$\myv \grad \cdot \myv E$

Drawing field lines

In problem 16, you sketched the vector field $\myv v= \frac{1}{\rr^2} \uv \rr$. The amplitude $|v|$ drops off as $1/r^2$ away from the origin, which most of you captured quite nicely with shorter arrows at greater distances.

This differs only by a constant from the electric field of a point charge $\myv E = \frac{q}{4\pi \epsilon_0}\frac{1}{\rr^2} \uv \rr$.

This way of drawing field lines was rather cluttered, so instead let's just draw connected lines in the local direction of the electric field.

Does that mean that we've lost the information related to the magnitude of the field?

Actually, that info is still preserved by considering that the density of field lines $\propto$ to the field strength.

So that in the orange region, the field lines are closer together $\Rightarrow$ the field has a greater magnitude in the orange region compared to the blue region, where the lines are less dense.

(Actually, we need to think in three dimensions: the field is dropping as $1/r^2$, and so the "density of field lines" metaphor only holds if we think of field lines spread out over a 3-d spherical surface, which has an area $a=4\pi r^2$, instead of a 2-d circle with circumference $c=2\pi r$.)

Further rules for drawing field lines

  • The number of field lines emanating from (or ending on) a charge should be proportional to the charge.
  • Close to any single charge, the lines should reflect the short-range radial symmetry around the charge (shown: two equal and opposite charges)
  • $\myv E$ lines point away from positive charges, and towards negative charges.
  • Indeed, field lines may never cross (How could $\myv E$ point in two different directions at the same time?),
  • and may only ever start or end on some chunk of charge.

Indeed, as intuited by our plumbing models, these pictures suggest positive charges as sources or "faucets" of water electric field lines, and negative charges as sinks or "drains" for electric field lines.

We shall further, define another very plumbing-like quantity which is the flux.

$$\Phi_E = \int_S \myv E \cdot d \myv a.$$

This is something like the "flow" of electric field vectors through the surface ${\cal S}$. (It corresponds to the fluid mass / unit time crossing the surface, if the vectors represent fluid velocity instead of electric field.)

Gauss'/Gauss's Law

Remember Gauss' theorem? Writing that 'fundamental law for divergence' out with $\myv E$ as the vector field:

$$\oint_{\cal S} \myv E \cdot d \myv a = \int_{\cal V} \myv \grad \cdot \myv E \,d \tau.$$

The interpretation was that the flow out of a closed surface was equal to the sum over the enclosed volume of all the "sources" of flow enclosed by the surface.

(This would also mean that if there are no sources or sinks inside a surface, then the total flux through the surface is zero, i.e. "Just as much flow is coming out as what's going in".)

Since charges are the sources (or sinks) of our electric field lines, this analogy suggests that the total flux of electric field through a closed surface is equal to the charge enclosed by the surface. Let's show this is so:

The electric field from a very general charge distribution was: $$\myv E(\myv r) = \frac{1}{4\pi \epsilon_0}\int_{\cal V} \frac{\rho(\myv{r}')}{\rr^2} \uv \rr\, d \tau'.$$

Using this in Gauss' theorem...

[Note that $\myv \grad \cdot$ is a derivative with respect to $\myv r$, not $\myv r'$.] $$\begineq \oint_{\cal S} \myv E \cdot d \myv a &=& \int_{\cal V} (\myv \grad \cdot \myv E) d \tau\\ &=&\int_{\cal V} \myv \grad \cdot \frac{1}{4\pi \epsilon_0}\int_{\cal V'} \frac{\rho(\myv{r}')}{\rr^2} \uv \rr\, d \tau' d \tau \\ &=& \frac{1}{4\pi \epsilon_0} \int_{\cal V} \int_{\cal V'} \myv \grad \cdot \left[\frac{\myv \rr}{\rr^3} \right] \rho(\myv{r}')d \tau'\, d \tau.\\ &=& \frac{1}{4\pi \epsilon_0} \int_{\cal V} \int_{\cal V'} \myv \grad \cdot \left[\frac{\myv r-\myv r'}{|\myv r-\myv r'|^3} \right] \rho(\myv{r}')d \tau'\, d \tau.\\ &=& \frac{1}{4\pi \epsilon_0} \int_{\cal V} \int_{\cal V'} 4 \pi \delta^3 (\myv r - \myv r') \rho(\myv{r}')\,d \tau'\, d \tau\\ &=& \frac{1}{4\pi \epsilon_0} \int_{\cal V} 4 \pi \rho(\myv{r}) d \tau\\ &=& \frac{1}{\epsilon_0} \int_{\cal V} \rho(\myv{r}) d \tau=q_{\text enc}/\epsilon_0.\endeq$$

There are two very important results here...

