0
$\begingroup$

What is a smart way to calculate current densities in 3D objects with unorthodox boundary conditions?

For example, J(x,y,z) in a cube with constant resistivity with applied voltages Va in a vertex and Vb in a plane?

enter image description here

$\endgroup$
2
$\begingroup$

Assuming these are constant voltages, you need to solve Laplace’s equation, $$\nabla^2 \phi = 0,$$ for the potential $\phi$, with appropriate boundary conditions. That gives you the electric field, $$\mathbf{E} = -\nabla \phi,$$ from which you get the current density by Ohm’s law in differential form $$\mathbf{J} = \sigma \mathbf{E},$$ where $\sigma$ is the conductivity, assumed constant throughout the volume. Boundary conditions may be Dirichlet $\phi = \phi_0$ (constant voltage) or Neumann $\partial \phi/\partial n = c$, in which the normal derivative of the potential is specified. In particular, when the medium adjacent to the object is non-conductive, no current can flow into it and the corresponding boundary condition is $\partial \phi/ \partial n = 0$.

To tackle the situation you depicted you could start with the vertex voltage defined on a small but finite part of the boundary that includes the vertex. For any finite area, the solution is unique and well-defined. The solution you seek would be obtained in the limit when the size of this area shrinks to zero.

$\endgroup$
  • $\begingroup$ Nice, thanks for the answer. And how would you deal with the conductivity when it is not uniform, but a known function of the spatial coordinates? $\endgroup$ – Gyromagnetic Oct 15 at 12:38
  • 2
    $\begingroup$ When the current flow is stationary, you have from the continuity equation div J = 0. Inserting into this equation Ohm’s law and $E =-\nabla \phi$, you obtain $\nabla \cdot (\sigma(r)\nabla \phi) = 0$, which is a generalization of Laplace’s equation. $\endgroup$ – Amit Hochman Oct 15 at 14:20
  • $\begingroup$ I think that in the limit indicated in the answer the field will be similar to that of a point charge at the vertex. This must be so because a finite current must pass through a single point, so the field must behave like 1/(distance from vertex). $\endgroup$ – Amit Hochman Oct 15 at 14:43

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.