Magnetic field of a spherical capacitor

[ShayanJ]

Homework Statement

A spherical capacitor with inner and outer radii a and b, contains a dielectric material with small conductivity $\sigma$ between its spheres.Find the vector potential and magnetic field of this configuration.

Homework Equations

The Attempt at a Solution

$\nabla^2\phi=0 \Rightarrow \frac{1}{r^2}(r^2\frac{d\phi}{dr})=0 \Rightarrow \phi=-\frac{p}{r}+q \\ I=\frac{\Delta \phi}{R}=\frac{p(\frac 1 a -\frac 1 b)}{ \int_a^b \frac{dr}{4\pi \sigma r^2 }}\Rightarrow I=4\pi \sigma p \Rightarrow \vec{J}=-\frac{\sigma p}{r^2}\hat{r}$
$\vec{A}= - \frac{\mu_0}{4\pi} \int_a^b \int_0^\pi \int_0^{2\pi} \frac{\sigma p \hat{r}}{r^{'2}}\frac{r^{'2} \sin\theta d\varphi d\theta dr}{\sqrt{ r^2+r^{'2}-2r r^{'} \cos\gamma }}= - \frac{\sigma p \mu_0}{4\pi r} \int_a^b \int_0^\pi \int_0^{2\pi} \hat{r} \frac{\sin\theta d\varphi^{'} d\theta^{'} dr^{'}}{\sqrt{ 1+(\frac{r^{'}}{r})^2-2 \frac{r^{'}}{r} \cos\gamma }}= - \frac{\sigma p \mu_0}{4\pi r} \int_a^b \int_0^\pi \int_0^{2\pi} \hat{r} \sin\theta (\sum_{n=0}^\infty P_n(\cos\gamma) (\frac{r^{'}}{r})^n) d\varphi^{'} d\theta^{'} dr^{'}$
And then,I expanded the Legendre functions using the following identity:
$P_n(\cos\gamma)=\frac{4\pi}{2n+1}\sum_{m=-n}^n Y^{m*}_n(\theta^{'},\varphi^{'})Y^m_n(\theta,\varphi)$
After some calculations,I arrived at:
$\vec{A}=\frac{\sigma \mu_0 p}{12 r^2}(b^2-a^2)\sin\theta e^{i\varphi}(\hat{x}+\hat{y})$
As you can see,its complex!
Another issue is about the symmetry of the problem. The spherical symmetry of the configurations suggests that the magnetic field should be radial but a radial magnetic field violates Gauss's law in magnetism($\oint \vec{B}\cdot \vec{dS}=0$). On the other hand,any other form for the magnetic field, doesn't respect the symmetry of the configuration.
I even thought that maybe the magnetic field is zero but as you can see easily,the vector potential isn't the gradient of a scalar field.
Please help!
Thanks

 

[TSny]

I don't think you need to get into detailed integrations. From your general expression for $\vec{A}$ in terms of $\vec{J}$, can you argue that $\vec{A}$ must be radial? What will that tell you about $\vec{B}$?

 

[Goddar]

I don't think you need to get into detailed integrations. From your general expression for $\vec{A}$ in terms of $\vec{J}$, can you argue that $\vec{A}$ must be radial? What will that tell you about $\vec{B}$?

Hi.
I don't think you can argue that $\vec{A}$ is radial only because we obtain it by integrating a radial function: in Cartesian coordinates, this logic would work because unit vectors $\hat{x}$ , $\hat{y}$ and $\hat{z}$ can be taken out of the integral but in spherical coordinates we can get any direction at the end of the integration...

Also, a radial B-field seems perfectly good to me as long as $\nabla$$\cdot$$\vec{B}$ = 0, as for example: $\vec{B}$ = $\frac{C}{r^{2}}$$\hat{r}$

So i would start with Ampere's law in integral form (here in Gaussian units):
$\oint$$\vec{B}$$\cdot$d$\vec{r}$ = $\frac{4\pi}{c}$$\int$$\vec{J}$$\cdot$d$\vec{a}$ + $\frac{1}{c}$$\frac{d}{dt}$$\int$$\vec{E}$$\cdot$d$\vec{a}$,
Around a circle centered at the origin on the left side, a hemispherical surface on the right and with the relation $\vec{J}$ = $\sigma$$\vec{E}$ (the electric field being easily found with Gauss' law and a charge Q(t) on the inner sphere). Then from the radial B-field, find a suitable vector potential...

