I am interested in a way to determine electric field and current in a region where some kind of conducting wire is placed in dielectric with some voltage applied to both of them. I am interested in general case but as a simple example consider two infinite conducting half-spaces under some voltage difference with dielectric possibly containing free charges between them connected with a straight conducting wire. Let's consider steady-state problem without any time dependence.

enter image description here

From one point of view there should be zero external electrostatic field outside of conducting conductor, but if I try to take it to the extreme an take very thin wire (single atom?) I cannot convince myself it will be enough to push all the field lines through (in case I try to visualize it somehow).

For dielectric alone the field can be determined from Gauss's law:

$$ \nabla \cdot \left(\varepsilon \nabla \phi\right) = -\frac{\rho_c}{\varepsilon_0} $$

This equation cannot be used for conductors as $\varepsilon$ for conductors is 0. And I don't know a reasonable boundary condition for boundary between wire and dielectric. So for conductors I have found another equation:

$$ \nabla \cdot \left(\sigma \nabla \phi\right) = 0 $$

As far as I understand this equation shouldn't be applied to dielectrics with possible free charges, even if they have a nonzero conductivity.

Perhaps I am overthinking it and you can simply add continuity condition for field on the boundary between the conductive wire and dielectric.

For the current after field calculations I can integrate it over arbitrary surface:

$$ I = \int_S \sigma \left(-\nabla \phi\right) ds $$


1 Answer 1


The electric field lines will not all pass through the conductive wire. I think you are mistaken with your statement about "zero external electrostatic field outside a conducting conductor."

I think it's easier to think of this in terms of equipotential lines. If we assume the top and bottom half-plane conductors are perfect conductors--zero resistivity, then those are each at a constant potential. Just to be concrete, let's say the top half-plane is at a potential $+V$ and the bottom is at zero.

In the conductive region, the potential drops linearly along the wire due to Ohm's law--current flows through a resistance. The equipotential lines here are horizontal, since the potential drops in the same way regardless of our position in the wire. There is an electric field within the wire, and it points the same direction as the current--down. If we work through the math, the potential decreases linearly from $V$ at the top to $0$ at the bottom.

In the dielectric region, the potential must also, somehow, go from $V$ to zero as we move from the top to the bottom electrode. If we focus on a spot far off to the right or left, this is basically a parallel plate capacitor. The equipotential lines are also horizontal here, parallel to the electrodes. An application of Gauss's law would show that the voltage here is also linear in the vertical position, in exactly the same way.

So even though the current is doing very different things in the two regions, the electric field and electric potential are doing exactly the same thing in both the conductive and dielectric regions. Equipotential lines are horizontal and continuous along the entire space, and the electric field is constant and points downward.

As soon as we start to incorporate some nonzero resistivity in the top and bottom half planes themselves, the current pattern will no longer be as symmetrical--current will be rushing into the wire's inlet from all angles (think a crowd of people trying to board a train through a narrow door) and the equipotential lines will curve around the inlet and outlet. That breaks the symmetries that my arguments above rely on, and you'd need to use some sort of finite element modeling to get an accurate answer for the current/field distributions.

Hope that helps!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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