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

Answer 1

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: $$ \lambda = i/v$$

Answer 2

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 3

The continuity equation is a good idea to use (but, to be warned here, won't really solve your problem. Here is why and what you still can state!): $\nabla \cdot \vec{j} = \partial_t \rho$. Two things you need to understand:

How to calculate the divergence

Let's look at your expression: $$\nabla \cdot \vec{j}=\nabla \cdot (I(t)\delta(x)\delta(y)\vec{e_z})$$

is actually quite easy. If you don't understand the following steps, please read about the divergence, gradient and nabla-operator itself.

As we only have a z-component in the current, the equation reduces to

\begin{align}\nabla \cdot \vec{j} &= \partial_z \cdot (I(t)\delta(x)\delta(y)\vec{e_z}) \\ &= \partial_z I(t) \end{align}

This is, for an infinite wire quite special, as there is no z-dependence. This of a smaller wire (connected to a capacitor for example) to make sense of the above.

How to get the charge density

This is equal to $-\partial_t \rho$. To get $\rho$, just integrate over time and we get: \begin{align} \int dt \partial_z I(t) = \rho + constant \end{align}

And this constant, from the integral, is the crucial part here (that others pointed out as well): You cannot determine the absolute charge, but only it's change. You will never be able to tell how much charge just sits there but does not move ($\partial_t \rho = 0$).

For your problem, as it is quite special with the infinite wire (actually, every closed wire has $\nabla \cdot \vec{j} = 0$), you will get $$\rho = constant$$

yeah, well, at least a statement... Determining that constant is not possible by only knowing the current (1. order ODE -> 1 initial condition required like $\rho(t=0)=...$)

To avoid any confusion with currents: there is no accumulation of charge if the current stays the same. It is true, that a current brings away charge, but also brings in the same amount. Only if the current brings away charge but none is brought in, the charge density will change with time. But this is a non-vanishing divergence of the current at that point (obviously, as a little bit before that point there is no current, after that point there is)

Answer 4

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.