The general term of the current reads
J_total=J_conduction+J_Displacement.
Where in most cases J_C obeys Ohms law and J_D=epsilon dE/dt.
Usually when a capacitor is disconnected from the electric circuit the the condition current is zero and so is J_t=0.
What will happened if I now intentionally adding a time dependent electric field between the capacitor plates (not sure if it's possible so it may be considered as a though experiment)?
My thought is that J_C will cancel somehow the displacement current in order to maintain J_t=0 since there can't be current outside the capacitor.
Is this analysis true or false? am I missing something? What is the meaning of conduction current in this case?
Hi Dor Gotleyb ,
I'm not sure about the source of the time dependent field: Do I understand correctly that the source is not connected to the plates of the capacitor? So, as an example, the distance of the plates is 10 mm, and you place a dipole of 5 mm length midway between the plates, for the sake of simplicity driven by a pointlike source inside the dipol. Does this picture comply with your intention?
Yes @Joerg Fricke, your example sums it well, thanks for the clarification.
I'm also not sure what will be the difference if instead of being restricted to the area between the plates the field will be presented also outside (maybe putting the capacitor between a much larger capacitor connected to an AC source, then the field is spatially uniform and varies in tome)
Eventually, I'm hoping to gain some insights and to better understand the relation or interplay between the total, conduction, and displacement currents.
Thanks Jinfeng Li for commenting.
I meant that just outside the capacitor the conduction currnet is zero and thus the total current. Since the current is continuous (or isn't it?) also inside it should be zero but there is a non zero displacement current. Or maybe a surface current that forms at the boundary?
Assuming the dipole axis aligned perpendicular to the plates of the capacitor, you will have a conduction current in the dipole, and a displacement current outside the dipole. If you imagine a plane parallel to the plates, cutting through the dipole, the sum of both currents through this plane is zero (neglecting for the moment the finite propagation velocity of the dipole field).
Basically, the field and the displacement current of the dipole reaches beyond the plates. But the field of the dipole induces conduction currents in the plates radial to the axis of the dipole, and the fields of this currents have two effects:
Inside the capacitor, the dipole is mirrored by both plates. Outside the capacitor, the superposition of the dipole field and of the radial current field (and hence the displacement currents) adds up to nearly zero everywhere.
If the plates were electrically shortened everywhere at the edges (thereby discharging the capacitor, of course, and completely enclosing the dipole), the superposition would result in exactly zero. (Provided the plates are sufficiently thick, depending on the spectrum of the dipole field.)
Taking the propagation of the dipole field into account, the capacitor acts as a parallel plate guide for the dipole field. Its wave impedance decreases with increasing distance from the dipole, so the dipole wave is partly reflected.
You cannot have displacement current unless the fields are changing. If the fields are changing then there will be currents in the capacitor, even if there are no external connections. The changing displacement current will result in changing magnetic fields which result in currents in the conductors (capacitor plates) next to the fields. These currents will be mainly along the capacitor plates, not through them. Currents will run along the plates because the changing fields will be associated with a wavelength, so the displacement current will not be constant across the capacitor.
Obviously: The total power a plate is the integral of the Poynting vector normal (S) to the plate with respect to the area (a) of the plate. In a capacitor power=i squared/C where C is capacitance and i current. Therefore i=square root of
(c a S). Note: each may see a different current. Remember the Poynting vector is power are density of the filed (power/area).
An electric current must flow in a closed loop. It is not possible to create an electric field between the plates of the capacitor without at least a temporary connection to an external voltage and current source.
So in the example of the point source above, that point source must be connected to an external source to charge up the plates of the capacitor, otherwise there would be no electric field between the plates. The same argument applies to putting the capacitor between the plates of a larger capacitor, although in this case, the external connection is capacitive.
In other words the original formulation of the problem is paradoxical.
@Ulrich Hugel, I've added a picture to the proposed system, I'm not sure why is it not possible, I'll be glad if you could clarify.
Nevertheless, the question is hypothetical, and it's intention is to understand the relation between conduction and displacement currents in situation where they sum up to zero.
It is perfectly possible to create an electric field between the plates of a capacitor with no external connection.
If there are two isolated plates for example 1 mm apart, they form a capacitor with no external circuit. If a radio wave of strength 1 mV/mm passes the plates travrelling in a direction parallel to the plates, some of it will pass between the plates and there will be an electric field of 1 mV/mm between the plates.
If the plates were PEC and perfectly thin then they would not affect the radio signal in any way.
- Badges
- Science topic
- Similar topics
- Physical Sciences
- Electromagnetism