I Gauss' law seems to imply instantaneous electric field

[Bob44]

Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire.

We wish to find an expression for the horizontal component of the electric field at a distance $\mathbf{r}$ from the sphere as it discharges.

By using the Lorenz gauge condition:
$$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$
we find the following retarded solutions to the Maxwell equations

If we assume that $\rho(\mathbf{r},t)=q(t)\delta^3(\mathbf{r})$ and $\partial_t\,\mathbf{j}(\mathbf{r},t)$, $\partial_t\mathbf{A}(\mathbf{r},t)$ are purely vertical then the horizontal component of the electric field is given by the retarded expression
$$\mathbf{E}(\mathbf{r},t)=-\nabla\phi=\frac{q(t-r/c)}{4\pi\epsilon_0}\frac{\hat{\mathbf{r}}}{r^2}.\tag{2}$$
Now let us compare eqn.$(2)$ with the result that we get from Gauss' law
$$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}.\tag{3}$$
We integrate both sides of eqn.$(3)$ and use the divergence theorem over a sphere centered at the origin to obtain
$$\mathbf{E}(\mathbf{r},t)=\frac{q(t)}{4\pi\epsilon_0}\frac{\hat{\mathbf{r}}}{r^2}.\tag{4}$$
Eqn.$(2)$ is a causal retarded expression whereas eqn.$(4)$ is instantaneous.

Why do we have this discrepancy?

It seems that we have artificially obtained a causal solution $(2)$ by choosing the Lorenz gauge condition. The non-local expression $(4)$ obtained from Gauss' law did not assume any gauge condition.

 

[PeroK]

Have you assumed a uniform field across the surface of the sphere?

 

[Bob44]

I’ve just used a spherical Gaussian surface with the same center as the charged sphere but with a larger radius so that it encloses the charged sphere. By Gauss’ law the total electric flux through the Gaussian surface is equal to the total charge inside divided by $\epsilon_0$.

By symmetry I assume that the electric field is uniform over the charged sphere and normal to both its surface and the enclosing Gaussian spherical surface.

 

[PeroK]

I’ve just used a spherical Gaussian surface with the same center as the charged sphere but with a larger radius so that it encloses the charged sphere. By Gauss’ law the total electric flux through the Gaussian surface is equal to the total charge inside divided by $\epsilon_0$.

By symmetry I assume that the electric field is uniform over the charged sphere and normal to both its surface and the enclosing Gaussian spherical surface.

The discharge breaks the symmetry.