Two concentric conducting spherical shells and resistor in between

[zenterix]

Homework Statement

A resistor consists of two concentric conducting spherical shells with the inner shell having radius $r_a$ and the outer shell having radius $r_b$.

The space between the two shells is filled with a material of resistivity $\rho$.

What is the resistance of this resistor?

Relevant Equations

$V=iR$

The first thing I thought about was the relationship $\vec{J}=\frac{\vec{E}}{\rho_r}$ which is a statement of Ohm's law. That is, current density is proportional to electric field and the constant of proportionality is the reciprocal of resistivity $\rho_r$, which is the same as conductivity, $\sigma_r$.

I'm not sure if this is the context in which to solve this problem, but one initial huge doubt I have is the following.

If $\vec{J}=\sigma\vec{E}$ and the current is steady (ie, stationary or constant in time everywhere) then the electric field is constant everywhere as well.

This assumption of steady current is used to find an expression for resistance on a conducting rod with well-defined geometric shape such as in the figure below

For such a conducting rod, we find that $R=\frac{L}{\sigma A}$.

Does any of this apply to the situation of the two spherical shells with the resistor in between?

 

[haruspex]

Think about current versus current density. What's constant here?

 

[Delta2]

@haruspex is trying to make you think of it but I will state this crucial fact explicitly.

No in the so called steady state , the electric field is not constant in the space between the two shells, it varies with the radius from the center $r$. How exactly it varies it relates to haruspex question as well.

I would define steady state here as the state that in which the total current through a conceptual spherical shell of any radius r, is independent of r.

 

[Orodruin]

Just because the current is steady does not mean it is the same everywhere. The requirement for no charge accumulation is $\nabla\cdot\vec J = 0$. You need to figure out what $\vec J$ is as a function of radius.

 

[Delta2]

Just because the current is steady dies not mean it is the same everywhere. The requirement for no charge accumulation is $\nabla\cdot\vec J = 0$. You need to figure out what $\vec J$ is as a function of radius.

Your statement is a bit ambiguous, the current must be the same everywhere(through every concentric shell) but the current density NOT, that's my view here.

 

[Orodruin]

Your statement is a bit ambiguous, the current must be the same everywhere(through every concentric shell) but the current density NOT, that's my view here.

It is not ambiguous in the least

 

[Delta2]

Then instead of saying the current steady you should say the current density steady, that is the current density independent of time, thats the best I can do from your statement.

 

[Delta2]

The current is independent of time and independent of the radius r (otherwise we would have charge accumulation). The current density is independent of time but depends on the radius r.

 

[Orodruin]

Then instead of saying the current steady you should say the current density steady, that is the current density independent of time, thats the best I can do from your statement.

The local quantity is obviously current density. This is the only thing that can vary from place to place so any other inference is impossible. To talk about a current yiu need a surface.

 

[Delta2]

I insist you must edit the first sentence of your post to "Just because the current density is steady, does not mean is the same everywhere"

 

[Orodruin]

Besides, the actual statement is $\nabla\cdot\vec J =0$. This is a mathematical fact and cannot be misconstrued.

That physicists are sometimes stating “current” when they are talking about current density is not really possible to misconstrue from context. We will many times omit “density”, eg, when we talk about Lagrange densities in field theories we will many times just say “Lagrangian”. Is it precise language? Perhaps not, but it is the jargon and what is actually implied is generally clear.

 

[Delta2]

It is not ambiguous in the least

Besides, the actual statement is $\nabla\cdot\vec J =0$. This is a mathematical fact and cannot be misconstrued.

That physicists are sometimes stating “current” when they are talking about current density is not really possible to misconstrue from context. We will many times omit “density”, eg, when we talk about Lagrange densities in field theories we will many times just say “Lagrangian”. Is it precise language? Perhaps not, but it is the jargon and what is actually implied is generally clear.

In that case you have to accept that your statement is just a bit ambiguous cause you said current while you meant current density e ehehe.

 

[zenterix]

I think part of the confusion on my part has to do with the transition from electrostatics to analyzing current.

Up to now, I've been analyzing electric fields and potential differences of different configurations of static charges (point charges, various continuous charges with different shapes, capacitors, conductors with no current).

One crucial important aspect in the transition to currents seems to be the realization that the focal point of the analysis are conductors.

