# Homework Help: Electric field in a resistive, shorted coaxial cable

1. Oct 7, 2015

1. The problem statement, all variables and given/known data
Consider a shorted coaxial line with a resistive inner conductor with radius a, and a perfect outer conductor located at radius b. A DC voltage is applied at the input end. Find the electric and magnetic fields in the dielectric region (a<r<b), assuming the battery sets up a potential that varies as ln(r) on the input surface.

2. Relevant equations
Not exactly sure if there is a better way, but I started by looking at Poisson's equation
$$\nabla^2 V = \frac{\rho}{\epsilon}$$

3. The attempt at a solution
Since there is no charge in the dielectric, Poisson's equation becomes

$$\nabla^2 V(r,z) = 0$$

where V has rotational symmetry. In cylindrical coordinates,

$$\frac{1}{r} \frac{\partial }{\partial r} \left( r \frac{\partial V(r,z)}{\partial r} \right) + \frac{\partial^2 V(r,z)}{\partial z^2} =0$$

The boundary conditions look like:

$$V(r=a) = V_0 - \frac{V_0 z}{L}\\ V(r=b) = 0\\ V(z=L) = 0\\$$

I'm not sure how to solve this differential equation. Can you use separation of variables to get:

$$V(r,z) = R(r)Z(z) \\ \frac{1}{R}\frac{1}{r} \frac{\partial}{\partial r}\left( r \frac{\partial R}{\partial r} \right) = - \frac{1}{Z} \frac{\partial^2 Z}{\partial z^2}\\$$

Even if I do that though, I'm not quite sure how to solve the R portion. Any nudge in the right direction would be appreciated. Thanks!

2. Oct 11, 2015

### Daz

Hi EEGrad. I notice it’s been quite a while since you posted your question - have you found a solution, yet? If not, here are some suggestions. You’ll have to think about it and decide for yourself whether this is a nudge in the right direction or not.

Firstly, you’re correct in that a full solution of Poisson’s equation would enable you to find the E field everywhere, including the fringing fields at the input face, but in real-world problems solving Poisson’s equation is often fiendishly difficult. So I like to look for an easier solution. When I see an electrostatics problem involving lots of highly symmetrical perfect conductors I think to myself “That looks like a problem which might be amenable to solution by the method of images.”

Reading the question a little more carefully I see: “assuming the battery sets up a potential that varies as ln(r) on the input surface.” That is the potential you would find near an infinite charged cylinder: C1 - C2 * ln(r), where C1 and C2 are constants. In other words the question is telling you to assume that the potential (and therefore the radial component of E) at the input face looks like you would find for an infinite cylindrical conductor. An invitation to ignore fringing fields at the input, perhaps?

What I would do is start with an infinite ohmic cylinder and calculate the field due to that ignoring everything else. Then wrap the outer conductor around it and find the field due to the outer conductor by the method of images.

Hint: it might help if you define your z-axis so that V(z) passes through zero at the origin - that may be handy when you come to “close off” the shorted end of the cable by the method of images.

Can you see how to calculate B?