# The total charge between two cylinders

1. Mar 27, 2015

### Vernes

1. The problem statement, all variables and given/known data
We have a discharge tube where the cathode is a cylinder with radius a and the anode is a coaxial cylinder with radius b, a<b. Both cylinders have length L, (Note that L can not be seen as large).The potential of the cathode is 0 and the potential of the anode is U>0. The electric field strength at the surface of the cathode is 0. If we assume that the electron cloud between the anode and cathode is uniformly distributed, what is then the total charge Q of the electron cloud?

2. Relevant equations
$$\oint \bar E \cdot d\bar s = \frac {Q_{inside}} {\epsilon _0}$$
$$\oint \bar E \cdot d\bar l = \Delta V$$

3. The attempt at a solution
My idea is to put a gaussian surface surounding the space between the anode and the cathode and then use Gauss's law to get the total charge inside. To be able to do this i need to know the electric field between the anode and the cathode and I can't seem to figure out how i can express this in a convenient way, mainly because L can't be seen as large and therefore I believe you need to consider the fringe effects? Maybe I should use the second equation to find the electric field but I can't seem to get my head around how this could be done.

2. Mar 28, 2015

### Staff: Mentor

Are you sure this has a nice answer if L cannot be seen as large compared to b-a?

This is a two-dimensional problem. Solving the Poisson equation analytically does not look easy.

3. Mar 28, 2015

### rude man

Been watching this thread, agree with mfb. The poisson equation in just r would have been easy, but if z is added as a second coordinate I don't see it either.

4. Mar 28, 2015

### Vernes

Ok, I got the right answer now. L could acctually be seen as large compared to a and b...
I'm so sorry for giving you guys the wrong information.
Still the mere fact that you said that you also found it hard to solve when L couldn't be seen as large made me rethink and get the right answer so thank you!