charge density in a wire

Question

if we have a infinitely thin and infinitely long straight wire on the z-Axis with given current I(t), how can i compute the charge density?

I figured out that the current-density is given by $\vec{j}(\vec{r},t)=I(t)\delta(x)\delta(y)\vec{e_z}$.

But how can i compute the charge-density? I thought about the continuity equation, but did not understand the term $div\vec{j}=div(I(t)\delta(x)\delta(y)\vec{e_z})$=?? On the other side i tried to find a "direct" relationship between the charge density $\rho$ and the current I(t) by $\rho=\frac{dQ}{dz}=I\frac{dt}{dz}$. But this seems absolutely wrong. Can you give me an advice?

Answer 1

The charge density will depend on the resistivity of the wire and on the potentials at the ends of the wire. The following is relevant: http://www.astrophysik.uni-kiel.de/~hhaertel/PUB/voltage_IRL.pdf

Answer 2

Localized charge density may change e.g. Surface charge, but volume charge density does not in any way depends on Current or voltage. Charge density is same whether wire is lying in backyard or being used in Light bulb. Charge simply moves, one electron goes forward right then new electron enters from left in the area being considered.

Answer 3

Net charge density is independent of current density:

1. There could be a set of fixed positive background charges that cancel the charge contribution of moving negative charges (that create current), so the net charge density is zero. (This case is exactly that of a current-carrying wire.)
2. There might be only the moving charges. If they all have speed $v$, and charge density $\lambda$ coulombs per meter of wire length, the current will be $i = \lambda v$, or: $$ \lambda = i/v$$