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.

Electrostatic Induction on a Conductor

Please login with a confirmed email address before reporting spam

Hi, I am very new to COMSOL and I was wondering if the problem I am interested in solving can even be solved in COMSOL.

I am trying to look at the induced charge (or voltage potential) of an isolated conductor that is in the presence of an electric field. Specifically what I want to do is to define a surface charge density on one face of a conductor, and then look at the potential that develops in a neighboring conductor that is not grounded and initially has no charge.

From my very limited experience with COMSOL, the simulation will not run (or will run forever) unless I specify some potential or charge value on every conducting surface in the model. So the only time I have ever gotten any results is when I have initially defined the values I am looking to solve for which makes the process completely redundant.

Can I do this in COMSOL, and if so, how to I go about constraining this problem?

1 Reply Last Post 2015年9月17日 GMT-4 23:32
COMSOL Moderator

Hello Mark Pallay

Your Discussion has gone 30 days without a reply. If you still need help with COMSOL and have an on-subscription license, please visit our Support Center for help.

If you do not hold an on-subscription license, you may find an answer in another Discussion or in the Knowledge Base.


Please login with a confirmed email address before reporting spam

Posted: 9 years ago 2015年9月17日 GMT-4 23:32
Hello,

I am also trying to simulate the same. could you find any solution?

Thanks
Hello, I am also trying to simulate the same. could you find any solution? Thanks

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.