Discussion Closed This discussion was created more than 6 months ago and has been closed. To start a new discussion with a link back to this one, click here.

Capacitor between 2 copper wired

Please login with a confirmed email address before reporting spam

Hi everyone, im a french student in mechatronics and i have a project where i have to find how a capacitor between 2 parallel-perpendicular copper wired ( as a ' + ') (but they don't cross each other) works. I found a formula but i have now to model on comsol the system and see how the value of the capacitance evolve.

Can you tell me what study do i have to choose and what physics ?

Thank you for your time


2 Replies Last Post 2018年10月10日 GMT-4 09:38
Robert Koslover Certified Consultant

Please login with a confirmed email address before reporting spam

Posted: 6 years ago 2018年9月26日 GMT-4 16:52
Updated: 6 years ago 2018年9月26日 GMT-4 16:53

Use the electrostatics module. You'll have to make your "wires" with finite lengths and non-zero radii, for this to actually work. So what you will really be modeling is the capacitance between two narrow cylinders that cross (as a '+') with specific lengths, radii, and a specific separation between them. (I presume that is acceptable to you.) Surround all that, centered, in a box or spherical space big enough that the boundary conditions on that box are of limited importance. You can set that box to ground if you wish. The "wires" should be PECs. Set the potential on one of the cylinders to +0.5V and the other to -0.5V. This gives you 1 V of potential difference and a symmetric configuration. Solve. In post processing, compute the integral over the surface charge density on one of the cylinders. Take the absolute value, if it is negative. So that's Q. The capacitance is now by definition C = Q/V, but V is unity (i.e., 1 volt), so your surface integral is C, in Farads. You can also use the other cylinder to find Q (with a different sign) and it will likely be a slightly different value (due to minor meshing differences). This gives you an idea of the numerical errors involved. You could use the average of the two results for C, if you prefer. If they differ by a lot, you did something wrong. Another thing you can do is compute the volume integral of the electric field energy density. Call that result U (in Joules). Now set that equal to 0.5CV^2. But since V (as before) is unity, C = 2U, numerically. You can compare this result for C to the result that you got using surface charge integration. Again, it should be very similar. But if there is a big difference, then (as before) it means that there is something wrong with your model. I hope that helps.

-------------------
Scientific Applications & Research Associates (SARA) Inc.
www.comsol.com/partners-consultants/certified-consultants/sara
Use the electrostatics module. You'll have to make your "wires" with finite lengths and non-zero radii, for this to actually work. So what you will really be modeling is the capacitance between two narrow cylinders that cross (as a '+') with specific lengths, radii, and a specific separation between them. (I presume that is acceptable to you.) Surround all that, centered, in a box or spherical space big enough that the boundary conditions on that box are of limited importance. You can set that box to ground if you wish. The "wires" should be PECs. Set the potential on one of the cylinders to +0.5V and the other to -0.5V. This gives you 1 V of potential difference and a symmetric configuration. Solve. In post processing, compute the integral over the surface charge density on one of the cylinders. Take the absolute value, if it is negative. So that's Q. The capacitance is now by definition C = Q/V, but V is unity (i.e., 1 volt), so your surface integral is C, in Farads. You can also use the other cylinder to find Q (with a different sign) and it will likely be a slightly different value (due to minor meshing differences). This gives you an idea of the numerical errors involved. You could use the average of the two results for C, if you prefer. If they differ by a lot, you did something wrong. Another thing you can do is compute the volume integral of the electric field energy density. Call that result U (in Joules). Now set that equal to 0.5*C*V^2. But since V (as before) is unity, C = 2U, numerically. You can compare this result for C to the result that you got using surface charge integration. Again, it should be very similar. But if there is a big difference, then (as before) it means that there is something wrong with your model. I hope that helps.

Please login with a confirmed email address before reporting spam

Posted: 6 years ago 2018年10月10日 GMT-4 09:38

Hello,

A big thanks for your help, it corresponds to what we were trying to found before experimentation.

We really appreciate what you did for us and we send you our best regards.

Hello, A big thanks for your help, it corresponds to what we were trying to found before experimentation. We really appreciate what you did for us and we send you our best regards.

Note that while COMSOL employees may participate in the discussion forum, COMSOL® software users who are on-subscription should submit their questions via the Support Center for a more comprehensive response from the Technical Support team.