Is electric field due to steady current conservative?

Question

It is the exemple problem in Griffiths' book

Example 7.1

A cylindrical resistor of cross-sectional area A and length L is made from material with conductivity σ. If we stipulate that the potential is constant over each end, and the potential difference between the ends is V, what current flows?

During the solution of Example 7.1, it was assumed that the electric field was uniform.

Example 7.3

I asserted that the field in Ex. 7.1 is uniform. Let’s prove it.

And, the solution of 7.3 is following:

Within the cylinder V obeys Laplaces's equation. What are the boundary conditions? At the left end the......

I can understand most of this solution, but I can't understand why V obeys the Laplacian equation.

I think the solution says "V obeys Laplace's equation" for the following reasons:

For steady current, the divergence of electric field $\vec{E}$: $\nabla \bullet \vec{E}=0$. Since $\vec{E}=- \nabla V$, $\nabla \bullet \vec{E}=0$ leads to $\nabla^2V=0$. But, as i know, For $\vec{E}$ to be $\vec{E}=- \nabla V$, $\vec{E}$ need to be conservative.

And, as i know, the electric field due to stationary charge is conservative not moving charge(current).

Is electric field due to steady current conservative?

Answer 1

From Maxwell's equations, we know that

$$ \nabla \times\mathbf E = -\partial_t \mathbf B $$

so that non-zero curl of electric field means also that magnetic field has to be changing in time. But it is not easy to think of a process where magnetic field is changing in time, while current is stationary. Usually, we think of magnetic field as due to (produced by) a distribution of current as given by the Biot-Savart formula, and in this formula, stationary current produces only static magnetic field, and and this implies conservative electric field. This "usually" is because we are very familiar with situations where the Biot-Savart law applies, and those have electric field that is very close to conservative. So we are usually in a consistent bubble of sorts, where conservative electric field is consistent with magnetic field being accurately given by the Biot-Savart law. But if the electric field is not conservative enough, the Biot-Savart law may no longer be accurate, and then we can see a possibility that magnetic field is changing in time, despite current being stationary.

In general, total current density constant in time implies (due to Maxwell's equations) this equation for the magnetic field: $$ \frac{1}{c^2}\frac{\partial^2 \mathbf B}{\partial t^2} - \Delta \mathbf B = 0. $$ This well-known equation has solutions that are waves, meaning any possible magnetic field pattern can keep its size and form while it travels through space with speed of light, or it can be an oscillating pattern that does not travel (a stationary wave). This kind of wave magnetic field can exist, and make magnetic field change in time, even if current is constant in time.

However, when this kind of field is present, we tend to look for its source, and find it in some other current that varies in time, nearby or more distant. A constant current is never regarded as a source of a wave field that is time-dependent, so we look further and always find some plausible source where the current is not stationary anymore (an inductor, an antenna, a distant hot gas or dust, etc.).

Total electric field around a stationary current need not be conservative, but electric field contribution due to usual bodies maintaining such stationary current (a discharging capacitor, or a battery) is very close to conservative, because it is due to the almost Coulombic field of almost stationary surface charges on those bodies (those charges may move slowly so there may be a small, usually neglected non-conservative component). So as long as there are no other sources of non-conservative electric field (such as an inductor nearby, or a distant radiating body), it is likely that total electric field is very close to conservative as well. In Griffiths' example, we really take the statement "electric field is conservative" as an independent additional assumption, which is realistic in most cases.

Answer 2

If $\frac{\partial \mathbf B}{\partial t} =0$, the electric field is lamellar, $\nabla \times \mathbf E = 0$, but from this one may deduce the existence of a $\phi$ such that $\nabla \phi = \mathbf E$ only locally, or stated differently, in a singly connected domain where the $\mathbf E$ is smooth.

In a magnetically homogeneous and linear matter, one having $\mathbf B =\mu \mathbf H$ with $\mu=$ constant, this also implies that $\frac{\partial \mathbf H}{\partial t} =0$, therefore $\mathbf H$ is time independent and we also get $\frac{\partial \mathbf D}{\partial t} =0$.

What remains is $\nabla \times \mathbf H = \mathbf J$ with $\nabla \cdot \mathbf J = 0$. The latter implies differentiability of the current $\mathbf J$ and also of $\mathbf E$ in a simple conductor defined by Ohm's law as $\mathbf J = \sigma \mathbf E$ the conductivity being $\sigma$. There then follows the smoothness of $\mathbf E$ everywhere as long as $\sigma$ is smooth. In practice, though, the conductivity is only piecewise smooth, as for a resistor embedded in an insulator, but that spatial discontinuity can be handled by a limit procedure at the interface.

In short, a stationary current has a conservative electric field.