Flux

  • The electric field doesn’t itself flow, but one can imagine the fieldlines represent a flow.
  • Imagine a constant field $\vec{E}$ passes through a surface with area $A$, at right angles to that surface. Then we say there is a total electric flux of $\Phi_E = \vert \vec{E} \vert A$.
  • If it passes through at an angle, use only the component of $\vec{E}$ perpendicular to the surface: $\Phi_E = \vec{E} \cdot \hat{\vec{n}} A$. This is like counting the number of fieldlines passing through the surface (if they are skimming along the surface, they don’t count).
  • Now make the area a closed surface; maybe the surface of a sphere, but possibly something much more complicated.
  • A really remarkable theorem called Gauss’s law says that the integral of the flux over the closed surface is directly proportional to the charge enclosed.1

To see why, try playing with the example below where there are two charges of equal and opposite magnitude. Here we are in 2D, so instead of an area/surface we just have a line/loop.

  • Drag to draw various closed contours of your own of any shape.
  • Count the number of fieldlines crossing out of your shape, and subtract the number of fieldlines crossing into your shape (if there are any).
  • You should find this number is exactly 8 times the charge enclosed; i.e., it’ll be either 0, 8 or -8 depending on the total charge enclosed.
  • Try then dragging the charges in and out of your shape to verify that this remains true.

Does this seem remarkable to you, or obvious? To me it seems a bit of both, but to see why it happens we need a bit more maths.

Surface integrals

To formulate Gauss’s law with mathematical precision, we need to introduce the idea of a surface integral.

  • Starting in 2D, as in the example above, imagine a closed loop and let’s try and calculate the flux of electric field through it, taking into account the direction of the fieldlines.
  • We start by splitting the loop into lots of little pieces of length $\Delta l$, each one centred at a point $\vec{x}_i$ and with a normal vector $\hat{\vec{n}}_i$ pointing outwards.
  • Then we define flux through a single piece of the loop as $\vec{E}(\vec{x}_i) \cdot \hat{\vec{n}}_i \Delta l$, and summing up gives us the total flux $\sum_{i=1}^{N} \Delta l \vec{E}(\vec{x}_i)\cdot \hat{\vec{n}}_i$.
  • Remember from calculus that when we have a sum of lots of little pieces like this, we can take the limit that $\Delta l \to 0$, turning it into an infinite sum of infinitesimal pieces while still traversing the same size of loop in total: $\oint \dd l \, \vec{E} \cdot \hat{\vec{n}}$. The circle over the integral sign just means we are integrating around a closed loop.
  • Now let’s move to the real world, which is 3D. Instead of lines, we actually should be imagining a surface enclosing one or more charges. Our sum should actually come from dividing up that surface into lots of little areas $\Delta A$, or in the infinitessimal limit, $\dd A$. Now we have the first rigorous definition:
\[\Phi_E = \oiint_S {\rm d} A \, \vec{E} \cdot \hat{\vec{n}}.\]
  • It is now a double integral ($\iint$) because the surface is two-dimensional, but just like the case of the line, you can imagine it as dividing the surface into lots of little mini-surfaces.
  • The $S$ near the bottom indicates that the integral is over a surface $S$ (this saves us from explicitly defining what that surface is using any particular coordinate system).
  • There is a circle over the integral symbol ($\oiint$) to indicate that the surface being integrated over is closed.
  • If one wants to actually calculate $\Phi_E$ explicitly, it is necessary to expand the expressions in a suitable coordinate system.

To make the expression even more compact, the unit normal vector to each infinitesimal area element can be built in, by defining $\vec{\rm{d} A} \equiv {\rm{d}} A\, \hat{\vec{n}}$. This turns out to be a very natural thing to do when you develop the maths a bit further, so let’s write the flux in this form for completeness:

\[\Phi_E = \oiint_S \vec{\mathrm{d}A} \cdot \vec{E}.\]

When you first see expressions like this they look quite abstract, but that is because they are designed to be general. In practice, you make use of them by choosing a good coordinate system and making them concrete. We will see examples of this below.

Gauss’s Law

Gauss’s law states that $\Phi_E \propto q$, where $q$ is the total charge enclosed by the surface $S$ over which the flux $\Phi_E$ has been calculated. To prove Gauss’s law formally requires a bit more mathematics (something called the divergence theorem, if you want to read ahead).

Instead of proving it rigorously, let’s work out what is the constant of proportionality. Imagine a point charge of strength $+q$ at the centre of a sphere. Then, according to Coulomb’s law, the electric field at every point on the sphere is $q / (4 \pi \epsilon_0 r^2)$, directed radially outwards. Now,

\[\Phi_E = \oiint_S {\rm d} A \, \vec{E} \cdot \hat{\vec{n}} = \oiint_S {\rm d} A \frac{q}{4 \pi \epsilon_0 r^2}\,.\]

But notice that the integrand is constant over the surface of the sphere, so can be taken outside the integral. All we are left with is $q / (4 \pi \epsilon_0 r^2)$ times the area of the sphere, which is $4 \pi r^2$. We conclude that

\[\Phi_E = \frac{q}{\epsilon_0}.\]
  • The size of the sphere has cancelled out, as it must do if Gauss’s law is valid (since the size of the sphere doesn’t change the charge enclosed)
  • This doesn’t prove Gauss’s law in general, because we used a special setup with the charge at the centre of a sphere.
  • But if you take the generality of the law on trust, this argument tells us the constant of proportionality exactly ($1/\epsilon_0$).

Why is Gauss’s Law useful?

Using Gauss’s law can simplify a lot of electric field calculations, when combined with suitable assumptions based on symmetries. For example, consider an infinite line of charges, $\lambda$ per unit length. In the lecture we will show that the corresponding electric field is directed radially away from the line, with strength

\[\vert \vec{E} \vert = \frac{\lambda}{2\pi \epsilon_0 R} \,.\]

Aside: it is often helpful to talk about charges distributed through space in some way. We can have line, surface, or volume charge densities:

Dimension Typical symbol Units
1 (e.g. charged wire) $\lambda$ ${\rm C\,m^{-1}}$
2 (e.g. charged sheet) $\sigma$ ${\rm C\,m^{-2}}$
3 (e.g. charged sphere) $\rho$ ${\rm C\,m^{-3}}$

Discussion point: does Gauss’s law also apply to Newtonian gravity?

Footnotes

  1. There are even more profound examples of this in physics, where behaviour on a surface can sometimes fully determine the behaviour of an interior region; see the holographic principle. ↩