Consider a conductor, we created a potential difference at the end of it . Therefore , The electric field is produced inside the conductor.
Now my question is: how is this electric field? Is the electric field uniform? Is the electric field constant? Is the electric field Linear?
The general case of a finite length conductor probably needs a full solution of the boundary value problem of the Laplace equation, which is rather involved. See this previous answer. If we consider the special case of an infinitely long cylindrical conductor, it is possible to show that the electric field and thus current density should be uniform over the cross section.
Assuming rotational symmetry around the cylinder axis, the potential $\phi$ in the wire is determined by the Laplace equation in cylindrical coordinates $$\frac {\partial \phi}{r \partial r}+\frac {\partial^2 \phi}{\partial r^2}+\frac {\partial^2 \phi}{\partial z^2}=0 \tag 1$$ In the infinite case you can assume that the electric field at a given $r$, $E_z(r)=-\frac {\partial \phi}{\partial z}$, is constant in z-direction, thus the last term of the LHS of equ. (1) is zero. Then it remains to solve the radial equation $$\frac {\partial \phi}{r \partial r}+\frac {\partial^2 \phi}{\partial r^2}=0$$ This gives the differential equation for the radial electric field $E_r(r)$ $$E_r+r\frac {\partial E_r}{\partial r}=0$$ This has either the trivial solution $E_r(r)=0$ or the solution $$E_r(r)=\frac {r_1 E_r(r_1)}{r}$$ where $E_r(r_1)$ is a boundary condition value of the electric field at a radius $r=r_1$. The radial electric field has to be zero at the surface of the cylinder (or at the axis) because the radial current density is zero there $J_r=\sigma E_r=0$. Therefore, the radial field has to be zero $$E_r(r)=0$$ From this follows that the potential $\phi(r,z)$ has to be constant over the cross section at any $z$ and consequently $E_z(r,z)=-\frac {\partial \phi}{\partial z}$ has to be constant over the cross section. Thus, in the infinitely long cylinder, you have a homogeneous electric field parallel to the axis and thus also a homogeneous current density over the cross section.
This reasoning should also hold for the middle cross section of a finite length cylinder with symmetric boundary conditions at the ends. It should also be a good approximation for no too short cylinders away from the contacts.
Assuming thin cylindrical shape: The differential of potential difference along the length gives electric field along the length. So assuming a linear variation of potential difference(Self resistance of wire is negligible) thus the electric field is thus constant along the length of conductor. The same concept can be generalized for a conductor of arbitrarily shape.
