# charge density in a wire

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 $\nabla.\vec{j}=\nabla.(I(t)\delta(x)\delta(y)\vec{e_z})$=?? 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?

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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: $$\lambda = i/v$$
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Thank you for your help! Lets say, it has no resistivity(infinitely thin) and no ends (since it is infinitely long). Maybe one can write analogously $\rho(r,t)=Q(t)\delta(x)\delta(y)$?? ps: I think the answer must depend only on I(t) like for the curren-density –  Braten Jan 4 '13 at 13:25