General Physics 27

Flux $\Phi$

Definition

A single loop: A single loop of area $A$ is placed in the stream, whose plane is perpendicular to the direction of the flow. Flux of the loop is the fluid volume passed through the loop per unit time.

$$\Phi = \vec{v}\cdot\vec{A} = vA\cos\theta$$

A closed area :

$$\Phi = \sum\Phi_i = \sum\vec{v}_i\cdot\vec{A}_i$$

A closed surface :

$${\rm d}\vec{A} = {\rm d}y{\rm d}z\vec{i} + {\rm d}x{\rm d}z\vec{j} + {\rm d}x{\rm d}y\vec{k},\quad \vec{v} = v_x\vec{i} + v_y\vec{j} + v_z{k}$$

$$\Phi = \oiint \vec{v}\cdot{\rm d}\vec{A} = \oiint v_x{\rm d}y{\rm d}z + v_y{\rm d}x{\rm d}z + v_z{\rm d}x{\rm d}y$$

Src or Sink within	$\Phi$
None	$\Phi = \oiint \vec{v}\cdot{\rm d}\vec{A} = 0$
Source	$\Phi = \oiint \vec{v}\cdot{\rm d}\vec{A} > 0$
Sink	$\Phi = \oiint \vec{v}\cdot{\rm d}\vec{A} < 0$

Flux of Electric Field/ Electric Flux

$$\Phi_E = \sum \vec{E}\cdot\vec{A},\quad \Phi_E = \oiint\vec{E}\cdot{\rm d}\vec{A}$$

The flux of electric field is the number of lines of electric field passing through the closed surface.

Specially, the integral over a closed surface is Electric Flux.

Note

• Electric Flux is a scalar quantity.
• ${\rm d}\vec{A}$ is normal to the surface and points out of the surface. Sign matters a lot!
• Physical meaning: Electric Flux is the sum of the normal components of the electric field all over the surface.
• For an open surface, the direction of $\vec{A}$ could be determined in two different ways.

Gauss’ Law

$$\oiint \vec{E}\cdot{\rm d}\vec{A} = \Phi_E = \frac{q_{enclosed}}{\epsilon_0}$$

The net electric flux through any closed surface is proportional to the charge enclosed by that surface.

Example: Dipole

Fig	Point a	Point b	Point c
	$\epsilon_0\oiint\vec{E}\cdot{\rm d}\vec{A} = +q > 0$	$\epsilon_0\oiint\vec{E}\cdot{\rm d}\vec{A} = 0$	$\epsilon_0\oiint\vec{E}\cdot{\rm d}\vec{A} = -q < 0$

How to Use

• Direction:

  $$\vec{E}\cdot{\rm d}\vec{A} = E{\rm d}A, \quad if \quad \vec{E}∥{\rm d}\vec{A}$$

  $$\vec{E}\cdot{\rm d}\vec{A} = 0, \quad if \quad \vec{E}⊥{\rm d}\vec{A}$$

• Magnitude: $E$ has same value everywhere.

With these two conditions, $E$ could be brought out of the integral. $i.e. E\oiint{\rm d}A = \frac{q_{enclosed}}{\epsilon_0}$.

NOTE $\vec{E}$ is the electric field generated by all charges (no matter within/without the closed surface), while $Q_{enclosed}$ just contains charges within the surface.

Calculating $\vec{E}$ by Gauss’

Uniform charged sphere

A solid sphere of radius $a$ is of uniform charge density $\rho$ . Calculate the electric field a point at distance $r$ from the center of the sphere.

Outside sphere $r > a$:

$$\oiint \vec{E}{\rm d}\vec{A} = 4\pi r^2E= \frac{q_{enclosed}}{\epsilon_0} = \frac{\frac{4}{3}\pi a^3\rho}{\epsilon_0}$$

$$E = \frac{a^3\rho}{3\epsilon_0r^2}$$

Inside sphere $r < a$:

$$\oiint \vec{E}{\rm d}\vec{A} = 4\pi r^2E= \frac{q_{enclosed}}{\epsilon_0} = \frac{\frac{4}{3}\pi r^3\rho}{\epsilon_0}$$

$$E = \frac{r\rho}{3\epsilon_0}$$

Conductors

Inside the conductors, $E = 0$. Therefore, $Q_{inside} = 0$, charges only reside on the surface.

isolated sphere	isolated conductor with cavity	charge in cavity

The outer E field:

$$\epsilon_0\oiint\vec{E}\cdot{\rm d}\vec{A} = \epsilon_0E\Delta A = \sigma\cdot\Delta A$$

$$E = \frac{\sigma}{\epsilon_0}$$

Here $\sigma$ is single sided compared to $\sigma$ in the disk equation $E = \frac{\sigma}{2\epsilon_0}$.

Infinite line of charge

Take a cylinder shell as the gaussian surface.

$$\epsilon_0\oiint\vec{E}\cdot{\rm d}\vec{A} = 2\epsilon_0\pi rhE = \lambda h$$

$$E = \frac{\lambda}{2\pi\epsilon_0r}$$

Infinite sheet of charge

Take a cylinder shell as the gaussian surface.

$$\epsilon_0\oiint\vec{E}\cdot{\rm d}\vec{A} = 2\epsilon_0AE = \sigma A$$

$$E = \frac{\sigma}{2\epsilon_0}$$

Similarly, E fields between outside infinite sheets of $\sigma$ charge density is $0$. E fields between is $\frac{\sigma}{\epsilon_0}$.

Summary For spherical, cylindrical, planar symmetry of electric fields, Gauss’ Law makes solving easy.

Gauss’ and Coulomb’s

Coulomb’s Law	Gauss’ Law (Fundamental Law)
Gives the force between two point charges.	Gives the relationship between $\vec{E}$ and charges.
	Offer much simple way in symmetry situation.
	Valid in rapidly moving charges, more general.

They give identical results.

By the spherical symmetry of $\vec{E}$ of a point charge,

$$\oiint \vec{E}\cdot{\rm d}\vec{A} = E\oiint{\rm d}A = \frac{q}{\epsilon_0}$$

$$E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$$