My goal is to understand whether or not there would be a source term $\rho$ in the Helmholtz (or Poisson) equation for the scalar electric potential, for the region inside an object with finite conductivity, hooked up to an alternating voltage source.

Consider an object with a finite conductivity $\sigma$, which is connected in series in some circuit, including a voltage source. So basically, current is allowed to flow in one end of the object, and out the other.

The object is electrically large at the frequency of operation, and electromagnetic effects must be accounted for.

Depending on the chosen gauge condition (Coloumb or Lorenz), the scalar potential inside the object will satisfy either the Poisson or Helmholtz equation. Let's assume a Lorenz gauge, so that the scalar potential $\Phi$ inside the object satisfies \begin{align} \nabla^2\Phi + k^2\Phi = -\dfrac{\rho}{\epsilon} \end{align} where $k$ and $\epsilon$ are the wave number and permittivity inside the object.

I want to set up a boundary integral equation for the inside of the object, and I'm trying to understand what exactly $\rho$ means here.

My understanding is that $\rho$ should represent a free volume charge density which exists in the relevant region of space (the bulk of the object, in this case).

For an isolated object with a finite conductivity $\sigma$, any excess free charges should eventually (per the relaxation time constant) make their way to the surface. So at steady state, the free volume charge density in the object should be zero. Therefore, $\rho=0$ in that case, and the Helmholtz equation becomes homogeneous.

However, since the object is hooked up to a circuit with an alternating voltage source, I'm not sure that $\rho=0$ applies anymore. Instead, I'm guessing we have to use the continuity equation somehow, \begin{align} \nabla\cdot\mathbf{J} + j\omega\rho = 0. \end{align} But if $\mathbf{J}=\sigma\mathbf{E}$, where $\mathbf{E}$ is the electric field induced in the object due to being hooked up to the AC circuit, then we would have \begin{align} \sigma\nabla\cdot\mathbf{E} + j\omega\rho = 0, \end{align} assuming the object is homogeneous. But then from the generalized Gauss' law, we can also write $\nabla\cdot\mathbf{D} = \epsilon\nabla\cdot\mathbf{E} = \rho$ (because $\rho$ represents a free charge density) so that \begin{align} \dfrac{\sigma\rho}{\epsilon} + j\omega\rho = 0, \end{align} which implies \begin{align} \left(\dfrac{\sigma}{\epsilon} + j\omega\right)\rho = 0. \end{align} But this just says that in steady state, $\rho=0$ again! But if that's true, then the continuity equation reads \begin{align} \nabla\cdot\mathbf{J} = 0 \end{align} which doesn't make sense to me - how is it possible that the current in the object is always divergence-free, no matter the frequency or permittivity?

The other option is that $\left(\frac{\sigma}{\epsilon} + j\omega\right) = 0$, but that doesn't make sense to me either, because $\sigma$, $\epsilon$ and $\omega$ can be anything (unless the $\omega$ here is not the same as the cyclical driving frequency of the voltage source?).

I'm also not sure if what I'm missing is an impressed current which is supplied by the circuit. But by the continuity of the normal component of the current, I would think that $\mathbf{J}=\sigma\mathbf{E}$ accounts for the current introduced in the object due to the circuit?

I think I am missing something fundamental here, or horribly mixing up some basic concepts. Where exactly am I going wrong?

Thank you!

  • $\begingroup$ In a given uniform conductor, the resistivity is a constant. $\endgroup$
    – R.W. Bird
    Commented Jun 20, 2021 at 14:20

2 Answers 2


The more you assume, the more you can derive. In your argument you assume:

  1. Ohm's law is valid with the same conductivity everywhere, which limits the argument to homogeneous metal body insides; on its the surface, Ohm's law is not valid because electric field has normal component there, but current can't have one. Also, if current flows, there are charges on the surface of the body, contributing to total electric field inside.

  2. electric field and time derivative of electric field are sinusoidal functions of time. This is often true but for some nonlinear materials (semiconductors?) this may not be true.

I think these assumptions are well valid for ordinary metallic bodies and the conclusion holds - there is no concentration of electric charge inside. Charge has non-zero distribution only on surface of the bodies.

Where the assumptions and conclusion may not be valid are inhomogeneous materials, where concentration of charge may happen in places of gradient of conductivity.

  • $\begingroup$ Does the fact that the object is connected to an AC circuit not play a role in the analysis then? If instead of the circuit, the object was in the presence of an applied incident field, wouldn't all the analysis stay the same? In the latter situation, charge is confined to stay in the object, but in the former, charge can exit and enter. So my intuition says there must be some difference in the two situations (circuit vs. incident field)? $\endgroup$
    – josh_eime
    Commented Nov 22, 2020 at 2:25
  • $\begingroup$ For AC current, the expression $J = \sigma E$ it is a good description if the Ohmic resistance is much more important than any wire inductance. $\endgroup$ Commented Nov 22, 2020 at 2:43
  • $\begingroup$ Depends on the incident radiation field. If it was constant harmonic oscillation, then it would be the same. If it was a pulse that dies out in an abrupt way, then depending on the time dependence of the pulse, time derivative of electric field may not be proportional to electric field. Abrupt changes usually stir up some transient phenomena. $\endgroup$ Commented Nov 22, 2020 at 2:47
  • $\begingroup$ @JánLalinský , yes, I'm considering time harmonic situations here. $\endgroup$
    – josh_eime
    Commented Nov 22, 2020 at 2:50
  • $\begingroup$ @ClaudioSaspinski By wire inductance are you referring to inductive effects due to the object? If so, I actually do not want to assume that there are no inductive or capacitive effects - in fact I'm interested in a full EM model of the object. But I was under the impression that $J = \sigma E$ is local, so I don't see why inductive effects would come into play? $\endgroup$
    – josh_eime
    Commented Nov 22, 2020 at 2:53

To maintain a constant current floe in a wire which can be formed into one or more loops, the E field must be nearly uniform along the length of the wire. This can only come from a charge gradient. The power supply pulls electrons from the positive end of the wire, and puts them into the negative end. The flow of current distributes the charge to give the required gradient. To satisfy Gauss's law, electric field lines must leave through the surface of the wire on the positive end and reenter toward the negative end. This in combination with the magnetic effects would effect the charge density on the surface of the wire.


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.