Could intensity be related to the flux of the electric field? Imagine a cable in which an intensity $\mathbf I$, is passing through. This intensity is caused by the moving particles (for example electrons), going with a certain speed, and with some density of charge, as shown in this picture:

So, this got me thinking. The more particles pass through a section of the cable, per unit time, the more intensity we would have, because the speed of particles is higher. The opposite also applies.
$$
\int_{\mathcal A}{\mathbf J \cdot \mathrm d \mathbf A} = I \\
\mathbf J = \rho \mathbf u \\
$$
But if individual charges creates their own electric fields, can $\mathbf J$ be expressed in terms of the electric field. I mean, somehow, the electric flux of all the charges at the cable, is proportional to the current density. The more flux of charges (hence more electric field), the more intensity you have.
If this relation could be expressed mathematically, Maxwell's Equations (expecially the Ampere-Maxwell Law) would be expressed, only, in terms of $\mathbf E$, $\mathbf B$. I don't know if my line of reasoning is correct, but how would I obtain this relation between electric flux/electric field and intensity?
The relationship must obey the following:
$$
\phi_E \propto I \\
\mathbf E \cdot \mathbf J = 0 \\
\mathbf E \times \mathbf B \propto \mathbf J
$$
 A: More current does not imply more charge, they are not proportional.
Maxwell ampere law:
$$\nabla × \vec{B} = \mu_{0} \vec{J} + \mu_{0} \epsilon_{0} \frac{\partial \vec{E}}{\partial t}$$
$$ \vec{J} = \frac{1}{\mu_{0}}\nabla × \vec{B} 
- \epsilon_{0} \frac{\partial \vec{E}}{\partial t} $$
Current density can be expressed in terms of $\vec{E}$ and $\vec{B}$
A: Playing around with this a little...
We know the Ampere-Maxwell law:
$$\nabla\times\vec{B} = \mu_0\vec{J}+\mu_0\epsilon_0\frac{\partial\vec{E}}{\partial t}$$
We can solve this for $\vec{J}$:
$$\vec{J} = \frac{1}{\mu_0}\left(\nabla\times\vec{B}\right)-\epsilon_0\frac{\partial\vec{E}}{\partial t}$$
Which is certainly expressed in terms of $\vec{E}, \vec{B}$, but I feel like this can be taken further. $\vec{B}$ is incompressible (i.e., its divergence is zero) so we can write it in terms of a vector potential:
$$\vec{B} = \nabla\times\vec{A}$$
When we plug it into our equation for $\vec{J}$, we end up with a triple vector product, which we can expand into
$$\nabla\times\left(\nabla\times\vec{A}\right) = \nabla\left(\nabla\cdot\vec{A}\right) - \nabla^2\vec{A}$$
We can choose the divergence of $\vec{A}$; I'll choose it for convenience to be equal to zero - in which case, the above becomes equal to $-\nabla^2\vec{A}$.
Now, separately, we know
$$\vec{E} = -\nabla\phi - \frac{\partial\vec{A}}{\partial t}$$
Putting it all together:
$$\vec{J} = \epsilon_0\left(\frac{\partial}{\partial t}\nabla\phi + \frac{\partial^2\vec{A}}{\partial t^2}\right)- \frac{1}{\mu_0}\nabla^2\vec{A}$$
Which basically tells us that $\vec{J}$ is related to the second derivative of $\vec{A}$ with respect to each dimension (x, y, z, time) and how the scalar electric potential changes over time.
An alternative approach:
The continuity equation gives $$\nabla\cdot\vec{J}=-\frac{\partial \rho}{\partial t}$$ so using the divergence theorem we get
$$-\iiint \frac{\partial \rho}{\partial t}dV = \iint\vec{J}\cdot d\vec{a} = I$$
Now that we have things in terms of charge density $\rho$, we can obtain the electric field via its definition/Gauss' Law, and similarly for the electric potential.
None of this is very clean though; I think you get better simplification of Maxwell's equations when you complexify the electric field than by trying to eliminate $\vec{J}$.
