# Intuitive, real-world explanation of displacement current?

1. May 7, 2013

### Nikitin

I know and fully understand the mathematical definition of it. But what's the physical explanation? Is it something like "preservation of current" or something?

2. May 7, 2013

### phyzguy

There's a great explanation of the displacement current and why it has to be there in "The Feynman Lectures on Physics" Vol 2, Chapter 18.

I attached a pdf below.

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3. May 7, 2013

### physwizard

There are pages written about this in intermediate physics textbooks. You can refer to Griffiths for instance. It can be viewed as a 'source' for the magnetic field inside and around a capacitor for eg. The idea is that magnetic fields can be produced not just by electric currents but by changing electric fields as well.

4. May 7, 2013

### Andrew Mason

Displacement current is a virtual current - a device that is added to Ampere's law to make Ampere's law work. Maxwell thought a vacuum behaved like a dielectric and imagined that there was something in the vacuum between the plates of a capacitor that behaved like little moving charges. But there isn't. So there is no physical explanation as a current.

AM

5. May 7, 2013

### Staff: Mentor

In particular note Figure 18-2. The displacement current between the plates of the capacitor makes the magnetic field around the gap "consistent" with the magnetic field around the wire on either side of the gap.

Recall that when calculating the current through an Amperian loop for Ampere's Law, you have to integrate the current density across a surface that is bounded by the loop. If you let the current density come from both "physical current" and "displacement current", it doesn't make any difference whether the surface of integration cuts through the wire or passes between the capacitor plates (see examples in the diagram).

6. May 7, 2013

### e.chaniotakis

Hello,
the problem that one has to see in order to intuitively comprehend displacement currents is the charging of a capacitor.
When one starts charging it current flows towards the high voltage plate, but it cannot proceed towards the low voltage one. In order the 'charging' to make sense, the circuit must be closed =>Therefore there must be current in it.
Therefore, for as long as the charging takes place a displacement current occurs which is equal to the current of the circuit : Id=I

7. May 7, 2013

### Jano L.

That is partially true, if the displacement current occurs inside a simple dielectric. Then part of the displacement current is due to changing polarization state of the dielectric, so electric current is involved, and this can be called 'source'.

In vacuum, however, the displacement current consists of changing electric field only.

This is bit problematic. Actually the changing electric and magnetic fields always occur together. Since they are two faces of one force acting on the charges, it is not good to say that one produces the other. Rather the electric charges and currents produce both fields - this corresponds better both to physical and mathematical usage of the term 'source'.

It is true that the Maxwell equation
$$\nabla \times \mathbf B = \mathbf j + \frac{\partial \mathbf E}{\partial t}$$
looks as if the displacement current played a role of additional electric current, so one may be inclined to include it into the Biot-Savart formula. However, that would not be correct, for the following reasons.

If the electric field is given accurately by gradient of potential, then the magnetic field is given correctly by the Biot-Savart formula with $\mathbf j$ only, even if the electric field changes in time (e.g. during the condenser discharge):

http://ajp.aapt.org/resource/1/ajpias/v31/i3/p201_s1?isAuthorized=no;

on the other hand, if the electric field has significant non-gradient contributions, then the Biot-Savart formula cannot be used, because we do not know the variation of the electric field any better than that of the magnetic field. In such cases, we have to deal with the displacement current in a different way. One can use Jefimenko's formulae, but in these the magnetic field is again a function of the currents only.