# Coaxial pair of infinitely long charged solid conductors

1. Jul 4, 2006

### ness9660

A) By definition linear charge density is Q/L. So for the inner conductor with 2λ I want to say it is 2Q/infinity but this cannot be right. Im sure using Guass's law produces a correct answer but I cannot see anyway to relate it to λ or length of the conductors for that matter

B) For this part I would assume you have to setup two integrals, one for the field produced by the inner conductor and for the outer conductor? Im thinking that dA would relate to cross sectional area of the conductors and not a differential square on the surface?

2. Jul 5, 2006

### Andrew Mason

The infinite length means that there is no horizontal component to the field (horizontal components are equal and opposite). So the gaussian surface to use is a ring of width dL centred on the axis of the cable.

Place a gaussian ring surface inside the outer conductor. What is the field inside the outside conductor (inside any conductor)? What must the total enclosed charge be? Since the charge density of the inner conductor is 2λ, what must the charge density on the inner surface be?

To do the outer surface, place a gaussian ring around the whole cable. What is the total enclosed charge? That and the previous answer should enable you to find the charge density on the outer surface.
Just apply Gauss' law: a) inside the inner conductor b) between the inner and outer conductor, c) inside the outer conductor and d) outside the outer conductor. It would help to plot the field on a graph as a function of r.

AM