi are over out. So here are the L's cancel and we are left with the linear density. Lambda is equal to sigma two pi R. And that is the same as cancer. D. Just rearranged a little bit. And so that is the answer to this problem. We'll see you in the next video