Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In electrostatic boundary value problems, usually the potential is set to some function or a constant on a certain boundary. It is never taught or told in textbooks how such assignment might be achieved experimentally.

For example how to set experimentally the potential on the boundary of a ball of radius $R$ to some constant $V_0$, where $V_0$ could be $-100 V$ or $0 V$ or $5 V$ say?

share|improve this question
    
For that you always imply some reference body, for example, the ground. Then you connect your ball and the ground to a battery with the necessary voltage. –  Vladimir Kalitvianski Nov 9 '11 at 15:53
    
@VladimirKalitvianski Can you show a pic for what you mean? –  Revo Nov 9 '11 at 16:21
    
I'm having trouble deciding if this problem is about reference voltage or about the constant potential boundary condition. Vladimir answered to the former. The constant potential comes from the fact that charges in an isolated conducting volume will reorganize along any electric field line until there is are no more electric field lines within the volume - leaving the entire volume at a constant electric potential. There are limitations to this statement of course. –  AlanSE Nov 9 '11 at 17:21
    
If you have two equal balls charged equally, there will not be a potential difference between them. If you connect them with a wire, the current will not appear. Normally if you have one charged ball, we say it has this potential and mean "with respect to infinity" where the potential is equal to zero (again, a convention). So, only a potential difference matters. To have a difference, you have to have another (reference) body. –  Vladimir Kalitvianski Nov 9 '11 at 17:45
    
@VladimirKalitvianski Okay potential on a boundary is 60 V say relative to 0. Now how to make the boundary of a ball to have that potential experimentally? I think the answer will depend whether the ball is a conductor or not. –  Revo Nov 9 '11 at 18:41

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.