A changing electric field delema

1. May 9, 2010

murad60

Hi, everyone

I'm a little bit confused. Is it implied by classical electromagnetism that electric field should not change instantaneously due a changing source at a distance, for example consider a parallel plate capacitor with a current entering its first plate this will lead to developing electric field at the other plate(and in between) but does this happen instantaneously?

thanks in advance

2. May 9, 2010

Livect

An electric field cannot travel faster than light.

3. May 10, 2010

murad60

thanks for reply but I think that for Gauss law to apply this must happent instantaneously. that is once the electric charge develop inside a closed surface electric field must change simultaneously.Am I right ???

4. May 10, 2010

Studiot

It is shown in elementary physics that this process cannot happen instantaneously. The rate of charge entry depends upon the supply circuit impedances.
The voltage between the plates and hence the electric field builds up at the same rate as the charge ie it starts from zero and build (usually exponentially with time).

5. May 10, 2010

murad60

Thanks studiot but I'm afraid you didn't get my point cause of my bad explanation. it is not in the rate of changing. my problem is that happens instananeously at the other plate of the capacitor regardless the distance between them. I meant should there be a travelling wave between the plates???

6. May 10, 2010

Studiot

It doesnt happen instantaneously at the other plate.

Suppose the plates were a long distance apart.

The connection to any apparatus measuring the voltage difference between the plates or the field at the second plate would have exactly the same distance to travel so could not report the voltage or state of the field before it arrived.

7. May 10, 2010

murad60

Thanks for quick reply, but if so then what about these two points

1. what about the direction of the propagation it seem that e-wave propagate along the direction. i.e. is not transverse???

• does gauss low hold instanteously the gaussian surface should feel the change instanteously, or must wait a steady state to hold

8. May 10, 2010

Studiot

I don't follow what you are asking in (1)

In (2) Livect has already mentioned that a disturbance in an electric field propagates at the speed of light. So if you create a disturbance somewhere it will reach your entire Gaussian surface at this speed, from the centre of the disturbance. If the distances are great the surface will be large and the local field correspondingly small.

When you get to large distances and long times in comparison to the speed of light the question of 'instantaneous' (known as the question of simultaneity) becomes relativistic.

9. May 11, 2010

murad60

In (1) I mean that the direction of propagation of the e-wave is form plate A to plate B. It seem tha E-wave travelling this way. along the direction of propagation.

In(2) from your reply should I conclude that gauss law does not hold instantaneously. I mean there is some developing charge inside the the gaussian surface while in the same instant gaussian surface retaining the old value and must wait for propagation to arrive???

thanks for reply and I would be grateful for further explanation.

10. May 11, 2010

Studiot

It is the wave-front that propagates in the direction of travel of the wave and yes this is from plate A to Plate B.

For a charging capacitor, yes the E vector is from plate A to plate B, the H vector is at right angle, tangential to the dielectric and the Poynting Vector directed inwards into the body of the dielectric. The current flow is all displacement current. All as in the sketch.

I am not quite sure what you mean by a 'Gaussian surface'? All surfaces that the flux passes through are 'Gaussian surfaces'. The increasing E field pases through a series of such surfaces at increasing distance from the source.
Gauss' law does not say that the flux must reach all these surfaces at once, in fact it says nothing about when. It is a conservation law relating the flux passing through to the strength of the generating charge. As the charge increases from zero so too will the E field flux.

Attached Files:

• capcharge.jpg
File size:
8.7 KB
Views:
76
11. May 11, 2010

matonski

I think I understand his questions.

1) Electromagnetic waves are transverse waves, meaning that the E-field points perpendicular to the direction of motion. However, in this case, it seems as if the E-field is pointing in the direction of motion since it is pointing from plate A to plate B and the wave is traveling from plate A to plate B. This seems like a contradiction.

2) Gauss' law states that the total flux of E through a closed surface is proportional to the charge inside. The question seems to be whether the flux and total charge are both calculated at the exact same moment of time, or whether there needs to be delays between when the charge is calculated and when the flux is calculated.

12. May 15, 2010

murad60

Thank you Studiot, I really appreciate it. But how comes that the wavefront travels from plate A to plate B and the E-wave travels that way?

matonski gave explanation to my questions thanks matonski, I am still confused

13. May 15, 2010

Studiot

I am a little unsure what level to answer this at.

Do you understand how current flows through a capacitor? In particular do you understand displacement current?

14. May 15, 2010

Staff: Mentor

Gauss' law does hold instantaneously, but recall that in its integral formulation it refers to the flux over the entire Gaussian surface and that the flux is in general not constant over the surface. So the fact that charge enters a Gaussian surface implies that the total flux over the whole surface will increase, but it in no way implies that the E-field at the far side of the surface will change instantaneously.

15. May 15, 2010

lugita15

Are you saying that the electric flux through a closed surface can change instantaneously? Can electric flux through a non-closed surface also change instantaneously? If the answer to the latter question were yes, that would mean that the displacement current could be infinite, which would mean that magnetic field could be infinite. If the answer to the latter question were no, then would punching an infinitesimal hole in a closed surface really change the response speed (I don't know the actual term for this) of flux from infinity to the speed of light?

16. May 15, 2010

Studiot

I think the original question was not about a single gaussian surface but a series of them, as the disturbance in the electric field proceeded from one plate to the other.

I also think the OP was confusing a disturbance in the electric field or even the introduction of one and a wave propagating through it.

Yes you can propagate a step function at/through a single gaussian surface. However it moves from one to the next at the local speed of light.

17. May 15, 2010

Staff: Mentor

Yes. Imagine a Gaussian surface consisting of a sphere and a single proton moving moving inertially from outside to inside. As the proton approaches the field lines crossing the sphere gradually become stronger and gradually point in a slightly different direction. As the proton crosses the sphere some of the field lines change from pointing in to pointing out. This causes an instantaneous jump in the flux despite the fact that the fields at the spherical surface are changing only gradually except right at the location where the charge enters.

Last edited: May 16, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook