So for current to flow, an EMF source such as a battery causes surface charges to form a gradient along the wire. These then have local fields which cause current flow by moving internal free charges.
My question is, does the battery field only affect surface charges, or both surface and inner charges. This is for a steady-state circuit, so I want to know if shielding stops the "external field" created by the battery getting "inside" of the main conducting wire.
In steady state, in a uniform conductor, the charge density within the body of the conductor is $0$. All net charge lies on the surface(s) of that conductor.
Proof:
In steady state, all derivatives with respect to time, vanish.
Thus, the continuity equation
$$\nabla \cdot \vec{J} + \frac{\partial \rho}{\partial t} = 0$$
becomes
$$\nabla \cdot \vec{J} = 0$$
The microscopic version of Ohm's law says
$$\vec{J} = \sigma\vec{E}$$
so
$$\nabla \cdot \sigma \vec{E} = 0$$
Expanding we get
$$(\nabla\sigma)\cdot\vec{E} + \sigma(\nabla\cdot\vec{E})=0$$
Wherever $\nabla\sigma=\vec{0}$, we have
$$\nabla\cdot\vec{E}=0$$
But from Gauss's Law
$$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$$
where $\rho$ is the charge density.
So, wherever $\nabla\sigma=\vec{0}$, we have the charge density must be $0$ i.e.
$$\rho=0$$
If we assume that the conductor is uniform, the only places where $\nabla\sigma \ne \vec{0}$ are on the surfaces of the conductor.
Thus, in steady state (all time derivatives vanish) for a uniform conductor, the only location where we will find non-zero charge density is on the surfaces of the conductor.
does the battery field only affect surface charges, or both surface and inner charges.
In stationary regime, inside a uniform conductor, whether the current is zero or not, net charge density vanishes. Connection of electrochemical EMF source (e.g. a battery) does not change this (except perhaps momentarily, until the stationary state is reached). It only changes charge distribution on the conductor surface; inside the conductor, this distribution vanishes.
This follows from Ohm's law and Maxwell's equations, see e.g. my answer here
