Gauss’s Law of Electricity

Gauss's Law of Electricity
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} \large (\Omega ).

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

\dpi{120} \large \varphi_{total}=\frac{\sum_{\Omega} Q_{internal}}{\varepsilon}=\frac{Q_1+Q_2+Q_3+\ldots+Q_n}{\varepsilon}

Gauss's Law of Electricity
Gauss’s Law of Electricity
Loads internal to a closed surface.

The unit of electrical flow is the\dpi{120} \large \frac{N\cdot 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} \large \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} \large (\Omega ).

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} \large \Delta Athis way:

Application of Gauss’s law to point charge.
\dpi{120} \large \fbox{$\varphi=E\cdot\Delta A\cdot cos\theta$}

Note that for all area elements,\dpi{120} \large \theta =0\degreeever. Therefore, the flow in the small area element is given by:

\dpi{120} \large \varphi=E\cdot\Delta A

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

\dpi{120} \large \varphi_{total}=\sum_{i=0}^{\infty}{E_i\cdot\Delta A_i}=E\cdot\sum_{i=0}^{\infty}{ \Delta A_i}

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

\dpi{120} \large \sum_{i=0}^{\infty}{\Delta A_i}=4\pi d^2

Therefore, the electrical flow is given by:

\dpi{120} \large \varphi_{total}=E\cdot4\pi d^2

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

\dpi{120} \large \varphi_{total}=\frac{\sum_{\Omega} Q_{internal}}{\varepsilon}=\frac{q}{\varepsilon}

So, we have to:

\dpi{120} \large E\cdot4\pi d^2=\frac{q}{\varepsilon}
\dpi{120} \large \therefore E=\frac{1}{4\pi\varepsilon}\cdot\frac{q}{d^2}

Stay tuned!

Which intertwines the theories already seen so far. If we had chosen another Gaussian surface\dpi{120} \large (\Omega )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} \large (\Omega ).

Given this, we can create a process to calculate the electric field\dpi{120} \large \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} \large (\Omega)closed that contains within it the distribution of charges and that has the symmetry of the problem:
  • \dpi{120} \large \left|\vec{E}\right|be constant at all points where\dpi{120} \large \vec{E}\cdot\Delta\vec{A}\neq0.
  • \dpi{120} \large \theta=0\degreeor\dpi{120} \large \theta=90\degreeor\dpi{120} \large \theta=180\degreeat all points of the chosen Gaussian.
  • Step 2: carry out\dpi{120} \large \varphi_{total}=\sum_{\Omega}{\vec{E}\cdot\Delta\vec{A}}=\frac{\sum Q_{internal}}{\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} \large (\Omega)Inside the conductor, the internal charge at this Gaussian is zero. Therefore, by Gauss’s Law, we have:

Drawing a Gaussian according to the objective of the problem.
\dpi{120} \large \varphi_{total}=\frac{\sum_{\Omega} Q_{internal}}{\varepsilon}=\frac{0}{\varepsilon}=0=\sum_{\Omega} {\vec{E}\cdot\Delta\vec{A}}

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

\dpi{120} \large {\vec{E}}_{int}=\vec{0}

Electric field created by a flat and unlimited distribution of charges

Uniformly loaded plate with uniform charge density σ.

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} \large \Delta Aand a load\dpi{120} \large \Delta 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:

Use of Gauss’ Law to calculate the field, due to the high degree of symmetry of the field. Note that there are two possibilities for the angle between the field and the area vector in the Gaussian.

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} \large \vec{E}\bot\hat{n}. On the upper surface and lower surface, the angle between\dpi{120} \large \vec{E}It is\dpi{120} \large \hat{n}is 0°. Thus, the flow is given by:

\dpi{120} \large \varphi_{inferior}=E\Delta AIt is\dpi{120} \large \varphi_{superior}=E\Delta A

By Gauss’s Law, we have:

\dpi{120} \large \varphi_{total}=\varphi_{inferior}+\varphi_{superior}+\varphi_{side}=
\dpi{120} \large \varphi_{total}=E\Delta A+E\Delta A+0=2E\Delta A=
\dpi{120} \large \varphi_{total}=\frac{\sum_{\Omega} Q_{internal}}{\varepsilon}=\frac{\Delta Q}{\varepsilon}

Therefore: 

\dpi{120} \large 2E\Delta A=\frac{\Delta Q}{\varepsilon}

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

\dpi{120} \large E=\frac{\sigma}{2\varepsilon}

As we saw previously through Integral Calculus.

Uniformly charged straight wire with linear charge density

Uniformly charged straight wire with 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} \large \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:

Gauss's Law of Electricity
Gauss’s Law of Electricity
Possible radial directions for the field generated by the infinite wire.

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:

Trace of the Gaussian on an infinite straight wire. The surface that best fits the problem is the outer surface of a straight cylinder.

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

Representation of the electric field generated by the wire and the drawn Gaussian surface.

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} \large \Delta qof lenght\dpi{120} \large \Delta l, we have that the flow can be given by:

\dpi{120} \large \varphi_{lateral}=E\Delta Acos\theta

By Gauss’s Law, we have:

\dpi{120} \large \varphi_{total}=\varphi_{lateral}+\varphi_{bases}
\dpi{120} \large \varphi_{total}=E\ 2\pi r\ \Delta l+0=\frac{\sum_{\Omega} Q_{internal}}{\varepsilon}=\frac{\Delta q}{\varepsilon}
\dpi{120} \large E=\frac{1}{2\pi\varepsilon r}\cdot\frac{\Delta q}{\Delta l}
\dpi{120} \large E=\frac{\lambda}{2\pi\varepsilon r}

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

Graph of the electric field as a function of distance for a straight, uniformly charged wire.

Insulating sphere with volumetric density

Uniformly charged insulating sphere.

Where the volumetric charge density can be given by:

\dpi{120} \large \rho=\frac{Q}{\frac{4\pi}{3}\cdot R^3},\ \rho=cte

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

  • Inside the sphere\dpi{120} \large (r<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:
Representation of the Gaussian, which is the external surface of a sphere concentric to the first sphere.

By Gauss’s Law, we have:

\dpi{120} \large \varphi_{total}=\sum{_{\Omega}}{\vec{E}\cdot\Delta\vec{A}}=\frac{\sum_{\Omega} Q_{ internal}}{\varepsilon}

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} \large \vec{E}It is\dpi{120} \large \Delta\vec{A}is 0°, therefore:

\dpi{120} \large E\sum{_{\Omega}}\Delta A=\frac{\sum_{\Omega} Q_{internal}}{\varepsilon}

Where the internal load is given by:

\dpi{120} \large \sum Q_{internal}=\rho\cdot\frac{4\pi}{3}\cdot r^3

Therefore, we have that the field is given by:

\dpi{120} \large E\cdot4\pi r^2=\frac{\rho\cdot\frac{4\pi}{3}\cdot r^3}{\varepsilon}
\dpi{120} \large E=\frac{\rho}{3\varepsilon}\cdot r

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} \large \sum Q_{internal}=\rho\cdot\frac{4\pi}{3}\cdot R^3
    \dpi{120} \large E\cdot 4\pi r^2=\frac{\rho\cdot\frac{4\pi}{3}\cdot R^3}{\varepsilon}
    \dpi{120} \large E=\frac{\rho R^3}{3\varepsilon r^2}\ ou\ E=\frac{1}{4\pi\varepsilon}\cdot\frac{Q} {r^2}

    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:

    Graph of the electric field as a function of distance from the center.

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