1. Electric Flux
Flux refers to the amount of field that "penetrates" in a direction perpendicular to a given surface.
Electric flux represents the electric field that penetrates in a direction perpendicular to a surface. It is easier to illustrate this by using a visual description of the electric field (i.e., depicting the electric field as lines).
With such a depiction of the field (lines), the electric flux can be described as the number of field "lines" that penetrate a surface.
The electric flux produced by a field E on a surface of area dA is
Electric flux
Electric Flux Formula
The direction of the area element dA is determined from the normal direction of the surface.
2. Gauss's Law
Electric flux is caused by the presence of an electric field, meaning that the presence of a charge creates electric flux. Consider a point charge q located at the center of a spherical surface.
In every part of the surface of the sphere E is perpendicular to the surface and has the same magnitude.
The total flux on the surface of the spherical shell is
Total Surface Flux of a Sphere
In general it can be extended that for any surface the results obtained are the same (independent of the surface chosen).
Thus it means that the total flux on a surface is proportional to the total charge enclosed by that surface. --> Gauss's law
Can be restated
H.Gauss's statement
Gauss's law can be used to calculate the electric field of systems that have high symmetry (for example, spherical, cylindrical, or box symmetry).
To use Gauss's law, it is necessary to choose an imaginary closed surface (Gauss surface). The shape of the closed surface can be arbitrary.
3. Conductors and Isolators
Conductors are characterized as materials that have free charges. If a conductor is given an electric field, the free charges will arrange themselves following the external electric field until a state of equilibrium is reached.
Conductors and Isolators
Thus, in a state of equilibrium (static) the electric field strength inside the conductor is equal to zero.
The electrons in insulating materials are tightly bound to their atoms so that they cannot move freely, meaning that the charges in the insulating material are spread throughout the object.
Some examples
Using Gauss's law, determine the electric field strength near a densely charged surface.
If the surface is large enough and the field near the surface is being considered, then E can be considered homogeneous in the direction perpendicular to the surface.
Surface Gauss Shell Cylinder
Use Gauss's law
Completion
Using Gauss's law, determine the electric field at a distance (ι) from a long wire having a charge density (λ).
The long wire has a cylindrical symmetry shape. Select a Gaussian surface in the form of a cylindrical shell with radius l and centered on the wire.
Since the wire is long, edge effects can be neglected and the electric field has a radial direction.
Gauss's Law
Completion
An insulating sphere is charged with a charge density that is a function of the distance from the center of the sphere, namely
Determine the electric field inside and outside the sphere!
Suppose the radius of the ball is a. To determine the electric field inside the ball, create a Gauss surface in the form of a spherical shell with a radius of r.
On the Gaussian surface, the direction of the electric field is parallel to the normal direction of the surface, so that
while the charge enclosed by the Gauss surface is
So, (in the sphere, r < a)
To determine the electric field outside the sphere, create a Gauss surface in the form of a spherical shell with radius r > a.
while the charge enclosed by the Gauss surface is
So, (outside the sphere, r > a)
A conducting sphere with radius a is given a charge of -Q. Determine the electric field inside and outside the sphere.
To determine the electric field inside the ball, create a Gaussian surface in the form of a spherical shell with radius r inside the ball
while the charge enclosed by the Gauss surface is equal to zero (because in a conductor the charge is only on its surface). So E = 0 (inside the ball).
To determine the electric field outside the sphere, create a Gauss surface in the form of a spherical shell with radius r outside the sphere.
while the charge enclosed by the Gauss surface is q (in) = -Q, so that
Two spherical conductor shells arranged at the center each with radii a and b (a < b). A spherical shell of radius a has a charge of +2q while a spherical shell of radius b has a charge of -q. Determine the electric field inside and outside the shells of the sphere.
For r < a
Create a Gaussian surface in the form of a spherical shell of radius r inside a spherical shell of radius a, so that, E = 0 (for r < a)
For a < r < b
Create a Gaussian surface in the form of a spherical shell whose radius r is between the two spherical shells
For r > b
Create a Gaussian surface in the form of a spherical shell with radius r with r > b
