**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!