Given a conductor, current is the surface integral of $\vec{J}$ over a cross section of the conductor. Therefore, $I$ is proportional to $\vec{J}$.

$\Delta V$ is the line integral of $\vec{E}$ on a path through the conductor from one terminal to the other. Therefore, $\Delta V$ is proportional to $\vec{E}$.

Therefore, $I$ is proportional to $\Delta V$.

Regarding "steady currents", Purcell says that

We speak of a steady or stationary current system when the current density vector $\vec{J}$ remains constant in time everywhere. Steady currents have to obey the law of charge conservation.

The simplest example of a conductor is a solid rod.

We assume that

- The conductors bringing charge in and out of the rod are such that there are no "end effects". That is, the current density is uniform right from the start to the end of the rod. This assumption can be relaxed if the length of the rod is very long compared to its width.

- The current density is uniform over a cross section of the rod. If this were not so, then $\vec{E}$ would not be path independent.

One of my doubts about the spherical resistor is that I can't really see how current is supposed to flow.

Does it flow from one end of the sphere to the other?

Does it flow radially outwards? But then where does it come from?

Does it flow radially inwards? But then where does it go?

What are the "terminals" for this spherical configuration?

 

[zenterix]

Think about current versus current density. What's constant here?

From my understanding, current density is constant in time, but it can vary based based on location.

Since current is the flux of current density then for any surface $S$, current through that surface is constant.

For any closed surface, charge conservation implies that current through the surface is zero, ie $\int_S\vec{J}\cdot d\vec{a}=0$.

By the divergence theorem, this implies that $\text{div}(\vec{J})=0$.

 

[PhDeezNutz]

Excuse my ignorance but why do we need current?

To me this entire resistor consists of a bunch of smaller resistors in series.

So it would seem that all we need to do is integrate

$\int_{r_a}^{r_b} \frac{\rho}{4 \pi r^2} \,dr$

Which is similar to the formula $R = \frac{ \rho L}{A}$ but put in integral form for this specific geometry.

 

[Delta2]

Excuse my ignorance but why do we need current?

To me this entire resistor consists of a bunch of smaller resistors in series.

So it would seem that all we need to do is integrate

$\int_{r_a}^{r_b} \frac{\rho}{4 \pi r^2} \,dr$

Which is similar to the formula $R = \frac{ \rho L}{A}$ but put in integral form for this specific geometry.

That is a clever shortcut, well done but in my opinion it doesn't give the insight that one might get by reading the other posts.

 

[PhDeezNutz]

That is a clever shortcut, well done but in my opinion it doesn't give the insight that one might get by reading the other posts.

Alright that’s fair. There are deeper aspects to this problem and OP is right to ask about them.

 

[zenterix]

@PhDeezNutz I did a google search and your answer coincides with what I found.

The solution is quite simple. However, I feel a bit confused about the problem setup itself.

Is this resistor purely theoretical?

It seems that if we consider the resistor as a sum of infinitesimal shell-like resistors then how is this resistor actually used? For current to pass through the resistors in series it must originate in the center of the spherical configuration?

 

[PhDeezNutz]

It sounds theoretical to me but I’m not sure. I’ll let others comment on its feasibility/application.

 

[zenterix]

In addition, it's not clear why we can apply the formula $R=\frac{\rho L}{A}$.

Let me try to reason about this.

Given a spherical shell that is a resistor, we would like to find an expression for $R=\frac{\Delta V}{I}$.

We need to find expressions for $\Delta V$ and $I$.

To get $I$ we compute the flux of $\vec{J}$ through a cross section of the shell, which is the shell itself seemingly.

But shouldn't this flux be zero?

 

[haruspex]

In addition, it's not clear why we can apply the formula $R=\frac{\rho L}{A}$.

Let me try to reason about this.

Given a spherical shell that is a resistor, we would like to find an expression for $R=\frac{\Delta V}{I}$.

We need to find expressions for $\Delta V$ and $I$.

That's backwards. You can't know I until you have figured out the resistance.
You have to start by showing, or realising, that $\Delta R=\frac{\rho \Delta r}{A}$.

 

[PhDeezNutz]

That's backwards. You can't know I until you have figured out the resistance.
You have to start by showing, or realising, that $\Delta R=\frac{\rho \Delta r}{A}$.

