How to measure electric and magnetic fields in asymmetric capacitors?

Summary: I need to build an asymmetric capacitor, but the mathematics of electromagnetics become too tough, do you have any info that can help? Appreciate it!!!

Hello,
I am an undergraduate student in engineering and I want to build an asymmetric capacitor, so I need electromagnetics which I know, but the problem with electromagnetics is that all the problems are of symmetrical bodies which simplify the problems. With asymmetric capacitors how is it possible to evaluate the potential distribution without having a given mathematical function, or the electric and magnetic fields, since there is no symmetry? I am an engineer, so if my question is kinda stupid, please don't laugh, plus I would post this question on the engineering part of the forum, and probably will also, but I think this is more of a theoretical mathematics problem.

Dale
Mentor
2021 Award
Summary: I need to build an asymmetric capacitor, but the mathematics of electromagnetics become too tough, do you have any info that can help? Appreciate it!!!

Summary: I need to build an asymmetric capacitor, but the mathematics of electromagnetics become too tough, do you have any info that can help? Appreciate it!!!

Hello,
I am an undergraduate student in engineering and I want to build an asymmetric capacitor, so I need electromagnetics which I know, but the problem with electromagnetics is that all the problems are of symmetrical bodies which simplify the problems. With asymmetric capacitors how is it possible to evaluate the potential distribution without having a given mathematical function, or the electric and magnetic fields, since there is no symmetry? I am an engineer, so if my question is kinda stupid, please don't laugh, plus I would post this question on the engineering part of the forum, and probably will also, but I think this is more of a theoretical mathematics problem.

I am not sure exactly what you mean by an asymmetric capacitor. Can you explain.

In general, to design anything you need a set of design criteria, your goals and any constraints. You would probably use those constraints to generate the mathematical functions that you need. Probably more than one will satisfy your constraints so then you judge the remaining ones based on how well they accomplish the goals.

Thank you for your answer. Basically, by asymmetric capacitor I mean, like two parallel plates as electrodes and the dielectric be a sphere, things like that.

berkeman
Mentor
Thank you for your answer. Basically, by asymmetric capacitor I mean, like two parallel plates as electrodes and the dielectric be a sphere, things like that.
Sounds like you will likely need to use numerical methods and FEA software to do your calculations. Which simulation packages are you familiar with so far?

http://people.ee.duke.edu/~drsmith/metamaterials/metamaterials_homogenization.htm • • Dale, vanhees71 and anorlunda
None, I didn't need them so far to make something. The only thing I needed till now was circuit theory.

Dale
Mentor
2021 Award
I agree with @berkeman. For something like this you will need to get some FEM modeling software capable of simulating electromagnetic or electrostatics, depending on the frequencies you are interested in. There isn’t a feasible way to do these calculations by hand.

• berkeman
berkeman
Mentor
None, I didn't need them so far to make something. The only thing I needed till now was circuit theory.
Well, it may be time to add FEA to your toolbox. You will likely use it in your work after school at some point, depending on what kind of work you end up doing.

In the mean time, are you just wanting to estimate the capacitance of such a capacitor? What physical size (ballpark) are you looking at? How big are the plates and their spacing, and what are the diameter of the sphere and its dielectric constant?

Depending on what kind of accuracy you need, you can estimate the capacitance of a capacitor like the one on the right shown in the figure above. You could start with just ratioing the volume of the sphere with the volume of the capacitor to get a ~30% estimate of the capacitance. If you need a better estimate than that, you could write your own Excel simulation to probably get it down to about a 5% accurate estimate, I would think...

Thank you all for your answers, they have been of great help!!! I will put them on the test. It's really amazing to see people from all countries come together and give advice to each other when it comes to making progress, despite the fact that most countries were in war a couple of years ago, thank you for your time to answer my questions!!!

berkeman
Mentor
It was a test? Did we pass? (If this was for schoolwork, please post such questions in the Homework Help forums next time. Thanks)

hahaha it wasnt homework

• berkeman
Hey stef3679. Take a look at Field and Wave Electromagnetics, 2nd Edition by David K. Cheng. In Chapter 3, Static Electric Fields, page 122, there is a procedure to be followed to find the capacitance of a two conductor system.

1) Choose an appropriate coordinate system. (Either rectangular, cylindrical, or spherical)
2) Assume charges +Q and -Q on the conductors.
3) Find the electric field, E, from Q by Gauss's Law or by other relations.
4) Find the arbitrary voltage, Varb, by evaluating Varb = -∫E dl.
5)Find the capacitance, C, by taking the ration Q/Varb. (C=Q/Varb).

In these steps Q is arbitary so it will always be represented by a variable. In step 5, when you divide into Q, the Q's will cancel leaving only a capacitance that relies ONLY on the physical dimensions of the two conductor system and on the electrical permittivity, ε, of the dielectric between those two conductors. If your equation has any other variables than those representing physical quantities like length and area and than the one representing electrical permittivity, you did something wrong.

Test this on a parallel plate capacitor until you are comfortable with the procedure. Remember, a parallel plate has a capacitance of C=ε(A/d) where A is the cross-sectional area of the dielectric between the two conductors and d is the distance between the two conductors covered by the dielectric.

Hope that helps.

Paul Colby
Gold Member
While FEM analysis would indeed yield the correct answer, one might try simply using a parallel plate capacitor with the volume averaged dielectric for an approximate answer. For relative dielectrics in the 1 to 4 I would expect reasonable estimates for capacitance. Just a thought.

• stef3679
Hey stef3679. Take a look at Field and Wave Electromagnetics, 2nd Edition by David K. Cheng. In Chapter 3, Static Electric Fields, page 122, there is a procedure to be followed to find the capacitance of a two conductor system.

1) Choose an appropriate coordinate system. (Either rectangular, cylindrical, or spherical)
2) Assume charges +Q and -Q on the conductors.
3) Find the electric field, E, from Q by Gauss's Law or by other relations.
4) Find the arbitrary voltage, Varb, by evaluating Varb = -∫E dl.
5)Find the capacitance, C, by taking the ration Q/Varb. (C=Q/Varb).

In these steps Q is arbitary so it will always be represented by a variable. In step 5, when you divide into Q, the Q's will cancel leaving only a capacitance that relies ONLY on the physical dimensions of the two conductor system and on the electrical permittivity, ε, of the dielectric between those two conductors. If your equation has any other variables than those representing physical quantities like length and area and than the one representing electrical permittivity, you did something wrong.

Test this on a parallel plate capacitor until you are comfortable with the procedure. Remember, a parallel plate has a capacitance of C=ε(A/d) where A is the cross-sectional area of the dielectric between the two conductors and d is the distance between the two conductors covered by the dielectric.

Hope that helps.

Thank you, i already own the book you suggested,but unfortunately Gauss' s law is only for conditions of symmetry, that's the first step in the book was to find an appropriate coordinate system,just for ease of calculations under symmetry conditions. I was covered by the above answers,thank you anyway. But i am curious,how big companies generally build things like complex electromagnetic devices or find mechanical stresses?