What is the physical significance of electric flux? [duplicate]

We know that the number of electric field lines crossing a unit area placed normal to the field at a point is a measure of strength of electric field $\mathbf{E}$ at this point. Therefore, the no. of electric field lines crossing this area element $dS$ is directly proportional to $\mathbf{E}\cdot dS$ .

The electric flux, d$\Phi$ through an area d$S$ is defined by$$d \Phi=\vec{E}.\text{d} \vec{S}.$$This is the same equation that gives the volume of a fluid flowing per second through an area $\text{d}\vec{S}$ if the fluid's velocity at that point is $\vec{E}$. That's how we get the name "flux". "Flux" is an old word for flow. Electric field lines (lines drawn so that their direction at any point along them is the direction of E at that point) are therefore exactly analogous to streamlines. But it is only an analogy. It might help you to get a feel for electric flux, but it shouldn't be taken to imply that anything is actually flowing!

Quite independently of the fluid analogy, if several electric field lines are drawn, starting, say, near a positive charge, then if we consider localities through which the lines pass, the number of these lines per unit area crossing an area normal to them will represent the relative strength of the field, so we can get an idea of both the magnitude and direction of the field in a whole region, from the pattern of field lines. Note that for an electric field due to charges that are stationary in our frame of reference, the lines start on positive charges and end on negative charges.

That's enough about field lines. They are an aid to visualisation.

Electric flux is important because it features in two of Maxwell's equations, when integrated. Here's the first of the two$$\int_{S} \vec{E}.\text{d} \vec{S}=\frac{1}{\epsilon_0} \int_{V} \rho \text{d} V$$ $\rho$ is the charge density at points inside a volume $V$ enclosed by a surface $S$. So the equation says that the total flux leaving a surface enclosing a volume is equal to $\frac{1}{\epsilon_0}$ times the net charge in that volume. This is Gauss's Law for electric fields. Putting it another way, the significance of the flux coming out of a closed surface is that it tells you the net charge inside that surface! Gauss's law is more general than Coulomb's law because it holds even if the charges are moving. A simple use of Gauss's Law is to calculate capacitance when there is a high degree of symmetry.

The second is$$\int_{C} \vec{B}.\text{d}\vec{\ell}=\int_S \left(\mu_{0} \vec{J}+\epsilon_{0} \mu_{0} \frac {\partial \vec{E}}{\partial t}\right) .\text{d}\vec{S}.$$ This gives the line integral of the magnetic field around the perimeter of an area S through which there is a current $\int_S \vec{J} .\text{d}\vec{S}.$ and an electric flux varying at a rate $\frac {\partial}{\partial t}\int_{S} \vec{E}.\text{d}\vec{S}.$ The varying flux term comes into its own when studying electromagnetic waves.

• I couldn't find the physical significance of electric flux. .... Jun 9 '18 at 14:58
• You have the definition (which you knew anyway) and two equations which involve electric flux. These equations give you its physical significance; its involvement in these equations is why it's significant! I also hinted at why the equations themselves were useful. And I tried to give you a feel for the idea of electric flux, using an analogy. What more would you like? Jun 9 '18 at 15:04
• I've now included another sentence in my answer, that may or may not help you in your quest. Jun 9 '18 at 15:35