We explain what is Gauss’s law, and its applications and we put solved exercises.
What is Gauss’s law?
The Gauss law states that the electric field flux through an imaginary closed surface is proportional to the net value of the charge of the particles found in the interior of said surface.
Denoting the electric flux through a closed surface as Φ E and the net charge enclosed by the surface by Q enc, then the following mathematical relationship is established: gauss’s law
Φ E = c ∙ Q enc
Where c is the constant of proportionality.
The light energy or light energy is the energy generated and conveyed by light waves. When light moves, it can act as an electromagnetic wave or as a particle, as it can interact with other materials. These particles are called photons.
Explanation of Gauss’s law
To understand the meaning of Gauss’s law, it is necessary to explain the concepts involved in its statement: electric charge, electric field, and electric field flow through a surface.
What is an Electric charge? gauss’s law
Electric charge is one of the fundamental properties of matter. A charged object can have one of two types of charge: positive or negative, although normally the objects are neutral, that is, they have the same amount of negative charge as positive.
Two charged objects of the same type repel each other even when there is no contact between them and they are in a vacuum. On the contrary, when each of the bodies has charges of a different sign, then they attract each other. This type of interaction at a distance is known as an electrical interaction.
In the international system of SI units, electric charge is measured in coulombs (C). The negative carrier is the elementary charge electron loaded -1.6 × 10 -19 C and the positive elementary charge carrier is the proton with a load value +1.6 x 10 -19 C . Typically charged bodies are between 10 -9 C and 10 -3 C . gauss’s law
What is Electric field?gauss’s law
An electrically charged body alters the space in its surroundings, filling it with something invisible called an electric field. To know that this field is present requires a test point positive charge.
If the test charge is placed in a place where there is an electric field, a force appears on it in a certain direction, which is the same as that of the electric field. Field strength is the force on the test charge divided by the amount of charge on the test charge. Then drives the electric field E in the International System of Units are newton between coulomb : [E] = N / C . gauss’s law
Positive point charges produce an outward radial field, while negative charges produce a radially inward field. Furthermore, the field produced by a point charge decays with the inverse of the square of the distance to said charge.
Electric field lines Properties
Michael Faraday (1791 – 1867) was the first to have a mental picture of the electric field, imagining it as lines that follow the direction of the field. In the case of a positive point charge, these lines are radial starting from the center outwards. Where the lines are closer together the field is more intense and less intense where they are further apart. gauss’s law
The positive charges are the sources from which the electric field lines emerge, while the negative charges are the sinks of the lines. gauss’s law
Electric field lines do not close in on themselves. In a set of charges the lines leave the positive charges and enter the positive ones, but they can also reach or come from infinity.
They also do not intersect and at each point in space the electric field vector is tangent to the field line and proportional to the line density there.
Electric field flux
Electric field lines resemble the streamlines of a gently flowing river, hence the concept of electric field flow.
In a region where the electric field is uniform, the flux Φ through a flat surface is the product of the normal component of the field E n to that surface, multiplied by its area A :
Φ = E n ∙ A
The component E n is obtained by multiplying the magnitude of the electric field by the cosine of the angle formed between the field and the unit normal vector to the surface area A . (see figure 4). gauss’s law
Gauss’s law applications gauss’s law
Gauss’s law can be applied to determine the electric field produced by charge distributions with a high degree of symmetry.
Electric field of a point charge
A point charge produces a radial electric field that is outgoing if the charge is positive and incoming otherwise.
Choosing as the Gaussian surface an imaginary sphere of radius R and concentric to the charge Q, at all points on the surface of said sphere the electric field is of equal magnitude and its direction is always normal to the surface. So, in this case the electric field flux is the product of the magnitude of the field and the total area of ​​the spherical surface: gauss’s law
Φ = E ∙ A = E ∙ 4πR 2
On the other hand, Gauss’s law states that: Φ = c ∙ Q, being the constant of proportionality c . When working in units of the international system of measurements, the constant c is the inverse of the permittivity of the vacuum, and Gauss’s law is formulated as follows: gauss’s law
Φ = (1 / ε o ) ∙ Q
Incorporating the result obtained for the flow to Gauss’s law, we have:
E ∙ 4πR 2 = (1 / ε o ) ∙ Q
And for the magnitude of E it results:
E = (1 / 4πε o ) ∙ (Q / R 2 )
Which fully agrees with Coulomb’s law of the electric field of a point charge.
Training gauss’s law
Exercise 1
Two point charges lie within a Gaussian surface S arbitrarily. One of them is known to have a value of +3 nC (3 nano-coulomb). If the net electric field flux through the Gaussian surface is 113 (N / C) m 2 , what will be the value of the other charge?
Solution gauss’s law
Gauss’s law states that
Φ E = (1 / ε o ) ∙ Q enc
Hence the net enclosed charge is:
Q enc = Φ E ∙ ε o
Substituting the data results:
Q enc = 113 (N / C) m 2 ∙ 8.85 x 10 -12 (C 2 m -2 N -1 ) = 1 x 10 -9 C = 1 nC.
But Q enc = + Q – q, where the positive charge has a known value of +3 nC, therefore the charge will necessarily be -2 nC.
Exercise 2
In figure 2 there is an arrangement (on the left) of two positive charges, each with a value + q and another arrangement (on the right) with a charge + q and the other -q. Each arrangement is enclosed in an imaginary box with all its 10 cm edges. If | q | = 3 μC, find the net electric field flux through the box for each arrangement.
Solution
In the first arrangement the net flow is:
Φ E = (1 / ε o ) ∙ (+ q + q) = 678000 (N / C) m 2
In the arrangement on the right, the net flow through the imaginary box containing the pair of charges is zero. gauss’s law
