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Let's say we have a cube of side length $a$ as a Gaussian surface in an electric field (in vacuum) and then we raise the strength of the electric field. As I understand, this rise in the field strength would travel through the field at the speed of light. This means that for a duration of $\frac{a}{c}$, the field strength, and thus the flux, would differ at the 2 opposing faces of the cube. This implies that the flux through the surface is non-zero and thus the charge inside it must also be non-zero by Gauss' Law. How is this possible?

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As the field begins to rise at one place, it must also be doing related things at other places. To get an intuition about this, try sketching field lines on a piece of paper. The equation $\nabla \cdot {\bf E} = 0$ (for a charge-free region) implies that the lines, when drawn in 3-dimensional space, have to be continuous. The spacing between the lines indicates the field strength. If you have a field that is weaker at one place than at another, then when moving from the weaker to the stronger field region the field lines have to curve in a bit, to end up closer together where the field is stronger. When you count the lines going into and out of a given volume (this is what indicates the total flux through the surface) you find, at each instant of time, that as many lines go in as come out.

The overall conclusion is that the field at one place cannot grow larger without this kind of modification of the field at nearby places. Using the cube you described as a Gaussian surface, if the field is initially uniform then initially there is no flux in or out through the sides of the cube that are parallel to the field. But if the field subsequently becomes larger at one end of the cube than the other, then there must now be a flux across those sides.

For further clarity, for electric fields $\nabla \cdot {\bf E} = 0$ always holds in charge-free regions, and it follows that $$ \oint {\bf E} \cdot d{\bf S} = 0 $$ for charge-free regions, and this equation is correct and exact at all times, including for time-varying fields. The fact that changes at one place don't immediately propagate to places a finite distance away is all accounted for correctly. As those changes propagate, $\nabla \cdot {\bf E}$ remains equal to zero at each and every local region at each and every moment, and therefore its integral over a charge-free volume of any shape or size also remains zero.

In the following three diagrams the rectangle is a Gaussian cylinder and the lines are electric field lines. The field has higher strength on the right than on the left. The diagrams show a change in the field propagating from left to right. The flux through any given edge of the rectangle is equal to the number of field lines crossing that edge.

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    $\begingroup$ Professor Steane, I too, have serious doubts about your answer, as @PhilipWood did. In time varying fields, we can use Jefimenko's equations to calculate electric and magnetic fields at every point at any time. The fields need not follow Gauss's law in its integration form. Your intuitive thinking (I think based on Purcell's approach with the field lines) may not be correct. Purcell used it without proof in his book. $\endgroup$
    – verdelite
    Mar 6, 2020 at 18:04
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    $\begingroup$ There is a problem with this argument. Suppose some charge is initially at rest at the origin, and starts accelerating at time $t = 0$. The disturbance in the electric field propagates as a spherical wave centered at $O$ with radius $ct$. Assume the box is centered on the $z$ axis with its bottom face parallel to the $xy$ plane, at distance $d$ from $O$. Then for $d < ct < \sqrt{d^2 + a^2/4}$, the disturbance only reaches the bottom face, so the flux on the other faces cannot change yet. One needs to consider the shape of the electric field from the Liénard-Wiechert formulas. $\endgroup$
    – Tob Ernack
    Mar 6, 2020 at 20:14
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    $\begingroup$ Moreover, if the charge starts accelerating along the $z$-axis, then the field must be rotationally symmetric around the $z$-axis, so there won't even be any cancellation on the surface itself. $\endgroup$
    – Tob Ernack
    Mar 6, 2020 at 20:26
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    $\begingroup$ @BlueRaja-DannyPflughoeft The diagrams show a wave front propagating from left to right. It could, for example, be part of the spherical wave indicated in the answer to which you provide a link, when that spherical wave has propagated out to a large size. The only difference is the direction of travel. I hope you will either argue back or else correct your comment. $\endgroup$ Jul 25, 2020 at 16:47
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    $\begingroup$ @BlueRaja-DannyPflughoeft (i) My diagram should not be completed in 3D in the way you suggest. (ii) Drawing continuous field lines in 3D is sufficient to ensure the field thus indicated (with the usual convention that line density represents strength of field) has zero divergence. $\endgroup$ Jul 27, 2020 at 10:18
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The Maxwell equation that is said to embody Gauss's law for electric fields, namely $$\nabla.\mathbf E=\frac{\rho}{\epsilon_0}$$ holds at every instant. That's alright even if the field changes with time, because it is a local equation, applying at a point. However the same need not be the case for the integrated version (arguably Gauss's law itself): $$\int_S \mathbf E.d\mathbf S=\Sigma Q$$ in which $\Sigma Q$ is the total charge in the volume bounded by surface S. This volume is finite, and, as you say, changes in field strength at one boundary couldn't propagate instantaneously to other parts of the boundary. This integrated version is strictly for electrostatics. The Maxwell version is more general, applying at any point at any time, even if charges are moving and fields are changing.

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    $\begingroup$ I have serious doubts about this answer. Surely the Divergence theorem (Gauss's theorem) applies at any instant, so if the Maxwell equation applies at any instant, whether or not fields are changing, so must the integrated version. $\endgroup$ Mar 6, 2020 at 17:35
  • $\begingroup$ Did you intend to comment on the answer by @AndrewSteane ? Your answer I think is correct and you surely did not have serious doubts about your own answer. $\endgroup$
    – verdelite
    Mar 6, 2020 at 17:59
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    $\begingroup$ @ verdelite No, I have serious doubts about my own answer! $\endgroup$ Mar 6, 2020 at 18:41
  • $\begingroup$ Thanks for your comment; I added a paragraph to my answer. I think that your answer is wrong when you say "this integrated version is strictly for electrostatics" and therefore the thrust of your answer is wrong. I would be interested to know if, on reflection, you agree. $\endgroup$ Mar 6, 2020 at 19:13
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    $\begingroup$ This answer is, as you suspect, quite wrong! The divergence theorem is a mathematical theorem, its validity has nothing to do with the physics of the vector fields it’s applied to. $\endgroup$
    – knzhou
    Mar 6, 2020 at 19:42
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Any disturbance in an electric field presumably propagates at the speed of light. Therefor a Gaussian surface some distance out, will lag in reflecting any changes in the enclosed charge.

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