12 julio, 2024

Gauss’s law: explanation, applications, solved exercises

We explain what Gauss’s law is, its applications and we put solved exercises

What is Gauss’s law?

The Gauss’s law It establishes that the electric field flux, through an imaginary closed surface, is proportional to the net charge value of the particles found inside said surface.

Denoting the electric flux through a closed surface as ΦE and to the net charge enclosed by the surface by qencthen the following mathematical relationship is established:

ΦE = c ∙ Qenc

Where c is the constant of proportionality.

Gauss’s Law Explanation

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 flux through a surface.

electric charge

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 objects are neutral, that is, they have the same amount of negative charge as positive.

Two charged objects with a charge 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 have charges of a different sign, then they attract each other. This type of interaction at a distance is known as electrical interaction.

In the international system of SI units, electric charge is measured in coulombs (C). The negative elementary charge carrier is the electron with load of -1.6 x 10-19C and the positive elementary charge carrier is the proton with a charge value +1.6 x 10-19C. Typically charged bodies have between 10-9C and 10-3C.

electric field

An electrically charged body alters the space around it, filling it with something invisible called an electric field. To know that this field is present, a point positive test charge is required.

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. The field strength is the force on the test charge divided by the amount of charge on the test charge. Then the units of the electric field AND in the International System of units are newton between coulomb: [E] = N/A.

Positive point charges produce a radially outward field, while negative charges produce a radially inward field. Also, the field produced by a point charge decays with the inverse square of the distance from that charge.

electric field lines

Michael Faraday (1791 – 1867) was the first to have a mental image 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.

Positive charges are the sources from which the electric field lines emerge, while negative charges are the sinks of the lines.

Electric field lines do not close 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 every point in space the electric field vector is tangent to the field line and proportional to the density of lines there.

electric field flow

Electric field lines resemble the stream lines of a smoothly 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 En to that surface, multiplied by the area TO Of the same:

Φ = In ∙ A

The component En 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 area surface. TO. (see figure 4).

Applications of 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 outward if the charge is positive and inward otherwise.

Choosing as 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 times the total area of ​​the spherical surface:

Φ = E ∙ A = E ∙ 4πR2

On the other hand, Gauss’ law establishes 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:

Φ = (1/εo) ∙ Q

Incorporating the result obtained for the flow into Gauss’s law, we have:

E ∙ 4πR2 = (1/εo) ∙ Q

And for the magnitude of AND result:

E = (1/4πεo) ∙ (Q/ R2)

Which fully agrees with Coulomb’s law of the electric field of a point charge.

Exercises

Exercise 1

Two point charges lie inside a Gaussian surface S of arbitrary shape. One of them is known to have a value of +3 nC (3 nano-coulomb). If the net flux of electric field through the Gaussian surface is 113 (N/C) m2, what will be the value of the other charge?

Solution

Gauss’s law states that

ΦE = (1/εo) ∙ Qenc

Hence the net charge enclosed is:

Qenc = ΦE ∙ εo

Substituting the data results in:

Qenc = 113 (N/C) m2 ∙ 8.85 x 10-12 (C2 m-2 N-1) = 1 x 10-9 C = 1 nC.

But Qenc = +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 value +q, and another arrangement (on the right) with one charge +q and the other -q. Each arrangement is enclosed in an imaginary box with all its edges 10 cm. 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:

ΦE = (1/εo) ∙ (+q + q) = 678000 (N/C) m2

In the arrangement on the right, the net flux through the imaginary box containing the pair of charges is zero.

References

Cosenza, M. Electromagnetism. University of the Andes.
Díaz, R. Electrodynamics: class notes. National university of Colombia.
Figueroa, D. (2005). Series: Physics for Science and Engineering. Volume 6. Electromagnetism. Edited by Douglas Figueroa (USB).
Jackson, JD Classical Electrodynamics. 3rd. Ed. Wiley.
Tarazona, C. Introduction to Electrodynamics. Editorial Manuela Beltrán University.

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