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feynman1

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feynman1

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cnh1995

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The current that flows inside a capacitor (through the dielectric) is called 'displacement current'. It is numerically equal to the drift current in the external wires attached to the capacitor plates.and 0 in capacitors

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hutchphd

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My physicist brain is delighted by the inner workings of the capacitor and the fomulation of the displacement current and all the richness of Maxwell's Equations. To my EE brain it is largely a distraction, and one needs to wear the correct brain for whatever task is at hand.

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It's indeed true that at a capacitor you need electrostatic approximation, because between the plates you need to take into account the socalled displacement current but can negelect ##\dot{\vec{B}}##, from which ##\vec{\nabla} \times \vec{E} \simeq 0##, which explains, why between the capacitor plates you can use this electrostatic approximation.

For both resistors and conductors you can always neglect the displacement current but not ##\dot{\vec{B}}##, which leads you to the magnetostatic approximation. For details see

https://itp.uni-frankfurt.de/~hees/pf-faq/quasi-stationary-edyn.pdf

The upshot of the analysis is that retardation effects are negligible and thus the condition is that the frequency of the AC fulfills ##\ell \omega \ll c##, where ##\ell## are the typical geometrical lengths of the circuit and ##\omega=2 \pi f## with ##f## the frequency of the applied AC.

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hutchphd

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In my vernacular when that becomes an issue I am talking about a waveguide, and not a simple circuit. On the very far side of that hill one is talking about a fiber optic device.

And neglecting B for even seemingly innocuous and straightforward circuits is responsible for untold mayhem.

Also sometimes E is the culprit. I once spent several unpleasant weeks tracking down spurious signals on a high precision medical spectrophotometer of my design. The device used photodiodes to look at various reflections from a rapidly rotating platter of samples. To my chagrin, the photodiodes could "see" the platter even in total darkness! It turns out that any charge on protuberances on the rotating platter created AC photocurrent in the diodes. Not a huge effect but nontrivially annoying.

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DaveE

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No. Just no. Every assumption you made here is wrong.

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feynman1

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What's the physical meaning (theoretical or experimental) of displacement current, apart from making Maxwell's equations consistent?The current that flows inside a capacitor (through the dielectric) is called 'displacement current'. It is numerically equal to the drift current in the external wires attached to the capacitor plates.

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feynman1

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What's the physical meaning (theoretical or experimental) of displacement current, apart from making Maxwell's equations consistent?

It's indeed true that at a capacitor you need electrostatic approximation, because between the plates you need to take into account the socalled displacement current but can negelect ##\dot{\vec{B}}##, from which ##\vec{\nabla} \times \vec{E} \simeq 0##, which explains, why between the capacitor plates you can use this electrostatic approximation.

For both resistors and conductors you can always neglect the displacement current but not ##\dot{\vec{B}}##, which leads you to the magnetostatic approximation. For details see

https://itp.uni-frankfurt.de/~hees/pf-faq/quasi-stationary-edyn.pdf

The upshot of the analysis is that retardation effects are negligible and thus the condition is that the frequency of the AC fulfills ##\ell \omega \ll c##, where ##\ell## are the typical geometrical lengths of the circuit and ##\omega=2 \pi f## with ##f## the frequency of the applied AC.

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First of all displacement current doesn't make only Maxwell's equation consistent but makes circuit theory consistent with its assumption (the assumption you mentioned that the current along the same branch is everywhere the same).What's the physical meaning (theoretical or experimental) of displacement current, apart from making Maxwell's equations consistent?

The physical meaning of the displacement current is that it is time varying electric field flux inside the capacitors plates and this time varying electric flux is like a current that generates a magnetic field as well. In the quasi-static approximation the displacement current equals the conduction current (that comes or goes to/from capacitors plates)(the math behind this equality are not hard at all). In the full dynamic scenario with very high frequency this is not the case and you ll have to read on that link to Feynman's Lectures provided by @rude man at that other thread you had made(about charges of capacitor being opposite and equal).

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feynman1

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Displacement current was introduced to explain magnetic field induced by current. But how to understandFirst of all displacement current doesn't make only Maxwell's equation consistent but makes circuit theory consistent with its assumption (the assumption you mentioned that the current along the same branch is everywhere the same).

The physical meaning of the displacement current is that it is time varying electric field flux inside the capacitors plates and this time varying electric flux is like a current that generates a magnetic field as well. In the quasi-static approximation the displacement current equals the conduction current (that comes or goes to/from capacitors plates)(the math behind this equality are not hard at all). In the full dynamic scenario with very high frequency this is not the case and you ll have to read on that link to Feynman's Lectures provided by @rude man at that other thread you had made(about charges of capacitor being opposite and equal).

