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Intuitive, realworld explanation of displacement current? 
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#1
May713, 06:49 AM

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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
May713, 08:12 AM

P: 2,179

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. 


#3
May713, 08:18 AM

P: 153

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
May713, 08:19 AM

Sci Advisor
HW Helper
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Intuitive, realworld explanation of displacement current?
AM 


#5
May713, 08:49 AM

Mentor
P: 11,778

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
May713, 08:54 AM

P: 72

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
May713, 03:22 PM

PF Gold
P: 1,165

In vacuum, however, the displacement current consists of changing electric field only. 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 BiotSavart 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 BiotSavart 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/ajpia...sAuthorized=no; on the other hand, if the electric field has significant nongradient contributions, then the BiotSavart 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. 


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