This^^^

## V = IR \Rightarrow I = \frac{V}{R}# tells you the current a resistor would have through it if a potential was applied to it. Even without a potential applied to it that resistor still has resistance.

Your light bulb still has resistance even when it’s off.

 

[haruspex]

This^^^

$I = VR$ tells you the current a resistor would have through it if a potential was applied to it. Even without a potential applied to it that resistor still has resistance.

Your light bulb still has resistance even when it’s off.

V/R, I believe.

 

[PhDeezNutz]

V/R, I believe.

okay that mistake on my part is inexcusable.

Fixed.

 

[zenterix]

That's backwards. You can't know I until you have figured out the resistance.
You have to start by showing, or realising, that $\Delta R=\frac{\rho \Delta r}{A}$.

It doesn't seem to be backwards. Consider the derivation of this formula from Purcell's Electricity and Magnetism

 

[Delta2]

how is this resistor actually used?

Engineering implementation details. I think they should make a small infinitesimal hole in the outer shell and down to the surface of the inner shell, they have to pass a infinitesimally thin externally insulated wire through that hole that connects to the surface of the inner shell and to one pole of the battery.

 

[zenterix]

Engineering implementation details. I think they should make a small infinitesimal hole in the outer shell and they have to pass a infinitesimally thin externally insulated wire through that hole that connects to the surface of the inner shell and to one pole of the battery.

Do you agree that the problem statement makes it difficult to even start this problem?

In every solution I've seen it seems people are on autopilot as to the integration involved. But that is the easy part. They are basically applying a formula for resistance.

 

[zenterix]

@haruspex do you know how to make sense of the formula $R=\rho L/A$ in the case of the spherical shell resistors that make up the integral that solves this problem? Or at least derive it for this case?

If we were to assume that the shell is made up of infinitesimally small rectangular resistors, they would all be in parallel.

 

[zenterix]

Here is a weird solution (the formatting is messed up but it is still readable).

This solution seems to assume the electric field is that of a point charge located at the center of the spherical configuration.

 

[Delta2]

Do you agree that the problem statement makes it difficult to even start this problem?

In every solution I've seen it seems people are on autopilot as to the integration involved. But that is the easy part. They are basically applying a formula for resistance.

The only thing missing from the statement is "Assume that when we apply voltage in the inner and outer surface, the resistor reaches the steady state where the current is the same through every concentric surface of the configuration"

 

[zenterix]

The only thing missing from the statement is "Assume that when we apply voltage in the inner and outer surface, the resistor reaches the steady state where the current is the same through every concentric surface of the configuration"

There would also need to be information about why we can disregard the assumptions violated by that setup.

The assumptions about $\vec{J}$ seem to be violated without explaining how charge gets into the spherical setup in the first place.

 

[Delta2]

There would also need to be information about why we can disregard the assumptions violated by that setup.

The assumptions about $\vec{J}$ seem to be violated without explaining how charge gets into the spherical setup in the first place.

Yes it seems like there should be charge accumulation in the surface of the inner shell if that's what you mean, unless we get an explanation of how the current circulates from the inner surface back to the outer surface.

But if you think that you think more like an engineer than a scientist, I classify this as an engineering detail.

 

[zenterix]

Yes it seems like there should be charge accumulation in the surface of the inner shell if that's what you mean, unless we get an explanation of how the current circulates from the inner surface back to the outer surface.

But if you think that you think more like an engineer than a scientist, I classify this as an engineering detail.

I think I have the capability to think like both.

I think you could also argue that disregarding such nuance would be the approach of some engineers, and that such details would be called academic by them.

In fact, I think that unless we go into the physics nuance this whole problem is useless unless you are practicing integration.

I still don't see the point of this problem really.

 

[haruspex]

It doesn't seem to be backwards.

It is not backwards in the context of defining resistance, but it is backwards in the context of the problem in this thread (where the resistivity is a given).

 

[haruspex]

@haruspex do you know how to make sense of the formula $R=\rho L/A$ in the case of the spherical shell resistors that make up the integral that solves this problem? Or at least derive it for this case?

I don't understand your difficulty. A shell radius r, thickness dr, with radial current, is a resistor length dr and area $4\pi r^2$. Plug that into the formula.