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First of all its not a missing term in Faraday's law but in Ampere's law.Displacement current was introduced to explain magnetic field induced by current. But how to understandtime varying electric flux is like a current? I can't think of this invention other than compensating a missing term in Faraday's law of induction.

Second displacement current is something real as I said it corresponds to the time variations of electric flux inside the capacitor. It is not just some mathematical term without physical meaning. It has physical meaning (unless you going to argue that electric field and/or electric flux do not have physical meaning).

- #12

feynman1

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Displacement current is only needed in capacitors and nothing else to remedy continuous current. Can you explain why this invention isn't unnatural?First of all its not a missing term in Faraday's law but in Ampere's law.

Second displacement current is something real as I said it corresponds to the time variations of electric flux inside the capacitor. It is not just some mathematical term without physical meaning. It has physical meaning (unless you going to argue that electric field and/or electric flux do not have physical meaning).

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Is electric field unnatural/unphysical? I think no. Same holds for displacement current, it is essentially (perhaps a bit of oversimplification here) the time derivative of the electric field.

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feynman1

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I don't mind about different philosophies. Then is 'magnetic field induced by changing electric field' merely an experimental find?

Is electric field unnatural/unphysical? I think no. Same holds for displacement current, it is essentially (perhaps a bit of oversimplification here) the time derivative of the electric field.

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feynman1

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Is there any theoretical support for displacement current?

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What do you mean by that, ofcourse there is theoretical support it appears in Maxwell's equations in Maxwell-Ampere law. It has been confirmed experimentally that the Maxwell-Ampere law holds.Is there any theoretical support for displacement current?

Maxwell "discovered" the displacement current term in a purely theoretical way (thinking along the lines of what happens with the magnetic field between the capacitors plates, does it vanish there because the conduction current vanishes there?)

There was no simple way of confirming experimentally that a time varying electric field creates a magnetic field back at 1860. Hertz experiments done around 1880 were the first that confirmed it experimentally.

- #18

feynman1

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I'm looking for theoretical support to make the introduction of displacement current into Maxwell's equations natural. What you say about things inside capacitors are just results about the introduction of displacement current.What do you mean by that, ofcourse there is theoretical support it appears in Maxwell's equations in Maxwell-Ampere law. It has been confirmed experimentally that the Maxwell-Ampere law holds.

Maxwell "discovered" the displacement current term in a purely theoretical way (thinking along the lines of what happens with the magnetic field between the capacitors plates, does it vanish there because the conduction current vanishes there?)

There was no simple way of confirming experimentally that a time varying electric field creates a magnetic field back at 1860. Hertz experiments done around 1880 were the first that confirmed it experimentally.

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DaveE

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First, I think it's great that you are asking about displacement current. Just asking puts you far ahead of many in understanding electronics. It is a strange concept at first, since we all started by thinking of electricity as tiny charged billiard balls flowing on wires, like water in pipes, perhaps. But that really isn't a good model for electricity; it works at first, but then is confusing and misleading later.Is there any theoretical support for displacement current?

I'm not going to explain displacement current, you can get that from teachers better than me. But what you want to focus on is a parallel plate capacitor in a vacuum; nothing between the plates, no wires, no molecules, no dielectrics, nothing. However current does flow through this capacitor, i(t)=C⋅(dv/dt), as you know. That is your experimental proof that it exists, you can build one yourself if you want. The theoretical support is that Maxwell's equations explains this as well as many other complex and subtle E&M phenomenon. There really hasn't been another theory of classical E&M that works as well, it's hard to imagine that there could be given the success of what you would have to replace.

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feynman1

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'current does flow through this capacitor, i(t)=C⋅(dv/dt)' How to measure the current through the capacitor? Normally current is rate of change of charge.First, I think it's great that you are asking about displacement current. Just asking puts you far ahead of many in understanding electronics. It is a strange concept at first, since we all started by thinking of electricity as tiny charged billiard balls flowing on wires, like water in pipes, perhaps. But that really isn't a good model for electricity; it works at first, but then is confusing and misleading later.

I'm not going to explain displacement current, you can get that from teachers better than me. But what you want to focus on is a parallel plate capacitor in a vacuum; nothing between the plates, no wires, no molecules, no dielectrics, nothing. However current does flow through this capacitor, i(t)=C⋅(dv/dt), as you know. That is your experimental proof that it exists, you can build one yourself if you want. The theoretical support is that Maxwell's equations explains this as well as many other complex and subtle E&M phenomenon. There really hasn't been another theory of classical E&M that works as well, it's hard to imagine that there could be given the success of what you would have to replace.

