Electric fields and flux

Faraday's idea with field lines

  • The number electric field lines is $ \frac{q}{\epsilon_0} $
  • The quantity$\frac{q}{\epsilon_0}$ is called the electric flux (number of lines exiting the surface)
  • (With some example shown in lecture) The electric
  • The number of lines getting out of a closed surface (flux) is denoted by $\Phi$
  • $ \Phi _{closed\ surface} = \frac{q _ {enclosed}}{\epsilon_0}$
  • In general, elecric field lines like to stay as far away from each other as possible

Electric fields

  • $ \vec{E} = \frac{\Phi}{A}$
  • The electric field can be thought of as field lines per unit area
  • The area vector $\vec{a}$ is a vector that is perpendicular to the surface and whose magnitube is equal to the surface area
  • Then, the electric flux $\Phi$ can b written as $\Phi = \vec{E} \cdot\vec{a}$
  • In the infinitessimal case, $$\ \Phi_S = \int_{S}{\vec{E} \cdot d\vec{a}}$$
  • Electric field is proportional density of field lines

Derivations from Gauss' law

  • For a spherically symmetric charge distribution: $$ field\ lines = \frac{q}{\epsilon_0}\
    \\area = 4\pi r^2\\ Field = lines \ / \ area \\ \vec{E} = \frac{q}{4\pi\epsilon_0 r^2 }$$
  • The same logic can be applied for thin, infinitely long lines
  • In general, the electric field for any surface can be calculated using the charge densities ($\lambda$ for the linear charge density, and $\sigma$ for the surface charge density)

Solid angle

The solid angle $\Omega$ is given by: $$ \ \Omega = \int \frac{dA}{r^2}$$

Usage of Gauss' law

  • Gauss' law is only useful when there is some form of symmetry (spherical, cylindrical etc.)

Spherical symmetry

  • For a spherically symmetric electric field: $$ \ \vec{|E|}\cdot 4\pi r^2 = \frac{Q_{in}}{\epsilon_0} \\ \vec{E} = \frac{\rho \vec{r}}{3\epsilon_0}$$
  • For the above equation, the difficulty lies in calculating $Q_{in}$
  • It is applicable even when the charge distribution is non-uniform. In which case, an integral over the distance would usually be involved (where the volumetric charge density is given as a function of the distance/radius)

Cylindrical symmetry

  • For a cylindrically symmetric electric field: $$ \ \vec{|E|}\cdot 2\pi rh = \frac{Q_{in}}{\epsilon_0} \\ \vec{E} = \frac{\rho \vec{r}}{2\epsilon_0} $$
  • The denominator is 2 instead of 3 (as in the shperical case) because of some properties of averages