Electric field between electrodes

In summary: 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,
  • #1
Shot
2
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



[tex]F = k\frac{Qq}{r^2}[/tex]
[tex]E = \frac{F}{q}[/tex]
[tex]V = \frac{E}{q}[/tex]


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.
 
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  • #2
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
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
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
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
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!
 
  • #7
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.
 
  • #8
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.
 
  • #9
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):
[tex]
T_{i+1,j} + T_{i-1,j} + T_{i,j+1} + T_{i,j-1} + 4T_{i,j} = 0[/tex] (assuming Δx = Δy)

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

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

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.
 

1. What is an electric field between electrodes?

An electric field between electrodes is a region in space where charged particles experience a force due to the presence of two or more electrodes with different electrical potentials. The electric field is created by the difference in charges between the electrodes and is responsible for the movement of charged particles.

2. How is the strength of an electric field between electrodes measured?

The strength of an electric field between electrodes is measured in volts per meter (V/m) or newtons per coulomb (N/C). These units represent the force exerted on a charged particle per unit charge in the direction of the electric field. The stronger the electric field, the greater the force experienced by the charged particles.

3. What factors affect the strength of an electric field between electrodes?

The strength of an electric field between electrodes is affected by the distance between the electrodes, the magnitude of the charges on the electrodes, and the dielectric constant of the material between the electrodes. The electric field strength decreases as the distance between the electrodes increases and increases as the magnitude of the charges on the electrodes increases. The dielectric constant of the material between the electrodes also plays a role in determining the strength of the electric field.

4. Can an electric field exist between two electrodes with the same charge?

No, an electric field cannot exist between two electrodes with the same charge. According to the principle of superposition, the electric field created by two or more point charges is the vector sum of the electric fields created by each individual charge. If the charges on the electrodes are the same, the electric fields will cancel each other out, resulting in no net electric field between the electrodes.

5. What are some real-world applications of electric fields between electrodes?

Electric fields between electrodes have a variety of real-world applications. They are used in electronic devices such as capacitors, batteries, and transistors. They are also used in medical devices, such as pacemakers, and in industrial processes, such as electroplating. Electric fields between electrodes are also utilized in particle accelerators and in the production of high-voltage power lines.

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