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DaveE

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These vacuum capacitors are, unfortunately, of low capacitance, so it's not easy to measure. The capacitor will charge quickly and then the current stops. But one basic set up is a battery, a high value resistor, and the capacitor all in series. Then you can monitor the voltage drop across the resistor, which is proportional to the capacitor current. You will probably need to borrow an oscilloscope or some other fast measuring device to see it. You also need to start your measurement with the capacitor voltage not equal to the battery voltage; for example, reverse the polarity of the battery somehow, or short out the capacitor first.'current does flow through this capacitor, i(t)=C⋅(dv/dt)' How to measure the current through the capacitor? Normally current is rate of change of charge.

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feynman1

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But this seems to measure voltage only not current in capacitors?These vacuum capacitors are, unfortunately, of low capacitance, so it's not easy to measure. The capacitor will charge quickly and then the current stops. But one basic set up is a battery, a high value resistor, and the capacitor all in series. Then you can monitor the voltage drop across the resistor, which is proportional to the capacitor current. You will probably need to borrow an oscilloscope or some other fast measuring device to see it. You also need to start your measurement with the capacitor voltage not equal to the battery voltage; for example, reverse the polarity of the battery somehow, or short out the capacitor first.

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feynman1

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That was the ingenious way Maxwell discovered the necessity for this term in his equations. It's, however, not a good interpretation in view of modern developments, particularly relativity. If you want to reveal the meaning of Maxwell's equations in a modern physical way, you should not consider this term as part of the current, but you should put it to the left side of the inhomogeneous Maxwell equation, so that the "true source", which is the current-density, is on the right-hand side, i.e., the Ampere-Maxwell Law should read (in Heaviside-Lorentz units, which are the most lucid choice to discuss fundamental relativistic properties of E&M)Displacement current was introduced to explain magnetic field induced by current. But how to understandtime varying electric flux is like a current? I can't think of this invention other than compensating a missing term in Ampere's law.

$$\vec{\nabla} \times \vec{B} -\frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}.$$

If you interpret Maxwell's equations in this way, you are automatically lead to the physical solutions in the sense of "cause and effect", i.e., Jefimenko's retarded solutions (or equivalently the retarded potentials in Lorenz gauge): The causal sources of the em. field are the charge density and the current density and not some parts of the electromagnetic field.

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I'm not sure what you are referring to, but I guess it's about the Röntgen-Eichenwald experiment, conducted by Röntgen in 1888 and refined later by Eichenwald in 1903. This answered the question, whether the surface-charge density at the edge of a dielectricum in an electric field can induce a magnetic field when set into motion. Röntgen's and Eichenwald's experiment showed that this is indeed the case. One must know that this was not so clear in these days, because the microscopic theory about the nature of matter as consisting of charged particles was in its infancy, and the question about electrodynamics in moving bodies was an enigma, which was finally only resolved with the discovery of special relativity some years later. Today it's clear that any kind of current density, i.e., moving charges, lead to a magnetic field, but the Röntgen current is now clearly seen as the convection current of a surface-charge distribution, it's clear that it is a true surface-current density and thus not something like the "displacement current".

You find a very good discussion on early experiments, among them the Röntgen-Eichenwald experiments, about electrodynamics in moving media in

A. Sommerfeld, Lectures on Theoretical Physics, vol. 3

Despite the sin to use the ##\mathrm{i} c t## convention for the pseudometric, it contains a very good treatment of relativistic electrodynamics.

One should note that Sommerfeld was among the rare physicists who got the radiation field of a particle in hyperbolic motion right, which even Pauli had wrong in his famous review article about relativity in 1921!

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Well, if you connect a battery to an uncharged capacitor (taking into account the finite resistance of the wires, because otherwise it's a singualr problem) you have to solve the EoM:These vacuum capacitors are, unfortunately, of low capacitance, so it's not easy to measure. The capacitor will charge quickly and then the current stops. But one basic set up is a battery, a high value resistor, and the capacitor all in series. Then you can monitor the voltage drop across the resistor, which is proportional to the capacitor current. You will probably need to borrow an oscilloscope or some other fast measuring device to see it. You also need to start your measurement with the capacitor voltage not equal to the battery voltage; for example, reverse the polarity of the battery somehow, or short out the capacitor first.

$$Q/C+R \dot{Q}=U_0$$

with the solution

$$Q(t)=C U_0 [1-\exp(-t/(RC))].$$

The current is

$$i=\dot{Q}=\frac{U_0}{R} \exp(-t/(RC)).$$

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