2.6. Stokes' theorem

Knowing the rotor of the velocity vector at each point of the surface S, the circulation of the vector on the contour G can be found.

Let us divide the surface on elements dS.

According to the equation (2.11) circulation velocity of the contour, bounded by the surface dS can be written as:

circulation ,

n – the positive normal to the surface element dS.

Let us summarize the expressions on the entire surface S and we will make limiting transition at which all values ∆S →0. We will get:

.

Indeed, adding up all , we will see that they are mutually destroyed.

For the area ΔS, which lies to the left of MN, area determining the circulation takes place in the direction from N to M, and for ∆S to the right of NM the same area takes place in the direction from M to N. Thus for adjacent areas differs only in their signs. Only thosewill be unrequited that lie outside the contourG.

We have: .

Stokes' theorem for the vector of tension:

.

Lecture 3

§ 3. Stress Evaluation of Field according to Gauss' theorem

3.1. Impossibility of stable equilibrium of charge in an electric field

Presume that a vacuum is a system of fixed point charges, which are in equilibrium. Let us consider a charge Q , which is in the closed surface S. Suppose that Q> 0, then for the stability of equilibrium, it is necessary that all points of S of field E, which is formed by all the other charges of the system, were directed to Q. Only in this case, any small displacement of Q from the equilibrium, it be attected by the turning force and the equilibrium will be stable. However, this configuration of the field E around Q contradicts the Gauss᾿s theorem. The flow F through S will be negative, and according to the theorem of Gauss, it must be equal to zero, since this flow was created by charges that are located outside of S. And the fact that the flow of intencity vector is numerically equal to zero means that at any point of the surface S, the vector E is directed inside, while others – outside. The conclusion is that the stable equilibrium of charge at any electrostatic field does not exist.

3.2. Field of homogenieosly charged plane

Presume the charge is equally distributed on the plane with surface density σ. For the reasons of symmetry it is obvious that the intensity vector can be only perpendicular to the charged plane and symmetrical with respect to this plane points; the vector E is equal in magnitude and opposite in direction.

This configuration of the field suggests that you need to choose the cylinder as a closed surface to the research of the field.

We believe that this surface density is positive: σ>0.

The flow through a circular surface is equal to zero: Ф_б=0.

So the total flux through the entire surface of the cylinder:

Ф_n=2E∆S,

where ∆S is the plane of each end of the cylinder.

The charge Q=σ∆S is located outside the cylinder. According to the theorem of Gauss, Ф_n=Q.

,

,

E_n - projection of the vector E on the normal n to the charged surface and the normal vector is directed to the surface.

According to (3.1), where σ >0 and the electric field intensity Е_n>0. This means that the vector E is directed from the charged plane.

If σ<0, then Е_n<0 (vector E is directed to a charged plane).

The fact that E does not depend on the distance from the plane means that the corresponding electric field is homogeneus, that is, equally to the left and to the right of this plane.

This result is valid only for an infinite homogeneous flat surface (only in this case symmetry is true), but it is approximately true for the adjacent part of the surface.