

§5 Flux of intensity We define the flux of vector through any surface dS,  normal to the surface, α  the angle between the normal and the force line of vector . You can enter a area vector . Flux of vector is called a scalar Ф_{Е} equal to the scalar product of the intensity vector on the area vector .
For a homogeneous field
For inhomogeneous field
Where  the projection on ,  the projection on the
In the case of a curved surface S it must be broken down into elementary surface dS, to calculate the flow through the unit area, and the total flux is equal to the amount or limit of integrals of elementary streams
where  the integral over a closed surface S (eg, sphere, cylinder, cube, etc.)
For a uniform field flow through a closed surface is equal to zero. In the case of inhomogeneous field
§ 6 Gauss's theorem and its application to the calculation of the electrostatic field I. Consider the electrostatic field generated by a single positive charge. We include it in the sphere of radius R. We define the flow of intensity through a spherical surface of radius R. We divide the surface of the sphere S at the elementary area dS. The normal to the area dS is directed through the sphere of radius and coincides with the direction : parallel so
Then the flow of the vector through the surface S is equal to the sum of flows through an area element dS and vanish dS we can write
Considering that the field of a point charge is
get
This result can be generalized to any surface. Given the principle of superposition can apply our results to any number of charge inside the surface.
II. Application of the Gauss theorem.
Lines of intensity are perpendicular to the surface and sent from it on both sides. Construct a cylinder with a base S, whose elements of cylinde are parallel lines to intensity .
The flow through the lateral surface of the cylinder is equal to zero, because perpendicular S cosα = cos90 ° = 0, so
2. The field intensity created by two parallel infinitely extended plates with surface charge density + σ and σ. We find the field E, using the principle superposition of the fields. In the region between the planes
Left and right of the plane fields deducted as lines of intensity are directed toward each other .
3. The field intensity created by an infinitely extended thread with linear charge density τ.
Required to determine the field intensity at a distance r from the filament. For this we construct a cylinder of radius r and height h, the axis of which passes the thread.
Lines of intensity are directed radially, moving away from the surface of the sphere at right angles. The surrounding area of ??this sphere of radius r and determine the intensity the flow through spherical surface of radius r. When r > R the entire charge q is inside the sphere r. Then, by the Gauss theorem
, becauseЕ_{n} = E.
When r < R, inside surface of radius r, and there are no charges so E = 0. Based on this screening  protection from external electric fields.
a) For r> R, under item 4 to determine the
b) For r < R