 

[ShayanJ]

I don't think you need to get into detailed integrations. From your general expression for $\vec{A}$ in terms of $\vec{J}$, can you argue that $\vec{A}$ must be radial? What will that tell you about $\vec{B}$?

As Goddar said,its not like that!
I wrote the cartesian unit vectors in terms of spherical ones and now we have:
$\vec{A}=\frac{\sigma \mu_0 p}{12r^2}(b^2-a^2)\sin\theta e^{i\varphi} \left[ (\cos\varphi+\sin\varphi)(\sin\theta \hat{r}+\cos\theta \hat{\theta})+(\cos\varphi-\sin\varphi)\hat{\varphi} \right]$
As you can see,all the components are present!

Hi.
I don't think you can argue that $\vec{A}$ is radial only because we obtain it by integrating a radial function: in Cartesian coordinates, this logic would work because unit vectors $\hat{x}$ , $\hat{y}$ and $\hat{z}$ can be taken out of the integral but in spherical coordinates we can get any direction at the end of the integration...

Also, a radial B-field seems perfectly good to me as long as $\nabla$$\cdot$$\vec{B}$ = 0, as for example: $\vec{B}$ = $\frac{C}{r^{2}}$$\hat{r}$

So i would start with Ampere's law in integral form (here in Gaussian units):
$\oint$$\vec{B}$$\cdot$d$\vec{r}$ = $\frac{4\pi}{c}$$\int$$\vec{J}$$\cdot$d$\vec{a}$ + $\frac{1}{c}$$\frac{d}{dt}$$\int$$\vec{E}$$\cdot$d$\vec{a}$,
Around a circle centered at the origin on the left side, a hemispherical surface on the right and with the relation $\vec{J}$ = $\sigma$$\vec{E}$ (the electric field being easily found with Gauss' law and a charge Q(t) on the inner sphere). Then from the radial B-field, find a suitable vector potential...

You have two mistakes in your post.

1-If $\vec{B}=\frac{C}{r^2}\hat{r}$,its divergence is not zero but $4\pi C \delta(r)$. Or, to put it in simpler terms, its surface integral on a spherical surface enclosing the origin isn't zero but $4\pi C$!

2-In the integral form of Ampere's law,the line element isn't radial,but in the direction of the curving of the integration so in case of a radial B field,the line integral on the left vanishes!

 

[Goddar]

For your first point the tentative B is valid in the region b > r > a and B = 0 everywhere at r < a since E = 0, so the origin is not a problem and the divergence is zero everywhere. Anyway radial B-fields happen and I'm not saying this is the right one, you can Google some examples.

I'll take your second point, i didn't pay enough attention.

Then i guess you're right to go after the vector potential but it seems strange that in Cartesian coordinates you obtain A only in terms of x and y directions, given the spherical symmetry, so there might be a mistake there...

 

[ShayanJ]

For your first point the tentative B is valid in the region b > r > a and B = 0 everywhere at r < a since E = 0, so the origin is not a problem and the divergence is zero everywhere. Anyway radial B-fields happen and I'm not saying this is the right one, you can Google some examples.

I'll take your second point, i didn't pay enough attention.

Then i guess you're right to go after the vector potential but it seems strange that in Cartesian coordinates you obtain A only in terms of x and y directions, given the spherical symmetry, so there might be a mistake there...

Yeah,that's right.
Anyway.I think its not a problem for the vector potential because we can always add the gradient of a scalar field and so we can make the potential have a z component that way.But even if we don't do that,the B field is going to have a z component and because B is observable and not A,again there is no problem.

Anyway,the curl of the vector potential I derived is:
$\vec{B}=\vec{\nabla}\times\vec{A}=-\frac{e^{i\varphi}\sigma mu_0 p (b^2-a^2)}{12r^3} \left\{ \cos\theta\left[ -\cos\varphi+\sin\varphi+i(\cos\varphi+\sin\varphi) \right]\hat{r}-\sin\theta \left[ i(\cos\varphi+\sin\varphi)+2(\cos\varphi-\sin\varphi) \right]\hat{\theta}+3\cos\theta\sin\theta(\cos\varphi+\sin\varphi)\hat{\varphi} \right\}$
Which is in no way radial!
And is complex!

