Electric field between electrodes

1. Jun 3, 2014

Shot

1. The problem statement, all variables and given/known data

Calculate electric field between two electrodes — one square and one circular. Dimesions of the electrodes are given (diameter a and length of the square side b) as well as their potentials. The square electrode is placed inside the circular one.

2. Relevant equations

$$F = k\frac{Qq}{r^2}$$
$$E = \frac{F}{q}$$
$$V = \frac{E}{q}$$

3. The attempt at a solution

I'm new to this forum and I'm not a physics guru (last time when I had touch with Physics was 3 years ago). Now I have to solve one problem in SciLab, but I can't figure out where to start. I'm asking for some help - some hints where I could start or some sketch solutions how to do this.

After the research I think I should solve two Laplace equations (in polar and cartesian coordinates) and then sum the results, but I am not sure about that. Also, I know that I need a grid N x N (where N = e.g. 60) to have nice computations.

2. Jun 5, 2014

BvU

Hello Shot, and welcome to PF.

Your 1 and 2 equations I understand. 3 is a mystery to me, also dimension-wise. You sure ?

Anyway, they apply to charges and can be used for charge distributions. Not what you have at hand here.

Your hunch that you are expected to solve Laplace equation is much more appealing. What boundary conditions do you think have to be imposed ?

If you are lucky, the exercise wants the circular electrode in the center of the square (you don't mention that!) and you can use symmetry.

What exactly are nice computations ? Are you asked to do this numerically or analytically ?

(PS can't help you further until next wednesday, so hopefully others jump in...)

3. Jun 6, 2014

bloby

In principle you could use the first two equations, but you would have to sum/integrate over the charges(Q) on the surfaces of the electrodes, thus having to solve for these charges using more equations.
In the third equation E is the electric potential energy, not the electric field. The relation between V and E is $E=-\nabla V$

4. Jun 6, 2014

rude man

Only way I see is solving Laplace's equation with the obvious boundary conditions.

Can't imagine a closed-form solution. A job for finite-element analysis software!

5. Jun 6, 2014

bloby

The first two equations give a relation between the eletric field and the charges. Gauss's law $\nabla \cdot E = \frac{\rho}{\epsilon_0}$ gives this in a more usefull way here.

6. Jun 6, 2014

bloby

I was thinking about something like: the Q's must be such that E = 0 inside the electrodes and normal to the surfaces and produce the given potentials on the surfaces. I would say there is only one solution, but I'm not absolutely sure.
Yes awful!

7. Jun 6, 2014

rude man

Why is E=0 inside the electrodes? I see the E field as zero only at the center, assuming the electrodes are concentric.

I believe the Uniqueness Theorem guarantees that there is only one solution.

8. Jun 6, 2014

bloby

I assumed they were conductors. (If E is non null, charges would move until E = 0 inside)

For dielectrics hem...

Thanks for the theorem.

9. Jun 6, 2014

rude man

Oh, sorry. Of course. I thought you meant inside the circle and square.

10. Jun 7, 2014

Shot

I've googled some knowledge about FDM method for solving Laplace equations (since I have to solve this problem numerically using SciLab) and for rectangular rod I've obtained the following equation (in Cartesian coordinates):
$$T_{i+1,j} + T_{i-1,j} + T_{i,j+1} + T_{i,j-1} + 4T_{i,j} = 0$$ (assuming Δx = Δy)

And for circular rod (which is in the center of the rectangle/square):
$$Aψ_{i,j} + Bψ_{i+1,j} + Cψ_{i-1,j} + Dψ_{i,j-1} + Dψ_{i,j+1} = 0$$
where:
h = Δr
k = Δθ

$$A = -2(\frac{1}{h^2} + \frac{1}{k^2(a+ih)^2})$$
$$B = \frac{1}{h^2} + \frac{1}{(a+ih)(2h)}$$
$$C = \frac{1}{h^2} - \frac{1}{(a+ih)(2h)}$$
$$D = \frac{1}{k^2(a+ih)^2}$$

Are these equations correct? They will give me a set of linear equations that I can solve using Gauss-Seidel method, right? What units should I use to describe results in the specific points in the grid? Volts?

11. Jun 7, 2014

rude man

Beyond me at this point. Sorry.