Current flowing between two concentric metal shells

[aftershock]

Homework Statement

Two concentric metal shells of radius a and b respectively ( a < b) are separated by weakly conducting material of varying conductivity σ(r) = kr where k is a constant and r is the distance from the common center.

If the metal shells are maintained at a constant potential difference V and a steady current flows between the shells, what is the value of this current?

Homework Equations

J=σE

I = ∫J da

The Attempt at a Solution

I = ∫σE da

Since the current is the same everywhere I set up a guassian surface inbetween the two shells. Here sigma is constant since r does not vary so I pulled it out of the integral. Now the integral is gausses law and can be replaced with q/ε

I = σq/ε

The final step is putting q in terms of the potential difference. That's easy enough however this seems to show that the current is dependent on sigma which is dependent on the distance from the center.

The problem specifically says steady current. Where did I go wrong?

 

[BruceW]

your working looks pretty good to me. Maybe they just meant steady as in 'constant with time'

 

[haruspex]

I = σq/ε

This doesn't seem right. The total current is independent of r, but σ = σ(r).
Edit: I don't think you can reduce ∫E da to q/ε here. That ignores what's going on in the conducting layer.
If the potential is V=V(r), write a differential equation for the current through the shell at radius r.

 

[aftershock]

Edit: I don't think you can reduce ∫E da to q/ε here. That ignores what's going on in the conducting layer.

I'm not sure I understand the error here. Can you elaborate on that a little?

 

[haruspex]

I'm not sure I understand the error here. Can you elaborate on that a little?

I was just trying to understand why your working was not producing the right answer. My suspicion fell on the equation.
Try the approach I suggested. Consider a shell of radius r. If the total current through that shell is a constant, I, and the conductance is σ(r)=kr, what is the voltage gradient there?

 

[aftershock]

I was just trying to understand why your working was not producing the right answer. My suspicion fell on the equation.
Try the approach I suggested. Consider a shell of radius r. If the total current through that shell is a constant, I, and the conductance is σ(r)=kr, what is the voltage gradient there?

I actually did solve it starting from another equation I found in my notes. I'll post what I used later if you're curious but I am on a tablet now and it would be difficult. I am more curious exactly why my original method is incorrect.

 

[haruspex]

Im more curious exactly why my original method is incorrect.

Fair enough. Gauss' law is for electrostatics. I do not believe you can apply it quite so simply to current electricity, particularly if resistance is non-uniform. My suspicion is that it breaks down because the field at radius r does not only come from the source at radius a. There are (moving) charges throughout the shell. But I'm only guessing.

 

[aftershock]

Fair enough. Gauss' law is for electrostatics. I do not believe you can apply it quite so simply to current electricity, particularly if resistance is non-uniform. My suspicion is that it breaks down because the field at radius r does not only come from the source at radius a. There are (moving) charges throughout the shell. But I'm only guessing.

I've seen an analogous example involving cylinders where my original method works. The only difference is the conductivity is constant. So I think it's not so much the current, but the fact that the conductivity is non constant that I can't use Gauss' Law.

Why should that matter though?

 

[haruspex]

I've seen an analogous example involving cylinders where my original method works. The only difference is the conductivity is constant. So I think it's not so much the current, but the fact that the conductivity is non constant that I can't use Gauss' Law.

Why should that matter though?

Not sure, but consider this example: 1D heat flow. If the conductivity is uniform then the temperature gradient bears a constant relationship to the local heat flow. It follows that wherever there's curvature in the plot of temperature against distance, the temperature is rising (convex) or falling (concave). This leads to Laplace's equation. In steady state, Δθ = 0. Now suppose the conductivity is non-uniform. The ratio between temperature gradient and heat flow varies with position, so curvature can arise even in steady state. Laplace's equation no longer holds.

 

[Dick]

I actually did solve it starting from another equation I found in my notes. I'll post what I used later if you're curious but I am on a tablet now and it would be difficult. I am more curious exactly why my original method is incorrect.

The current through each shell of radius r a<=r<=b is constant. That doesn't mean that the current density is constant. If the electrons are flowing from the outer shell to the inner shell then they are going to become more concentrated as they approach the inner shell. That means the charge density is not constant between the two shells. Charge enclosed by a shell is not a constant q, it's some function of r, q(r). I'm glad you solved it using another equation you've got because this method isn't going to work.

