Suppose I have a non uniform conductor which is kept in a uniform electric field maintaining a constant potential difference across its ends.Both electric field and current density are a properties of points in space.Since current density is defined as current per cross sectional area, in order to maintain same current through out the conductor the current density changes with cross sectional area. Now according to ohm's law $J=sigma*E$, as current density changes electric field should also changes with cross sectional area. How is this possible if electric field is a property of point in space and we maintain a uniform field across the conductor.How electric field changes with cross sectional area?
Electric Circuits – Does Electric Field Vary with Cross Sectional Area in a Non-Uniform Current Carrying Conductor?
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Related Solutions
It is a analogic useful way to understand the resistive circuit, in the meaning that the electric expression $V = RI$ has the same form as the mechanical $F = kv$, in an environment where the drag force is proportional to the velocity.
In the case of a conductor, it is important to note that even whithout any applied field, the free electrons have momentum to all directions, and of different magnitudes. But there is no net flow without an E-field. The effect of the E-Field is to increase the fraction of electrons to one direction, and decrease the fraction to the opposite. The scattering with the lattice limits that net flow, and is responsible for the Joule effect of the electrical resistance.
A uniform current density $\vec{J}$ is just as much an abstraction as a uniform charge density $\rho$.
Even if a "real" charge density $\rho$ is made up of a bunch of different point charges at discrete points in space, averaging the charge over a sufficiently large volume (which is still going to be pretty small on macroscopic scales) will let us think of it as being approximately uniform so we can solve problems about it using Gauss's Law & the like.
Similarly, even if a "real" current density $\vec{J}$ is made up of many different electrons with different speeds, we can define an averaged current density by looking at the amount of charge per time that crosses a particular surface over a particular period of time, and defining the magnitude of $\vec{J}$ to be $Q/(\Delta A \Delta t)$. This averages out all the microscopic fluctuations in current that you're worried about and lets us treat $\vec{J}$ as nice and uniform on larger scales.
The situation is analogous to water in a pipe or an air current, where on a microscopic level the density is not uniform and the molecular velocities vary according to the Maxwell-Boltzmann distribution with a small additional "drift" velocity. Moreover, as with water flows and current flows, the current density $\vec{J}$ doesn't have to be uniform in space or in time; it can vary over space or time either in a "microscopic sense" or in a "macroscopic sense" (think of turbulent water flow or a storm with gusts of wind.) The only reason you usually see uniform $\vec{J}$ in introductory E&M classes is that it makes the calculations easier for beginning students.
Similarly, the equation $\vec{E} = \sigma \vec{J}$ (which applies inside a conductor) is an equation that really only applies to an "averaged" electric field. In general, when we talk about the electric field "inside a medium", we are really referring to the average of an electric field over some volume that is large compared to the size of the molecules that comprise it. This issue is discussed briefly in ยง4.2.3 of Griffith's Introduction to Electrodynamics, in the context of linear dielectrics.
Best Answer
We can't maintain uniform electric field across such a conductor, because its resistance is non-uniform.
If you have cross section $A$ depending on distance $d$ from the origin as
$$A = A(d)$$
the resistance will depend on distance too:
$$\textrm dR(d) = \frac 1 \sigma \frac{\textrm d \ell}{A(d)}$$
where $\textrm d\ell$ is a linear element of the conductor and $\textrm dR(d)$ is its resistance.
Since
$$V = IR$$
on a linear element $\textrm d\ell$ there will be a drop of potential $\textrm d\varphi(d)$
$$\textrm d\varphi(d) = I \frac 1 \sigma \frac{\textrm d \ell}{A(d)}$$
Since
$$\int\limits_a^b E\,\textrm d\ell = V$$
we can write for the difference of potentials on an element $\textrm d\ell$
$$E\, \textrm d\ell = \textrm d\varphi(d)$$
Therefore
$$E = I \frac 1 {\sigma {A(d)}}$$
Hopefully this shows the intuition behind the electric field not being uniform.