3
$\begingroup$

According to D. Griffiths in Introduction to Electrodynamics, the second uniqueness theorem says

in a volume $V$ surrounded by conductors and containing a specified charge density rho, the electric field is uniquely determined if the total charge on each conductor is given.

Is it correct to think of this as a Neumann problem?


Recall that in a Neumann problem knowledge of the directional derivative of the unknown function on the boundaries is required to solve the problem uniquely. However, the second uniqueness theorem says that knowledge of the total charge on each conductor is sufficient. Now, the total charge on a conductor is not simply the directional derivative of the potential, but rather the surface integral of the directional derivative of the potential over the surface of the conductor.

$\endgroup$
2
$\begingroup$

No, the problem is not a Neumann type boundary problem. For it to be such a problem, you would need to know the directional derivative which is not given - it can just be calculated once the problem is solved. I would not call it a Dirichlet problem either because even though we know that the boundaries have constant potential, the different conductors should be able to be at different potentials in my opinion. It is still sufficient to uniquely determine the problem as is shown by Griffiths. Note that it is important that Griffiths says that the whole volume is bounded on the outside by only one conductor or infinity.

For future reference, this is from p. 118, section 3.1.6 of Prentice Halls 1999 print of Griffiths Introduction to Electrodynamics.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I understand how knowledge of surface charge density gives the directional derivative of potential. What I don't understand is how knowledge of the total charge on conductors (without any information on how it is distributed) is sufficient to solve Laplace's/Poisson's equation. $\endgroup$ – terahertz Aug 12 '16 at 10:39
  • $\begingroup$ Yes, I also think that the second uniqueness theorem as given by Griffiths is not a Neumann Problem because knowing the total charge on a conductor tells us nothing about how it is distributed. (If we knew how charge was distributed on the surface of a conductor, that is, if we knew the surface charge density, then it would become a Neumann problem). Now the question is, what kind of a problem is it, mathematically speaking? The problem with Griffiths' proof of the second uniqueness theorem is that it only shows why the solution is unique, not how to get it! $\endgroup$ – terahertz Aug 13 '16 at 5:49
  • $\begingroup$ @terahertz the proofs for Dirichlet/Neumann bc are also non-constructive if I remember correctly. I am unfortunately not aware of a name for this kind of boundary condition in mathematical terms either. $\endgroup$ – Sanya Aug 13 '16 at 9:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.