 

[ShayanJ]

There is an alternative approach to the above one.The observation point can be arbitrarily assumed to be on the z axis(which doesn't harm the generality). The only change to the vector potential integral in the first post is that $\gamma$ becomes $\theta$ which means after expansion using Legendre functions, the addition theorem is not needed.The result I got with this method is:
$\vec{B}=-\frac{\sigma \mu_0 p \sin\theta \hat{\varphi}}{2r^3} \sum_{n=0}^\infty \frac{f(n)}{r^{2n}}$
Where $f(n)=\frac{b^{2(n+1)}-a^{2(n+1)}}{2^n} \sum_{k=0}^n \frac{(-1)^k [2(2n-k+1)]!}{k! (2n-k+1)![2(n-k)+1]![2(n-k)+3]!}$
This solves the problem of magnetic field being complex but still doesn't seem right from the point of view of symmetry.This magnetic field is azimuthal but this can't be right because we can choose the z axis arbitrarily and the magnetic field can't be rotating around all directions of space!
Damn it...Its too confusing.In fact in this case we have current wires in all directions of space (radii of the spherical capacitor) and each has a magnetic field circulating around it so the above field seems reasonable in some sense(or not?!)but what is it like?
Too confusing!

 

[phyzguy]

The Feynman lectures on physics, Vol 2 Chapter 18, has an analysis of just this situation. It is available for free online here. Understanding this problem will give you a key insight to electromagnetism.

 

[ShayanJ]

The Feynman lectures on physics, Vol 2 Chapter 18, has an analysis of just this situation. It is available for free online here. Understanding this problem will give you a key insight to electromagnetism.

Thank you very much.
I should have known it,but...shame!
It happens sometimes...you expect something so much that you can't think about it somehow else!
I think one thing that should be solved for a novice physicist like me,is thinking out of structure and free from patterns.That takes time and practice though!

 

[phyzguy]

Thank you very much.
I should have known it,but...shame!
It happens sometimes...you expect something so much that you can't think about it somehow else!
I think one thing that should be solved for a novice physicist like me,is thinking out of structure and free from patterns.That takes time and practice though!

Why should you have known it? It took the best minds of the human race more than a century to figure this all out. Don't feel bad if you didn't grasp it overnight.

 

[ShayanJ]

Why should you have known it? It took the best minds of the human race more than a century to figure this all out. Don't feel bad if you didn't grasp it overnight.

Well, I knew that moving charges have changing electric field and I knew Maxwell's equation.But looks like I just was saying to myself "this is a problem which should be solved using the formula giving the vector potential of a current density".
Anyway,I don't feel that bad,I'm just trying to remember it so I don't make the same mistake and think about problems using real physics, not that its this kind of problem or that kind of problem!

 

[ShayanJ]

I still see another issue. In this problem,we have$\vec{\nabla}\times\vec{B}=0$. But we can write $\vec{B}=\vec{\nabla}\times\vec{A}$ which,along with the vanishing of curl gives us:
$\vec{\nabla}\times\vec{\nabla}\times \vec{A}=0 \Rightarrow \vec{\nabla}(\vec{\nabla}\cdot\vec{A})-\nabla^2\vec{A}=0$
Using Coulomb gauge we'll have:
$\nabla^2\vec{A}=0$
So we have the potential obeying Laplace equation which has non-trivial solution and its not at all obvious that the magnetic field coming out of such a potential is zero!

 

[Goddar]

Yes but boundary conditions are crucial in solving these equations and here they don't leave you much choice...
Interesting problem

 

[ShayanJ]

But what are the boundary conditions?
Zero B-field at r=a and infinity?

 

[Goddar]

r = a and r = b

 

[ShayanJ]

r = a and r = b

Ahh...yes...Charges on the conductors shouldn't accelerate and so there should be no Lorentz force parallel to the spheres' surfaces.But B is radial and $\hat{r}\times\hat{\theta}=\hat{\varphi}$ and $\hat{r}\times\hat{\varphi}=-\hat{\theta}$ and so B should be zero at the surfaces.
Very beautiful!

 

[phyzguy]

I still see another issue. In this problem,we have$\vec{\nabla}\times\vec{B}=0$. But we can write $\vec{B}=\vec{\nabla}\times\vec{A}$ which,along with the vanishing of curl gives us:
$\vec{\nabla}\times\vec{\nabla}\times \vec{A}=0 \Rightarrow \vec{\nabla}(\vec{\nabla}\cdot\vec{A})-\nabla^2\vec{A}=0$
Using Coulomb gauge we'll have:
$\nabla^2\vec{A}=0$
So we have the potential obeying Laplace equation which has non-trivial solution and its not at all obvious that the magnetic field coming out of such a potential is zero!

