# What does Gauss' law mean?

What does Gauss' law means?

$$\nabla\cdot\mathbf{E} = \dfrac{\rho}{\epsilon_0}$$

Divergence doesn't need a direction because it's the average outflow of bananas in every direction around a point $p$. You draw a closed curve around $p$, add up the total number of bananas passing through the curve (positive for outgoing bananas, negative for incoming) and then divide by the area enclosed by the curve. And then the divergence is what happens to that quantity when the curve is very small, enclosing $p$ and not much else

This means I should take a positive point charge at $p$ and then draw closed curve around $p$, add up the total number of Electric field passing through the curve and then divide by the area enclosed by the curve.This should give me the divergence of Electric field at point $p$

Again divergence is the it's the average outflow of Electric field in every direction around a point $p$.

I cannot connect these two definitions or really understand what this equation really tells us.

A simple answer without complicated mathematics(perhaps using analogy) would be highly appreciated.

• What in the whole wide universe has maxwell got to do wih bananas? Ireally didnt understand the banana resemblance here – Lelouch Sep 17 '16 at 16:45

Gauss Law

$$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$$ is telling us that the charge density enclosed in an infintesimal region is equal to the total electric flux in/out of it

More rigorously, the divergence can be written as

$$\nabla \cdot \vec{E} = \lim_{V\rightarrow 0}\frac{1}{V}\int_{\partial V} \vec{E}\cdot d\vec{a}$$

Therefore it is basically counting the number of field lines passing through the volume $V$ (which is enclosed by a surface $\partial V$). The surface can be as simple as a sphere, but any closed surface even as irregular as a potato is also ok.

Therefore if you have more positive charge, then based on the notes you learn in high school, the electric field lines $\vec{E}$ will be more densely packed thus there will be more field lines going out of the region $V$. Likewise if you have more negative charges, then there will be more field lines going into the region

Now we take this volume infinitesimally small, thus this will mean at the vicinity of the charges , on average how many field lines go in/out from the charge density. The volume being infinitesimal also ensures any field lines that basically be going in/out in all directions in space from the charge density will pass through the enclosed surface, and hence be counted towards the average number of field lines (since at a charge density, the field lines are radiating out/in in all directions)

A simple analogy that illustrate the maths almost accurately

Consider the RHS of the equation as like a fountain. If the fountain is strong, then at its vicinity, there is going to be a lot of water flowing out in all directions. Likewise, if the fountain is weak, there won't be much water going out at all. Now if we consider a drian instead, then the stronger the suction of the drain, at its vicinity, more water is going to rush into it, likewise if the drain is weak, the amount of water going through is little.

Therefore the amount of water flowing in/out at the vicinity of the drain/fountain is the LHS, while how strong they are is the RHS

First, in 3-D space you have to consider a closed surface containing charges in the interior. Then add all the electric flux elements 𝜀0·En·dA normal to each small surface element dA on this closed surface to obtain the total flux. Then add all small charge elements dQ=𝜌dV in the volume enclosed by this surface to obtain the total charge Q. Then Gauss law states that the total electric flux going through the surface is equal to the total charge Q enclosed by this volume. The mathematical divergence theorem equates the total flux of a vector field (here E) over a closed surface to its volume integral of the divergence div E. When you consider the closed surface contracting to a point, the Maxwell equation div (𝜀0·E) = 𝜌 follows. The meaning of Gauss law is that you can visualize the electric field as a flow of an incompressible liquid. The total flux of this liquid leaving the closed surface must sum up to zero when no sources or sinks (negative sources) of the flux in the volume exist or be the sum of all sources minus sink flows in this volume. The charges can therefore be considered as sources or sinks of this electric flux. The equation div (𝜀0·E) = 𝜌 expresses this for an infinitesimal small volume element, when you multiply it by dV. The flux leaving an infinitesimal small volume has to be equal to the source strength (positive or negative), which is the charge in the infinitesimal volume 𝜌·dV.