Question about electrostatic equilibrium for a conductor

[shoestring]

At equilibrium, a conducting object will have no electric field, current or excess charge in its interior. There's a fairly simple derivation for how an initial (non-equilibrium) interior charge density d₀ falls to zero exponentially with time t when the conductor approaches equilibrium. For a homogeneous and isotropic conductor with conductivity g and permittivity epsilon the interior charge density d can be written as

d(x,y,z,t) = d₀(x,y,z)e^(-gt/epsilon)

(I tried to include the derivation of the formula, but haven't mastered the math notation yet, so I gave up. The derivation makes use of the continuity equation, Ohm's law and Gauss' law and can be made in four lines.)

So far so good, the interior charge distribution falls exponentially to zero, with a known relaxation time epsilon/g (i.e. the permittivity divided by the conductivity).

Now a conductor hasn't necessarily reached equilibrium because it has no interior excess charge. We could have started with zero interior charge density, and had a non-equilibrium surface charge distribution instead. A non-equilibrium surface charge distribution will cause an interior electric field and current until equilibrium is reached.

Now to the questions:

Is there any easy way to calculate, or at least estimate, the time it takes for the internal electric field and current density, not just the interior charge distribution, to drop to zero?

Will the result, like for the interior charge density, be independent of the shape of the conductor?

 

[shoestring]

Perhaps it was just the preveiw that didn't work for me when I tried to use the tex notation. I'll give it a new try, now with $\rho$ for the volume charge density. For every interior point $(x,y,z)$,

$$\frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{J} = 0$$

$$\frac{\partial \rho}{\partial t} + g \ \nabla \cdot {\mathbf{E}} = 0$$

$$\frac{\partial \rho}{\partial t} + \frac{g}{\epsilon} \rho = 0$$

which is a differential equation with the solution

$$\rho (x,y,z,t) = \rho_0 (x,y,z) e^{-(g/ \epsilon)t}$$

where $\rho_0$ is the initial interior charge density.

The first equation is the equation of continuity, the second follows from application of Ohm's law, and the third from Gauss' law.

 

[Delta2]

erm this might sound silly but why these equations don't hold at the surface of the conductor? For sure Gauss Law and Ohms law seem to hold for me. And i think continuity equation should hold too since it can be derived from maxwell-ampere's law.

PS my bad! Ohms Law doesn't hold at surface.

 

[shoestring]

Yes, I agree that at the surface the step

$$\nabla \cdot \mathbf{J} = \nabla\cdot (g\mathbf{E}$$)

is the problematic one, because outside the conductor $g$ vanishes, so at the surface $g$ is no longer constant but a function of $(x,y,z)$. It's even discontinuous, so it isn't differentiable at the surface. Because of that, $\nabla g$ in

$$\nabla \cdot (g\mathbf{E})= (\nabla g)\cdot \mathbf{E} + g (\nabla \cdot \mathbf{E})$$

has no value at the surface.

 

[shoestring]

It's only the divergence of the electric field, not the field itself, that depends only on the local charge. The field itself will depend on all charges, so I doubt a local equation can describe the approach to equilibrium for the surface charge, even if $g$ is smoothed out to be made differentiable at the surface. It'd only make the $(\nabla g)\cdot \mathbf{E}$ term defined, but not zero, so the term with $\mathbf{E}$ itself wouldn't disappear.

 

[Delta2]

Well in the case we have to do with interior charge density, one solution to the electric field is $$\mathbf{E}(\mathbf{r},t)=\mathbf{E}(\mathbf{r})e^{(-g/ \epsilon)t}$$ as long as

$$\nabla \cdot \mathbf{E}(\mathbf{r})=\rho_0(x,y,z)$$. And ofc $$\mathbf{J}=g\mathbf{E}$$ so electric field and current seem to have the same time dependence as $$\rho$$.

 

[shoestring]

Yes, the interior electric field from the interior charges disappears with the interior charges. But when they turn up at the surface, they may still affect the interior field until equilibrium is reached. They may have to redistribute themselves on the surface ... :-/

If we start with a sphere with a spherically symmetric interior charge (varying only with the distance from the center) then I imagine equilibrium is reached at the same rate as the interior charge disappears. In a less symmetric situation I'm not sure.

 

[Delta2]

Hm maybe somehow the charges come up at surface in an "equilibrium ready" way . But intuitevely this doesn't seem correct unless there is some obvious symmetry as you said with your example. But then again i look at the equation and i don't see where is the mistake... E(r) seems to depend only on the initial charge density.

 

[Delta2]

Ok find out what the problem is. It is all about the equation

$$\nabla \cdot \mathbf{X}=y$$

This equation can "fool" us in two ways:

1) If y doesn't depend on some variable (like time) makes you thing that X won't depend either, which isn't true in the general case.
2) if the equation is defined in an open region makes you thing that X won't depend on what is happening on the boundary of that region which again isn't true in the general case.

 

[shoestring]

We can think of the vector field $\mathbf{X}$ as having a divergence free term, $\mathbf{X} = \mathbf{X}_1 + \mathbf{X}_2$, where $\mathbf{X}_2$ is divergence free in the open region we're looking at. Then

$$\nabla \cdot \mathbf{X} = \nabla \cdot (\mathbf{X}_1 + \mathbf{X}_2) = \nabla \cdot \mathbf{X}_1$$

and it's obvious that the equation

$$\nabla \cdot \mathbf{X}=y$$

gives no information about the divergence free part.

If $\mathbf{X}$ is an electrostatic field it means that the equation says nothing about the part of the field that is generated by charges outside or on the boundary of the open region.

 

[JohnV]

This seems to make mathematical sense, but there's something strange about it. Suppose we move the surfaces of the conductor out to infinity; the charges start in a localized region near the origin in this infinite conduction medium. Now the equations are true everywhere. But notice that the solution does not conserve charge. That is, if you integrate rho(x,y,z) over all space, you get a total charge that starts (we can imagine) at some finite nonzero value but decays to 0. What am I missing?