For the magnetic field, the currents are one source of the magnetic, but this problem is more linked to the source of the current in the wire. For a conductor with finite conductivity, an electric field is needed in order to drive a current in the wire.
If we assume your wire is straight, this required field is uniform. One way to realize this field is by taking two oppositely charged particles and send them to infinity while increasing the magnitude of their charge to maintain the correct magnitude of electric field. In this limit, you will obtain a uniform electric field through all space.
Now, put your conductor in place along the axis between the voltage sources--a current will flow. In the DC case, this gives rise to the magnetic field outside of the wire. As for the electric field, a conductor is a material with electrons that can move easily in response to electric fields and their tendency is to shield out the electric field to obtain force balance. Because the electrons can't just escape the conductor, they can only shield the field inside the conductor and not outside the conductor. With this model, we see that the electric field is entirely set up by the source and placing the conductor in the field really just establishes a current. Note here that if you bend the wire or put it at an angle relative to the field, surface charges will form because you now have a field component normal to the surface.
For the limit of an ideal conductor, no electric field is needed to begin with to drive the current and so there isn't one outside the wire.
For the AC case, solving for the fields becomes wildly complicated very fast as now the electric field driving particle currents has both a voltage source and a time-varying magnetic source through the magnetic vector potential. The essential physics is the same, though, as the source will establish the fields (in zeroth order), and the addition of the conductor really just defines the path for particle currents to travel. In the next order, the current feeds back and produces electromagnetic fields in addition to the source(s) and will affect the current at other locations in the circuit.
I guess a short answer to your question is that there are always fields outside of the current-carrying wire and the electric field outside disappears only in the ideal conductor limit. Conductors generally do not require very strong fields to drive currents anyway so that the electric field outside is usually negligible, but don't neglect it for very large potentials in small circuits.
At a fixed time $t$, the wire is the line $(0,vt,z)$ where $z$ can be any value and $s$ can be your instantaneous distance to this line. Specifically, we can consider $\vec{A}(x,y,z,t)=-\frac{\mu_0I}{2\pi}\ln s\hat{z}$ where $s$ is nothing more than a shorthand for $\sqrt{x^2+(y-vt)^2}$.
Thus, taking the quasistatic approximation, $$\vec{E} = -\frac{\partial}{\partial t}\left(-\frac{\mu_0I\hat{z}}{2\pi}\ln s\right)= \frac{\mu_0I\hat{z}}{2\pi}\frac{\partial}{\partial t}
ln(\sqrt{x^2+(y-vt)^2}).$$
Just take the partial by treating $x$ and $y$ as constants and evaluate at $t=0$ and the sign works out fine because the very last chain rule gives a factor of $-v$ because the wire moves in the positive $y$ direction.
The quasistatic approximation is handled by the fact that we are just ignoring how or why the charges move in a steady way down the wire, joule losses and what compensates for them, etc. We have a nice vector potential that instantaneously gives a magnetic field around an axis just as if there had always been a steady current there, and we ignore any scalar potential that might help the charge in the wire move along. If you had a wire that moved slowing compared to how long it was, those things would approximately be true very quickly, so we did it by setting up a vector potential that acts like vector potentials act in the quasistaic limit. It doesn't change how we take the partial, and the wire is still moving, giving the position of the wire as $(0,vt,z)$ from the beginning definitely has the wire moving. It's ignoring any electrostatic forces and setting up an $\vec{A}$ that at every moment acts like the charge has always been where it is now that is the quasistatic approximation.
I don't know if you've studied retardation yet, but here are details about quasistatics. A real and finite wire with a steady current that moves slowly has the wire moving so slowly relative to its size that the wire acting due to the parts of the wire it sees is approximately the same as it acting due to what the wire is actually doing. Our vector potential is based on giving a $\vec{B}$ like what you'd get from looking at where the wire is now, as if it had been there and steady all along. In that sense it is a quasistaic type field.
Best Answer
I assume you are referring to the method of electrostatic images : http://en.wikipedia.org/wiki/Method_of_image_charges
If so then a charged particle over a perfect conductor can be simplified for calculations
If the particle has a charge $+q$ and is at a distance $d$ from the surface of the perfect conductive plane then the image charge will be at a distance of $-d$ for the perfect conductor and its charge will be $-q$
So yes it depends on the distance $d$, if the charged particle goes farther away from the conductive plane its image will also go away from it
The image charge of a thin wire will be a charged thin wire as seen through a mirror across the conductive interface.
If the charge of your wire is $q=0$ then it's current is also $I = 0$ as:
$$I = \frac{dQ}{dt}$$
In that case, the image has also $Q' = 0$