# 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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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

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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 at 13:25
@Braten: Not sure I agree. Let us imagine for a moment that the wire is extremely long, but still finite (an infinite wire just makes no sense). The wire can have arbitrary total charge, which can have a static configuration (with a (linear) charge density constant everywhere but near the ends). The current is zero for static configurations for any total charge. Therefore, current doesn't determine the charge (density) uniquely. You should define the potentials (with respect to infinity) at the ends (or at least at one end, if you know resistivity, the length, and the current). –  akhmeteli Jan 5 at 16:51
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: $$\lambda = i/v$$