0
$\begingroup$

Suppose we have a large area where the net electric field is zero.
We place a charge q somewhere in this area.
How long will it it take for another positive charge q' standing at a distance d from q to feel the effect of the electric field?
Both charges are static and the net electric field is just the one generated by the charge q.
My guess is that the effect will be observed by q' after a time t=d/c.

$\endgroup$
  • 4
    $\begingroup$ Yes you are correct, the influence travels at the speed of light. $\endgroup$ – Andrew Steane Jun 29 at 11:24
  • 2
    $\begingroup$ I'm afraid this question doesn't make much sense as written at least because a static electric field does not change with time (by definition) but the problem you describe is quite time dependent. Indeed, it sounds like you're describing creating a charge q because, otherwise, q' would 'feel the effect' of q at all times since the field of q extends to infinity. Also, is q' a test charge? $\endgroup$ – Alfred Centauri Jun 29 at 11:48
  • 3
    $\begingroup$ So you are asking if we violate the continuity equation do things still propagate with speed $c$? $\endgroup$ – Aaron Stevens Jun 29 at 12:22
  • 1
    $\begingroup$ @AaronStevens I've not the knowledge to answer this question. I'm probably oversimplifying the topic. Is what I'm asking unrealistic? Even if being unrealistic, could it provide a simplified intuitive example just to grasp the idea given that we should be willing to revise it once strenghened our knowledge? $\endgroup$ – Gabriele Scarlatti Jun 29 at 12:42
  • 1
    $\begingroup$ Gabriele, it seems to me that what you're thinking of is so-called monopole radiation which does require that electric charge is created and/or destroyed within a volume. As far as we (I) know, electric charge is absolutely conserved which is why we say there is no electromagnetic monopole radiation. But, if there were, it might not propagate at speed $c$ since it wouldn't be transverse. I'll have to think about that sometime. $\endgroup$ – Alfred Centauri Jun 29 at 18:27
4
$\begingroup$

At the present time of physics models of nature , we expect that Lorentz transformations are exact and are the way the world is working. They have been tested in innumerable particle experiments, cosmological observations are consistent and there has been no experimental falsification.

Thus as far as our theories go, nothing can move faster than light because this has not been falsified (up to now and probably will not be).

Let us take your specific experiment:

Suppose we have a large area where the net electric field is zero.

OK

We place a charge q somewhere in this area.

Placing means motion, non- inertial frame, so it will generate electromagnetic waves.

How long will it it take for another positive charge q' standing at a distance d from q to feel the effect of the electric field?

So there exists a test charge q'. It will "see" the electromagnetic waves of q at the time you give , $t=d/c$.

Both charges are static and the net electric field is just the one generated by the charge q.

That cannot be true since q' is charged. You might say that q' is small, an electron, and q is large.

When we reach the level of electrons, we enter the realm of quantum electrodynamics (QED). QED has a very precise and accurate way of calculating the interaction between charged quantum particles, the Lorenz transformation is inherent, and no interaction between the electron test charge and the collective charges of electrons on q will happen faster than light, by construction of the theory, and the theory is continually validated.

$\endgroup$
1
$\begingroup$

Your assumption is correct.

You may be asking this because in the Coulomb gauge the Coulomb potential obeys the Poisson equation, which predicts instantaneous propagation of the potential. However the electric field is equal to $\vec \nabla V$ only if the charges are truly static. If not the full Maxwell equations are needed from which a wave equation for E can be derived. This wave equations sets the speed of propagation at $c$.

$\endgroup$
  • $\begingroup$ This would be more convincing if you could show that Maxwell's equations are still valid even if the continuity equation is no longer true. $\endgroup$ – Aaron Stevens Jun 30 at 10:38
  • $\begingroup$ I don't think the continuity equation is invalid in the example sketched by the OP. He places a charge q somewhere. There is no mention of the charge being created out of nothing, unless you want to interpret his wording "static" very strictly. This in my opinion would be besides the point. Btw Maxwell's equations are only valid under charge conservation. $\endgroup$ – my2cts Jun 30 at 10:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.