Since j has only a radial component, A will have only a radial component. Also, because of the spherical symmetry, the solution can only depend on r. So the solution for A must have the form:
$$A = A(r) \overrightarrow{e_r}$$

this does have non-trivial solutions, but the curl of this is zero, so B = 0.

 

[vanhees71]

I'm not so sure about that, because the magneto-static Biot-Savart law reads
$$\vec{A}(\vec{x})=\int_{V} \mathrm{d}^3 \vec{x}' \frac{\vec{j}(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
Note that you have to evaluate the integral with cartesian components, i.e.,
$$\vec{j}(\vec{x})=\frac{C}{\vec{x}^2} \vec{e}_r=\frac{C}{\vec{x}^2} [\cos \varphi \sin \vartheta \vec{e}_x + \sin \varphi \sin \vartheta \vec{e}_y + \cos \vartheta \vec{e}_z].$$ The volume element you can, of course write in spherical coordinates,
$$\mathrm{d}^3 \vec{x}'=r'^2 \sin \vartheta' \mathrm{d} r' \mathrm{d} \vartheta' \mathrm{d} \varphi'.$$
However, I guess the integral is not so easily calculated. Perhaps, it's easier to find the answer by directly solving the local equation for $\vec{A}$ with the appropriate boundary conditions for $\vec{B}$.

Note that in spherical coordinates, you cannot directly evaluate $\Delta \vec{A}$, which is only defined in Cartesian components. You have to use
$$\vec{\nabla} \times \vec{B}=\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\frac{1}{c} \vec{j}.$$
Note that you can use one constraint on $\vec{A}$ due to ("static") gauge invariance, e.g., the Coulomb-gauge condition
$$\vec{\nabla} \cdot \vec{A}=0.$$

 

[phyzguy]

Yes, clearly it's not correct to say that since j is radial, A must be radial. I apologize for muddying the water.

 

[TSny]

Since j has only a radial component, A will have only a radial component. Also, because of the spherical symmetry, the solution can only depend on r. So the solution for A must have the form:
$$A = A(r) \overrightarrow{e_r}$$

I think this is correct. If $\vec{J}$ is radial of the form $\vec{J}(\vec{r}) = J(r)\hat{r}$ then you can show that the integral of $\frac{\vec{J}(\vec{r'})}{|\vec{r}-\vec{r'}|}d\tau'$ is also radial.

this does have non-trivial solutions, but the curl of this is zero, so B = 0.

Right.
---------------------------------------------------------------------------------------

Yes, clearly it's not correct to say that since j is radial, A must be radial. I apologize for muddying the water.

Is this because you are no longer assuming a gauge where A is the integral of J/|r-r'|?

I think it would be odd to choose a gauge that would make A non-radial. It would disrespect the spherical symmetry of the problem. But, you could do that if you wanted.

 

[TSny]

You have to use
$$\vec{\nabla} \times \vec{B}=\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\frac{1}{c} \vec{j}.$$

For this problem, you'll need the full Maxwell equation $$\vec{\nabla} \times \vec{B}=\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\frac{1}{c}( \vec{j} +\frac{\partial \vec{E}}{\partial t}).$$
B = 0, but j is not zero.

 

[vanhees71]

This is very difficult. As I understood the question, it's supposed to be a stationary current (DC).

 

[TSny]

As I understood the question, it's supposed to be a stationary current (DC).

Oh. In that case the capacitor would need to be connected to a voltage source to keep a fixed charge on the plates as charge leaks between the plates. Then you would need to worry about the fields produced by the currents entering and leaving the capacitor.

 

[vanhees71]

Of course, there must be a constant potential on the capacitor. As usual, one neglects the perturbations of the current conducting wires connecting the capacitor with the battery. Otherwise you can't do the problem analytically. :-)).

 

[TSny]

Of course, there must be a constant potential on the capacitor. As usual, one neglects the perturbations of the current conducting wires connecting the capacitor with the battery. Otherwise you can't do the problem analytically. :-)).

Assume J is radially outward and constant in time. Integrating the flux of J over a spherical surface between the plates would gives the rate at which charge is leaving the volume enclosed by the surface. So, if J is non-zero and constant in time, the net charge inside the volume must be decreasing in time.

But if J is constant in time, then E is constant in time because J = σE. If you apply Gauss' law to the same spherical surface, then constant E would imply that the charge enclosed in the surface must remain constant. A contradiction.

