# Gauss’s Law of Electricity

Let’s consider a distribution of electric charges and any closed imaginary surface that surrounds these charges. We call this imaginary surface the Gaussian surface$\dpi{120}&space;\large&space;(\Omega&space;)$.

Gauss’s Law determines that the total flow$\dpi{120}&space;\large&space;(\varphi&space;_{total})$that crosses the Gaussian$\dpi{120}&space;\large&space;(\Omega&space;)$ is equal to the total internal load$\dpi{120}&space;\large&space;\sum&space;Q_{interna}$divided by the permittivity of the medium$\dpi{120}&space;\large&space;(\varepsilon)$.

The unit of electrical flow is the$\dpi{120}&space;\large&space;\frac{N\cdot&space;m^2}{C}$.

To exemplify the use of Gauss’s Law, we will calculate the electric flow generated by a point electric charge with charge +q, in a medium whose electrical permittivity is$\dpi{120}&space;\large&space;\varepsilon$.

Let’s take a Gaussian surface so that we can have the greatest symmetry in the problem when we calculate the sum of the fluxes of each area element in the Gaussian$\dpi{120}&space;\large&space;(\Omega&space;)$.

Without much effort, we see that if we take a sphere centered on the point charge, we will simplify our sum of the small fluxes. Then we can calculate the flow in the area element$\dpi{120}&space;\large&space;\Delta&space;A$this way:

Note that for all area elements,$\dpi{120}&space;\large&space;\theta&space;=0\degree$ever. Therefore, the flow in the small area element is given by:

The total flux on the spherical surface is given by the sum of the fluxes on all area elements in the Gaussian$\dpi{120}&space;\large&space;(\Omega&space;)$, remembering that the magnitude of the electric field is even given the distance d:

But,$\dpi{120}&space;\large&space;\sum_{i=0}^{\infty}{\Delta&space;A_i}$corresponds to the surface area of ​​the sphere (surface area of ​​the Gaussian$\dpi{120}&space;\large&space;(\Omega&space;)$, then:

Therefore, the electrical flow is given by:

But, according to Gauss’s Law, the total flow is:

So, we have to:

## Stay tuned!

Which intertwines the theories already seen so far. If we had chosen another Gaussian surface$\dpi{120}&space;\large&space;(\Omega&space;)$any involving the point charge, the theorem would remain valid, since the total flux piercing this other surface is the same flow that pierces the spherical surface.

Note: Gauss’s Law presented here is valid as long as there are no charges distributed along the Gaussian surface$\dpi{120}&space;\large&space;(\Omega&space;)$.

Given this, we can create a process to calculate the electric field$\dpi{120}&space;\large&space;\vec{E}$of charge distributions with a certain degree of symmetry:

• Step 1: from the point where you want to calculate the field modulus, idealize a Gaussian$\dpi{120}&space;\large&space;(\Omega)$closed that contains within it the distribution of charges and that has the symmetry of the problem:
• $\dpi{120}&space;\large&space;\left|\vec{E}\right|$be constant at all points where$\dpi{120}&space;\large&space;\vec{E}\cdot\Delta\vec{A}\neq0$.
• $\dpi{120}&space;\large&space;\theta=0\degree$or$\dpi{120}&space;\large&space;\theta=90\degree$or$\dpi{120}&space;\large&space;\theta=180\degree$at all points of the chosen Gaussian.
• Step 2: carry out$\dpi{120}&space;\large&space;\varphi_{total}=\sum_{\Omega}{\vec{E}\cdot\Delta\vec{A}}=\frac{\sum&space;Q_{internas}}{\varepsilon}$along the drawn Gaussian.

## Some applications of Gauss’s Law

### Electric charge distribution of an electrified conductor in electrostatic equilibrium

We can verify that the electric field inside an electrified conductor in electrostatic equilibrium is zero using Gauss’ Law. As we well know, given the electrostatic equilibrium condition of the conductor, the charges on the conductor are distributed along the external surface.

