Magnetic field of a spherical capacitor

[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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[Gigel]

I understand it is DC voltage applied to the plates.

If B=0 then what is the Poynting vector, especially in the case with increasing voltage and σ=0? Normally it is S=E x B/μ0, but if B=0 that would make S=0 so the capacitor is not charging? 0.o

Also, the wires powering the plates don't seem to be important. You can simply replace them with radioactive sources of equal activity, but emitting oppositely charged particles. Place one inside, the other outside of the capacitor. The plates collect the charges and charge up. No current goes between the plates, yet E increases in time, so there is just displacement current between the plates.

I have to admit, spherical capacitors are pretty interesting for having B=0; they have no inductance (except for the power wires) so they would charge and discharge pretty fast. But what do you do with the Poynting vector? By the same reasoning as above for B, S should be radial and thus 0. But then how does energy flow into the capacitor?

 

[phyzguy]

Charge is conserved, so if you have a radioactive source inside the inner sphere which emits charged particles which are collected by the inner sphere, the total charge inside the inner sphere does not change. So the total charge inside the Gaussian surface (green dotted line in the attached drawing) does not change. So the E-field between the plates does not change.

 

[rude man]

Charge is conserved, so if you have a radioactive source inside the inner sphere which emits charged particles which are collected by the inner sphere, the total charge inside the inner sphere does not change. So the total charge inside the Gaussian surface (green dotted line in the attached drawing) does not change. So the E-field between the plates does not change.

If there is discharge then there is a (displacement) current inside the capacitor, and so a changing E field and a B field by ∇ x H = ∂D/∂t. But only if an external discharging resistor is used. If the capacitor is discharged via its own internal resistance then the net current is zero because the displacement current and the conductive current are equal in magnitude and opposite in direction.

 

[Gigel]

If there is discharge then there is a (displacement) current inside the capacitor, and so a changing E field and a B field by ∇ x H = ∂D/∂t. But only if an external discharging resistor is used. If the capacitor is discharged via its own internal resistance then the net current is zero because the displacement current and the conductive current are equal in magnitude and opposite in direction.

That is Feynman's argument too.

phyzguy is right in the post above. I was trying to eliminate the wire powering the inner spherical plate that crosses the space between plates and thus gives an ugly conduction current between plates. I was trying to hold only a displacement current. So far the only way I could do this is to make the charged plates approach each other; or collapse the outer plate towards the inner one. Then apparently (but is it so?) one can choose a Gaussian surface between plates that is not crossed by a conduction current at a given time, just by a displacement current. So then ∇ x H = ∂D/∂t and there would be rotating H around each D line that would cancel with neighbouring Hs, thus yielding a total H=0. And then the Poynting vector would also be 0, so no flow of energy towards the collapsing capacitor. o.0 Weird!

 

[phyzguy]

I think if the whole configuration is spherically symmetric, then we must have B = 0, since there is no spherically symmetric non-zero B field which has zero divergence everywhere.

 

[Gigel]

phyzguy, I agree with you, but still can't get rid of all the weirdness.