Magnetic field of a spherical capacitor

[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.