Why is current density proportional to electric field strength, and why does this relation seem contradictory to Maxwell (and others)?

I've got three physics equations in mind which seem (to me) to contradict eachother, using a simple case of charge(s) in a static electric field. If someone can give an explanation as to what I'm missing that would be much appreciated. The equations:

Maxwell-Amperes circuital law:

$$\nabla \times B = \mu_0 (J + \epsilon_0 \frac{\partial{E}}{\partial{t}})$$

Ohms law:

$$J = \sigma E$$

And finally, Lorentz Force Law (in absence of magnetic field) (along with F = ma):

$$F = qE$$

Starting with the equation for Coloumb force, this tells me charges should accelerate in the presence of an electric field. Assuming current is proportional to velocity of charges, this suggests that the time derivative of current would be proportional to electric field strength.

Now for ohms law, it clearly states that current (density) is proportional to Electric field strength. I'm guessing this refers to a steady state due to material properties?

Finally, for Maxwell-Amperes law, assuming no magnetic field it suggests current density is proportional the time derivative of electric field, so we have another apparent discrepancy - assuming my above reasoning applies.

The apparent discrepancy between Maxwell and Coloumb laws are what I'm most interested in. So, what am I missing in my logical reasoning? Any insights will be very much appreciated.

• Do not insert your equations as pictures. Especially when your equations are this simple. Please just use MathJaX instead. Commented Jun 26, 2023 at 15:21
• @naturallyInconsistent OK, can you explain why? Commented Jun 26, 2023 at 15:23
• The image links very easily cause link rot. And then nobody can read what your question was asking. MathJaX is much more able for search engines to find. It is a win on all the important fronts. Commented Jun 26, 2023 at 15:24
• @naturallyInconsistent OK I will edit it. Commented Jun 26, 2023 at 15:25
• There is no contradiction with Maxwell’s equations. Commented Apr 6 at 16:08

Coloumb

The name is Coulomb.

$$\vec F=q\vec E$$

This is NOT called the Coulomb force. You should call it the Lorentz force law in the absence of magnetic field.

And this is actually the wrong force law to use here. More later.

Now for Ohms law, it clearly states that current (density) is proportional to Electric field strength. I'm guessing this refers to a steady state due to material properties?

Yes.

Finally, for Ampère-Maxwell's law, assuming no magnetic field it suggests current density is proportional the time derivative of electric field, so we have another apparent discrepancy - assuming my above reasoning applies.

You correctly noted that Ohm's Law is really a steady state condition. This means that, in Newton's 2nd Law, it would correspond to the terminal velocity case: $$\tag1\text{N2L}:\qquad\vec F=q\vec E-\vec f_\text{collisions}=\vec0$$ And in this way, because you were using the wrong force law, your correct reasoning led you to a wrong rabbit hole.
Finally, what you really know, is that once the steady state conditions are reached, $$\partial_t\vec E=\vec0$$, so that in Ampère-Maxwell's law, you cannot have the current density be proportional to the time derivative of the electric field. Instead, what it says is that the steady state circuit loop must produce a magnetic field. Which is necessary for correct operation and understanding of the circuit.