You could keep the charge constant on the inner plate by feeding in charge to the plate from an outside source at the same rate charge is leaking off of the plate. So, there must be a net input current of the same magnitude as the net leaking current. Thus, the fields of the input current would be of the same order of magnitude as the fields of the leaking current.

If you do away with the external source and just let the capacitor discharge slowly through leakage, then you can show that the charge on the inner plate decreases exponentially with time with a time constant $\tau = \small RC$ where C is the capacitance and R is the resistance of the material between the plates. Then, J, E, and A decrease exponentially with time during the discharge. B = 0 during the discharge.

 

[vanhees71]

Of course the charge is constant. You must have
$$\vec{\nabla} \cdot \vec{j}=0$$
if
$$\partial_t \rho=0.$$
That means it flows as much charge into the volume between the two spheres as flows out of it.

I guess the most simple solution of the problem is to evaluate the integral. In Coulomb gauge we have
$$\vec{A}(\vec{x})=\int_{V} \mathrm{d}^3 \vec{x}' \frac{\vec{j}(\vec{x}')}{4 \pi c |\vec{x}-\vec{x}'|}.$$
As I said earlier, you can use spherical coordinates for $\vec{x}'$ and $\vec{x}$ but you should write out $\vec{j}(\vec{x}')$ in cartesian components:
$$\vec{j}(\vec{x}')=\frac{C}{r'^2}(\cos \varphi' \sin \vartheta' \vec{e}_x+ \sin \varphi' \sin \vartheta' \vec{e}_y + \cos \vartheta' \vec{e}_z).$$

 

[TSny]

Of course the charge is constant. You must have
$$\vec{\nabla} \cdot \vec{j}=0$$
if
$$\partial_t \rho=0.$$
That means it flows as much charge into the volume between the two spheres as flows out of it.

But what about the charge on the inner spherical conductor? Does it remain constant? If so, then current must be supplied to the inner conductor by an external source if there is a leakage current. I don't see how you can neglect the fields of the current supplied by the source.

If you wanted to maintain perfect spherical symmetry, then the current going from the supply to the inner sphere would need to be a radially inward current of the same magnitude as the outward radial leakage current. Seems to me the two currents would cancel. Maybe I'm overlooking something.

Anyway, it would be nice to know if Shyan was considering an isolated discharging capacitor or a capacitor with an external source maintaining constant charge on the plates.

 

[Gigel]

The Feynman lectures on physics, Vol 2 Chapter 18, has an analysis of just this situation. It is available for free online here. Understanding this problem will give you a key insight to electromagnetism.

I have tried to solve the same problem as in the first post with Ampere's law, with some results which are probably wrong. Feynman considered an example which I don't think is very relevant here. He considered a radial distribution of both currents and electric fields, so that the current j is canceled at any point by the displacement current 1/c^2 dE/dt (in SI units).

Now I think the 2 currents can be separated so that there is only one of them on the spherical surface that is considered. The cleanest way I can think of is the following: the spherical capacitor is charged and is not leaking; then the plates are pushed one toward the other and the surfaces are modified continuously so that they remain spherical with the same initial center. Then the plate movement would give an electrical current for each plate, while the field between plates would change in time, thus giving a displacement current. The 2 currents would not appear in the same region. A spherical surface between the plates would see only the displacement current.

Then how could one prove that B=0? This would be true because the spherical symmetry means B is radial and since div B = 0, B must be 0.

 

[Gigel]

Now about my results, I considered a capacitor in the form of a spherical cap (plates extend for 0<θ<θ₀, so they are not full spheres). The two plates have very close radii, so that the fringe effect can be neglected (but can it?). A took a surface S between the plates (say in the middle) and applied Ampere's Law to determine the magnetic field on its contour (loop L). I considered B to be circular here due to the figure's symmetry wrt. the vertical axis (also think how the magnetic field produced by each j_d looks like - it should be along L). The capacitor's plates are not represented in the figure.

Now as the plates approach but keep their spherical shape, the electric field between varies as ∂E/∂t. Then by Ampere-Maxwell Law on L:

∫_LB⋅dl = μ₀ ∫∫_S (j + j_d)⋅dS,

where j is the current density j=0 and j_d = 1/c² ∂E/∂t is the displacement current density, the only one counting on S. Now if j_d is constant, this can be integrated easily. The surface element is dS = R dθ r dφ = R² sin(θ) dθ dφ, where r = R sin(θ) and θ goes from 0 to θ₀ and φ from 0 to 2π. It follows:

2πr B = μ₀ j_d S,

2πR sin(θ₀) B = μ₀ j_d 2πR² (1-cos(θ₀))

and thus:

B(θ₀) = μ₀ j_d R (1-cos(θ₀))/sin(θ₀) = B1.