  1. Divergence of $\myv E$--Referring back to first and last lines: $$\myv \grad \cdot \myv E = \frac{\rho}{\epsilon_0},$$ at each and every point in space.
  2. Gauss' Law -- that the surface integral of the electric field flux is equal to the total enclosed charge inside the surface: $$\oint_{\cal S} \myv E \cdot d \myv a = \oint_{\cal S}d\Phi_E = q_{\text enc}/\epsilon_0.$$

Gauss' law is "always true, and occasionally useful". The useful situations are mostly problems with a high degree of symmetry.

Problem 2.7 revisited

Here was our vector integration solution to Problem 2.7 .

Find the electric field a distance $r$ away from the center of a spherical surface of radius $R$ that has a total charge $q$ on it, spread uniformly over its surface.

This time, imagine a second, "Gaussian" spherical surface ${\cal S}$ of radius $z$ centered on the the charged sphere. Symmetry considerations: $\myv E(r, \theta, \phi)=E(r) \uv r$.

If $z>R$: $$\oint_{\cal S} \myv E \cdot d \myv a = \oint_{\cal S} |E| da = |E| 4\pi z^2 = q/\epsilon_0.$$

Show that this gives us... $$|E| = \frac{q}{4\pi \epsilon_0} \frac{1}{z^2}.$$

If $z < R$ there is no charge enclosed by the Gaussian surface ${\cal S}$, then $E(r) = 0$.

Field near an infinite plane of charge

Imagine an infinite flat sheet of charge with a two-d charge density $\sigma$. What is $\myv E$ at a height $z$ above the sheet?

By symmetry, it seems that the field can only have a $z$-component: $\myv E = E_z\uv{z}$. So, when we take the flux integral over the "pillbox" pictured, $\myv E$ and $d \myv a$ are at right angles on all the sides of the box. On the top and bottom, $\myv E \cdot \myv{a}= E_z da$ on both. Thus...

Show that, $$\begineq \oint_{\cal S} \myv E \cdot d \myv a &=& q_\text{enc}/\epsilon_0\\ 2E_zA &=& \frac{\sigma A}{\epsilon_0}.\endeq$$

Show that: $$E_z = \frac{\sigma}{2 \epsilon_0}.$$

The field above an infinite sheet of charge is constant--independent of height $z$!

A Gaussian cylinder, used to find the field around a line charge. Use symmetry to convince yourself that $\myv E(z,s,\phi) = E(s)\,\uv s$,

You'll find this progression:

  • Spherical symmetry: Field of a point charge: $E \propto 1/z^2$,
  • Cylindrical symmetry: Field of a (infinite) line charge: $E \propto 1/z$,
  • Planar symmetry: Field of a (infinite) surface charge: $E = $ constant.

Problem 2.12

Find the electric field inside a uniformly charged sphere of radius $R$, with charge density $\rho=Q/(4 \pi R^3/3)$

The field is spherically symmetric.

The field at a distance $r$ (where $r<R$) from the center is the same as that of a point charge at the origin with the total charge enclosed by a sphere of radius $r$:

$$\begineq\myv E &=& \frac{1}{4\pi \epsilon_0} \frac{q_{\text enc}}{r^2}\uv{r}\\ &=&\frac{1}{4\pi \epsilon_0} \frac{(4/3)\pi r^3 \rho}r^2\uv{r}\\ &=&\frac{\rho}{3 \epsilon_0}\myv r=\frac{Q}{4\pi \epsilon_0 R^3}\myv r.\endeq$$

For $r\gt R$, the total enclosed charge is constant, $Q$, and the field is

$$\myv E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\uv{r}.$$

Graphing the radial component of the electric field, you should find...

Slightly more complicated...

Consider the field in and around two spheres, each with constant charge density $\rho$.

Even though this problem does not have spherical symmetry, because of the principle of superposition the resulting field will be equal to the sum of the fields from the two spherically symmetric bodies, and we can use Gauss' Law to find the solution of each of those in isolation.