# Displacement currents, aren't really currents?

I'm confused with this definition of displacement currents within capacitors via Wikipedia:

However it is not an electric current of moving charges, but a time-varying electric field.

It's a time-varying electric field, but there isn't any actual flow of current, while at the same time... it produces a magnetic field. It's counter-intuitive.

If air was in the center between the plates, it's possible to have some "current flow" due to atoms within the gas molecules, likewise, for a dielectric.

Yet, displacement currents aren't considered "real/free" current. Could someone clarify this confusion for me.

How significantly different are displacement currents from normal/free currents?

• IMO It is mostly a confusing historical name, there is no charge transport (which is the usual definition of electrical current), but it has the same units as a current density and it is a source of the magnetic field. (And it has to be added to make the whole theory conserve the total charge). – Sebastian Riese May 8 '19 at 19:42
• It is not a "source of magnetic field". – my2cts May 8 '19 at 21:48
• A displacement current generates a time varying magnetic field - which when coupled with a time varying electric field, produces an electromagnetic wave - which in turn, allows a cell phone tower to transmits voice and data to a cell phone. The antenna on the cell phone converts the electromagnetic wave to an electron current which the cell phone uses to decode voice and data. Without the displacement current there would no cell phones or light. It's not a real current - it's "magically" current. – Cinaed Simson May 9 '19 at 8:06
• @my2cts If the displacement currents aren't, is it the propagating EM field? See I'm confusing that propagating EM-field with current flow in a conductor that Biot Savart's law & Amperes law describes. – UnkownConstants May 9 '19 at 12:03
• @SebastianRiese If you check the answer below, and my comments... I think we agree. They are two different phenomena that lead to the same result to maintain the conservation of total charge. – UnkownConstants May 9 '19 at 12:15

The effect you are describing was discovered theoretically by considering a though experiment (at least that is how many texts present it). Ampere's law relates the line integral of the B field around a loop to the number of currents that pass through any surface bound by the loop. The mathematical theorem is Stokes' theorem.

When there is current in a wire there is a B field circulating around it. The integral of this around a circle enclosing the wire is proportional to "I" poking through the surface bound by the curve. The thought experiment goes something like this. Consider a charging or discharging capacitor through a wire loop with some resistance. For simplicity consider discharging. The circuit seems to be a closed loop and while current is flowing there would be a B field around the circuit. Now, calculate the circulation of B around a circular loop that surrounds a segment of the wire and consider the flat plane surface bound by it. Clearly the current in the wire pokes through this and Ampere's law holds. But mathematically speaking to should hold for ALL surfaces bound by the curve. Now consider a surface that is bound by the curve but morphed to pass through the space between the capacitor plates. Clearly there is NO current poking through this surface but there is a circulation of B along the curve! This lead to the realization that the changing E field acts as current and creates a B field. Notice the choice of words "acts like". The changing E field makes up for the discontinuity in I and the two work together to generate B.

This was not an easy thing to arrive at. I agree with the comment that it is perhaps a poor choice of name but it has stuck over time.

To answer your question directly, yes it is possible that with air or some other medium in the space a small current is set up but this is really more of a polarization effect that changes the capacitance. It does not account for the phenomenon. Also, in principle the effect has to be there in a vacuum and your proposed explanation would break down in that case.

(sorry for not posting a figure. The description of Stokes' theorem and Ampere's law require 3-dim, at least).

• Regarding the diagram it really isn't required, because while reading your post I remembered the countless lectures I've watched and books I've ready regarding the topic. But the conclusion that cleared my confusion is your great description: "This lead to the realization that the changing E field acts as current and creates a B field. Notice the choice of words "acts like". The changing E field makes up for the discontinuity in I and the two work together to generate B." – UnkownConstants May 9 '19 at 12:09
• Sure, the motion of charges(i.e conduction current) does yield a magnetic field that curls around it, likewise(from your answer), a time-varying electric field does the same exact thing but! There isn't any motion of charges(regardless of a medium present or not, they don't "move" they just adjust hence why they are dipoles). Ultimately, they are two different phenomenon that lead to the same result :) – UnkownConstants May 9 '19 at 12:11
• Well throughout response, thank you. – UnkownConstants May 9 '19 at 12:11
• You should include a diagram – BioPhysicist May 9 '19 at 12:22
• > "This lead to the realization that the changing E field acts as current and creates a B field." It acts as current in the sense it contributes in the Ampere law as additional term besides electric current, as if static current density $\partial_t \mathbf E$ was there even if it isn't. But contribution of this changing electric field to magnetic field anywhere is completely negligible in most quasistationary processes, including charging / discharging of a capacitor. Magnetic field there is given by the Biot-Savart formula, which attributes magnetic field to real current only. – Ján Lalinský May 9 '19 at 21:30

In the absence of current the Maxwell equation reads $$\partial \vec E = \vec \nabla \times \vec B /c^2$$. This equation means that both sides denote the very same thing. It does not mean that E induces B or vice versa.

• Could you clarify this just a bit further? – UnkownConstants May 9 '19 at 12:13
• This answer does not really address the original question. – ggcg May 9 '19 at 12:22