Edit: Can someone check my answer and possibly complete my task at the end?
The helmholtz theorem states that any vector field can be decomposed into a purely divergent part, and a purely solenoidal part.
What is this decomposition for $\vec{E}$, in order to find the field produced by its divergence, and the induced $\vec{E}$ field caused by changing magnetic fields.
The Potential Formulation:
$$\vec{E} = -\nabla V - \frac{\partial \vec{A}}{\partial t}$$
Is often transformed as $$\vec{E} = - \frac{\partial \vec{A}}{\partial t}$$
For showing induced fields, where charge density is not important ( and subsequently the scalar potential is zero).
There are a number of issues with this however using current density without including charge density violates $\vec{J} = \rho \vec{V}$
with that being said, in general, although it is a good approximation for the induced part of the field.
When not modeling $\rho$ as zero, From the lorenz gauge condition
$\nabla \cdot \vec{A} = - \mu_0 \epsilon_0 \frac{\partial V }{\partial t}$
We know the divergence of
$- \frac{\partial \vec{A}}{\partial t}$
Is non zero.
And thus that component of the $\vec{E}$ field cannot "just" be caused by the induced part, it is caused by the $\vec{E}$ fields divergence.
So what is an expression for the purely solenoidal part of the E field?
Edit:
artificially removing $V$ and then choosing the coulomb gauge to show its zero divergence is also incorrect as artifically removing $V$ with approximations should remove $A$ as well. Instead do the same with the full equation. Also, the components each terms represents is gauge dependent as well.
Perhaps the helmholtz decomposition is :
$$\vec{E} = -\nabla V - \frac{\partial \vec{A}}{\partial t}$$
Only under the coulomb gauge. As in the coulomb gauge, the first term has zero curl, but has divergence. But the second term has curl, but zero divergence?
What each part represents is gauge dependant, the full E field however being gauge independant.