Now extend this to θ₀=π and it follows that B=something/0=∞, where it should be 0 I think.

Another approach would be to take a full spherical capacitor, so that loop L sees 2 displacement currents (one above, one below L). The two currents produce opposing magnetic fields on L so that:

B(θ₀) = μ₀ j_d R [ (1-cos(θ₀))/sin(θ₀) - (1-cos(π-θ₀))/sin(π-θ₀) ],

B(θ₀) = -2 μ₀ j_d R cos(θ₀)/sin(θ₀) = B2.

Now B=-∞ at θ₀=0 and B=∞ at θ₀=π and B=0 at θ₀=π/2 (on the equator). Again B is not 0 everywhere. Instead B is along the sphere, such that div B=0, but it lacks spherical symmetry. Moving the vertical axis obliquely would change the direction of B at the same points, which cannot be.

Am I missing something? Does the fact that the last formula works well on the equator mean something? Maybe loop L has to be composed of 2 loops symmetrical wrt. to the equator and knotted on a line perpendicular to the equator so that B cancels on it.

 

[phyzguy]

Gigel - Why do you think Feynman's analysis is not relevant? The OP has a spherical capacitor with a conductive dielectric between the two concentric spheres. Since everything is spherically symmetric, the current will be radial, just like in Feynman's analysis. The current in the conductive dielectric will be canceled by the displacement current, and the B-field will be zero. How can it be otherwise? In a spherically symmetric problem, in what direction could the B-field point?

 

[Gigel]

Sorry, I forgot that his capacitor was leaking. You are right, it is Feynman's case there.

Anyway, there is always the possibility of a non-leaking spherical capacitor. In that case there is no current, yet the electric field may vary. That produces a magnetic field generally. In a spherically symmetric setting, B would point only radially, and since its divergence is 0 by Maxwell's equations, it should be that B=0. Still I can't prove that. When a spherical cap capacitor is completed to a fully spherical capacitor, its B should become 0, yet in my case it becomes infinite. I think I am missing something.

 

[phyzguy]

In the case of a non-leaking spherical capacitor, the electric field can't change unless there is a current to change the charge on the capacitor. If the current isn't radially symmetric as in Feynman's case, there has to be a wire or something to source the current. This breaks the spherical symmetry and B can become non-zero.

 

[EricGetta]

I was confused by the problem at first but I see now that its implied the capacitor is initially charged and that is discharging through the conductive path causing the magnetic field.

 

[Gigel]

In the case of a non-leaking spherical capacitor, the electric field can't change unless there is a current to change the charge on the capacitor. If the current isn't radially symmetric as in Feynman's case, there has to be a wire or something to source the current. This breaks the spherical symmetry and B can become non-zero.

Well in the case I presented the current is not brought by any wire. It is indeed a radially symmetric current and it appears because the plates move one towards the other, while PRESERVING the spherical shape with respect to the center of the system. They change their radius as they do so, but they remain spherical. It is as the plates were formed of elastic infinitesimal elements dS which are moved radially, while at the same time they are stretched a bit to fit on a smaller sphere.

But this current appears only in the plate area, not between the plates. Between the plates there is just the displacement current, i.e. the variation in time of the electric field. No charge moves between the plates, so there is no need of the continuity equation (both Q and j are 0).

Yet there should be no B. But by starting with a capacitor in the shape of a spherical cap (it is not completely spherical, just the upper part), that one should have a B, which is along the surface of the sphere at the cap's edge. Now by extending the spherical cap to a fully spherical structure, B becomes infinite instead of 0. There is something wrong in my assumptions I think.

 

[rude man]

Homework Statement

A spherical capacitor with inner and outer radii a and b, contains a dielectric material with small conductivity $\sigma$ between its spheres.Find the vector potential and magnetic field of this configuration.

Zero for both? No applied voltage given, ergo no potential distribution. And if the applied voltage is dc there's still no mag field. What sort of applied voltage is assumed?
And I don't see the parallel between the Feynman example and this. In his case there is a net reduction of charge density over time; with an applied dc voltage here there isn't. In particular, Feynman's eq. 18-5 would be zero with applied dc. Applied ac? Clue me in folks.
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