 

[andrien]

just determine the resistance using formula
R=l/σA,where l=dr ,A=4∏r²,σ=kr,integrate from a to b.evaluate current by
I=V/R

 

[BruceW]

Ah, Dick has the right idea. aftershock's working is right, but as Dick said, he can't assume the enclosed charge is independent of radius from the centre.

 

[aftershock]

Ah, Dick has the right idea. aftershock's working is right, but as Dick said, he can't assume the enclosed charge is independent of radius from the centre.

Then why can we in this example http://i.imgur.com/9xRkOCM.jpg?1

 

[BruceW]

yeah. To be honest, I think they are not giving enough detail in the question. In the problem in this thread, they have told you that there is a steady current. So from this, you can infer the charge distribution in the weakly conducting material.

In the example you have just given the image link to, you pretty much just have to guess that there is supposed to be no charge distribution in the weakly conducting material, because it usually is zero in these kinds of problems, and they do not tell you anything that would mean it is non-zero.

 

[Dick]

Ok, let me retract my previous bogus explanation and give you the real one. You know the current through each sphere is a constant independent of r. That's given by the current density integrated over the sphere. The E field over the sphere is given by the current density times the resistivity. In the case of constant resistivity that means that since the current density integrated over the sphere is a constant independent of r, that the E field integrated over the sphere (your 'enclosed charge') is also a constant independent of r. On the other hand, if the resistivity is a function of r, it is no longer true. It's the variable resistivity that causes charges to pile up.

 

[TSny]

I think andrien's method gets the answer in a nice manner.

You can also use the method of the link except without assuming beforehand an expression for E(r). Instead, find E(r) by considering an arbitrarily chosen spherical surface of radius r (a < r < b). The current through the surface is

I = ∫j[itex]\cdot[/itex]dA = ∫σ(r)E(r)dA.

From this you can get an expression for E(r) in terms of I and r. Then you can integrate E(r) between the conductors to relate I to V.

Gauss' law is always valid. But there will be a charge density ρ(r) in the material between the conductors.

 

[haruspex]

Gauss' law is always valid. But there will be a charge density ρ(r) in the material between the conductors.

Quite so, and we can work out how that depends on conductance.
At radius r (ignoring the 4pi's and epsilons), the voltage due to shells radius s < r is ##\frac1r\int_0^r \rho(s)s^2 ds##, and that due to shells of greater radius is ##\int_r^b \rho(s) ds##. Differentiating, the voltage gradient is ##-\frac1{r^2}\int_0^r \rho(s)s^2 ds##, the other two terms cancelling. Since the total current through shell radius r is a constant I, we can also write that this gradient varies as ##-\frac1{r^2 \sigma(r)}##. Hence ##\rho(r)\propto -\frac{\sigma'}{\sigma^2r^2}##.
In the special case of uniform conductance, the charge density outside the source shell is zero. In consequence, the potential function is the same as in the static case. Even stranger, at first sight: if the conductance increases with radius then the charge density goes negative. But on reflection, that seems reasonable.

 

[Dick]

Quite so, and we can work out how that depends on conductance.
At radius r (ignoring the 4pi's and epsilons), the voltage due to shells radius s < r is ##\frac1r\int_0^r \rho(s)s^2 ds##, and that due to shells of greater radius is ##\int_r^b \rho(s) ds##. Differentiating, the voltage gradient is ##-\frac1{r^2}\int_0^r \rho(s)s^2 ds##, the other two terms cancelling. Since the total current through shell radius r is a constant I, we can also write that this gradient varies as ##-\frac1{r^2 \sigma(r)}##. Hence ##\rho(r)\propto -\frac{\sigma'}{\sigma^2r^2}##.
In the special case of uniform conductance, the charge density outside the source shell is zero. In consequence, the potential function is the same as in the static case. Even stranger, at first sight: if the conductance increases with radius then the charge density goes negative. But on reflection, that seems reasonable.

Sure, nonuniform conductance induces regions of charge overpopulation and underpopulation. See my last post. If the conductance is uniform the potential will satisfy Laplaces's equation. If not, it's a Poisson equation.

 

[aftershock]

Ok, let me retract my previous bogus explanation and give you the real one. You know the current through each sphere is a constant independent of r. That's given by the current density integrated over the sphere. The E field over the sphere is given by the current density times the resistivity. In the case of constant resistivity that means that since the current density integrated over the sphere is a constant independent of r, that the E field integrated over the sphere (your 'enclosed charge') is also a constant independent of r. On the other hand, if the resistivity is a function of r, it is no longer true. It's the variable resistivity that causes charges to pile up.

This makes sense. Thank you!