Does a static electric field travel at the speed of light?

Question

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.

Answer 1

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.

Answer 2

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 $c$.