$\myv \grad \times \myv E$

Curl of $\myv E$

Stoke's theorem said: $$\int_{\cal S} \myv \grad \times \myv v\cdot d \myv a = \oint_{\cal P} \myv v\cdot d\myv{l}.$$So, to calculate $\myv \grad \times \myv E$, this suggests that we should look at how path integrals of $\myv E$ behave.

For a single point charge at the origin (radial symmetry coming): $$\myv E = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}\uv{r}.$$

If we place our "paddle wheel" in this field, does that suggest places where the curl of this field is non-zero?

The path integral from some point $\myv a$ to another $\myv b$ is $$\int_a^b \myv E \cdot d\myv{l}.$$

A step along our path can be decomposed into components:

$d\myv{l} = dr\uv{r} + r \,d \theta \uv\theta + r \sin \theta\, d\phi \uv\phi$.

But the field of our charge at the origin has only a radial component, so

$$\myv E \cdot d\myv{l} = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} dr.$$

The path integral depends only on the radial component as you step along, and the integral is...

$$\begineq\int_{ a}^{ b} \myv E \cdot d\myv{l} &=& \frac{1}{4 \pi \epsilon_0} \int_{ a}^{ b} \frac{q}{r^2} dr = \left.\frac{-1}{4 \pi \epsilon_0}\frac{q}{r}\right|_{r_a}^{r_b}\\ & =&\frac{1}{4 \pi \epsilon_0}\left(\frac{q}{r_b}-\frac{q}{r_a}\right).\endeq$$

Oh!-- so it only depends on the radial coordinate of the initial and final position.

This might remind you of the work done as you move an object of mass 'm' between two points in a gravitational field: The work only depends on the height of the initial and final positions, and not the details of the path taken.

So, the path integral around any closed path should vanish (since the radial coordinate of the first=last point is the same...): $$\oint \myv E \cdot d \myv l = 0.$$

Holds for a charge anywhere, any path...

This calculaton was for a point charge at the origin. But if we translate the point charge and the path to some other location, the physics should be the same. So this result should hold no matter where the charge or the path are.

Because of superposition, we could take the field of many particles apart into the fields of each point charge... $$\myv E_\text{net} = \myv E_1 + \myv E_2 +....$$

Since the path integral of the field due to any one charge vanishes, the sum should also vanish, so...

the path integral vanishes for any configuration of charges:$$\oint \myv E_\text{net} \cdot d \myv l = 0.$$

So, now we bring in Stokes' theorem, which holds for any surface bounding what we've now established (for the electric field) to be any closed path, $$\int_{\cal S} \myv \grad \times \myv E_\text{net}=\oint \myv E_\text{net} \cdot d \myv l=0.$$

This is only possible if

$$\Rightarrow \myv \grad \times \myv E = 0$$

at every point in space.

[You did problem 1.62 in which you showed that $\myv \grad \times(r^n\uv{r}) = 0$, and using that, you could also demonstrate this directly from our expression for the electric field.]


$$ \myv \grad \cdot \myv E = \frac{\rho}{\epsilon_0}$$

$$ \myv \grad \times \myv E = 0$$