Therefore, if we idealize a Gaussian$\dpi{120}&space;\large&space;(\Omega)$Inside the conductor, the internal charge at this Gaussian is zero. Therefore, by Gauss’s Law, we have:

The only way to ensure that$\dpi{120}&space;\large&space;\sum&space;{_{\Omega&space;}}\vec{E}\cdot\Delta&space;\vec{A}$ is if the electric field is zero. This way, we confirm the result that we already knew:

### Electric field created by a flat and unlimited distribution of charges

Given the symmetry of an unlimited (infinite) plate with uniform electric charge distribution, we know that the electric field is zero at a point on the plane and perpendicular to it at a point outside it.

Indeed, when we take a charge q located to the left of the desired point, there is a charge q to the right, the same distance from the point, in such a way that the lateral component is zero.

If we pass a Gaussian across a charge element on the plate, we can determine the electric field at a point outside the charge plane. Therefore, the Gaussian must involve an area$\dpi{120}&space;\large&space;\Delta&space;A$and a load$\dpi{120}&space;\large&space;\Delta&space;Q$.

As the horizontal field is zero, an excellent Gaussian to be taken is the external surface of a straight cylinder. Taking the front view, we have:

In this way, we can see that the fluxes on the lateral surface of the cylinder are zero, as we will always have$\dpi{120}&space;\large&space;\vec{E}\bot\hat{n}$. On the upper surface and lower surface, the angle between$\dpi{120}&space;\large&space;\vec{E}$It is$\dpi{120}&space;\large&space;\hat{n}$is 0°. Thus, the flow is given by:

$\dpi{120}&space;\large&space;\varphi_{inferior}=E\Delta&space;A$It is$\dpi{120}&space;\large&space;\varphi_{superior}=E\Delta&space;A$

By Gauss’s Law, we have:

Therefore:

Since the uniform surface charge density of the plate can be written as$\dpi{120}&space;\large&space;\sigma=\frac{\Delta&space;Q}{\Delta&space;A}$, we concluded that:

As we saw previously through Integral Calculus.

### Uniformly charged straight wire with linear charge density

Let’s calculate the electric field generated by a straight wire as long as we want, with a linear charge density$\dpi{120}&space;\large&space;\lambda$, such that:

To do this, let’s take a load element situated at one length element of the wire. Similar to the case of an infinite plate with uniform charge distribution, we can see without much effort that the electric field generated by the wire will have the same modulus for points at a distance in the radial direction as follows:

Given this set, the best Gaussian surface is again the surface of a cylinder, where the straight wire coincides with the axis of rotation of the straight cylinder, as follows:

Taking a frontal view of the side surface of the cylinder, we have:

Thus, on the surfaces that form the bases of the cylinder there will be no flow of the electric field, only on the lateral surface of the cylinder. For the load element$\dpi{120}&space;\large&space;\Delta&space;q$of lenght$\dpi{120}&space;\large&space;\Delta&space;l$, we have that the flow can be given by:

By Gauss’s Law, we have:

This result shows that the electric field decreases as we take points further and further away from the wire. Graphically, we have:

### Insulating sphere with volumetric density

Where the volumetric charge density can be given by:

For this case, we have two regions to be analyzed:

• Inside the sphere$\dpi{120}&space;\large&space;(r:
We note that the Gaussian that meets our conditions of symmetry and constant field is also the surface of a sphere with the same center, but radius r. This way, we have the following configuration:

By Gauss’s Law, we have:

Given the symmetry in the Gaussian with radius r, we have that the field has the same module and the angle between the vectors$\dpi{120}&space;\large&space;\vec{E}$It is$\dpi{120}&space;\large&space;\Delta\vec{A}$is 0°, therefore:

Where the internal load is given by:

Therefore, we have that the field is given by:

Note that the field inside an insulating sphere, with uniform volumetric charge density, grows linearly as we reach points further away from the center of the sphere.

• Sphere exterior\dpi{120}&space;\large&space;(r>R)” alt=”\dpi{120} \large (r>R)” align=”absmiddle”>:
Repetindo o processo para um ponto externo, encontrar que:
$\dpi{120}&space;\large&space;\sum&space;Q_{internas}=\rho\cdot\frac{4\pi}{3}\cdot&space;R^3$

In other words, for a point outside the sphere, everything happens as if the charge were punctual and located in the center of the sphere. Thus, we have the following graph of the electric field as a function of distance for the two regions: