How to Calculate the Electric Field Between Two Different Shaped Electrodes?

Shot
Messages
2
Reaction score
0

Homework Statement



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.

Homework Equations



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


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.
 
Physics news on Phys.org
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...)
 
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##
 
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!
 
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.
 
rude man said:
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!

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!
 
bloby said:
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!

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.
 
rude man said:
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.

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

For dielectrics hem...

Thanks for the theorem.
 
bloby said:
I assumed they were conductors. (If E is non null, charges would move until E = 0 inside)

For dielectrics hem...

Thanks for the theorem.

Oh, sorry. Of course. I thought you meant inside the circle and square.
 
  • #10
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):
<br /> 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):
<br /> Aψ_{i,j} + Bψ_{i+1,j} + Cψ_{i-1,j} + Dψ_{i,j-1} + Dψ_{i,j+1} = 0<br />
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
Beyond me at this point. Sorry.
 
Back
Top