Theorem on the circulation of the tension vector. Theorem on the circulation of the strength vector Theorem and circulation of the electrostatic field strength vector
Circulation theorem
Previously, we found out that a charge (q) that is in an electrostatic field is acted upon by conservative forces, the work ($A$) of which on any closed path (L) is equal to zero:
where $\overrightarrow(s)$ is the displacement vector (not to be confused with the area), $\overrightarrow(E)$ is the field strength vector.
For a unit positive charge we can write:
The integral on the left side of equation (2) is the circulation of the intensity vector along the contour L. A characteristic property of the electrostatic field is that the circulation of its intensity vector along any closed contour is zero. This statement is called the circulation theorem of the electrostatic field strength vector.
Let us prove the circulation theorem on the basis that the work of the field to move a charge does not depend on the trajectory of charge movement in the electrostatic field, which is expressed by the equality:
where $L_1\ and\ L_2$ are different paths between points A and B. Let us take into account that when replacing the integration limits, we obtain:
Expression (4) is represented as:
where $L=L_1+L_2$. So the theorem is proven.
A consequence of the circulation theorem is that the electric field strength lines are not closed. They start on positive charges and end on negative charges or go to infinity. The theorem is true specifically for static charges. Another consequence of the theorem: the continuity of the tangential components of tension (as opposed to the normal components). This means that the tension components that are tangent to any selected surface at any point have equal values on both sides of the surface.
Let us select an arbitrary surface S, which rests on the contour L (Fig. 1).
In accordance with the Stokes formula (Stokes theorem), the integral of the rotor of the tension vector ($rot\overrightarrow(E)$), taken over the surface S, is equal to the circulation of the tension vector along the contour on which this surface rests:
where $d\overrightarrow(S)=dS\cdot \overrightarrow(n)$, $\overrightarrow(n)$ is a unit vector perpendicular to the section dS. The rotor ($rot\overrightarrow(E)$) characterizes the intensity of the “swirling” of the vector. A visual representation of the vector rotor can be obtained if a small, lightweight impeller (Fig. 2) is placed in the fluid flow. In those places where the rotor is not equal to zero, the impeller will rotate, and the speed of its rotation will be greater, the greater the projection module of the rotor projection onto the impeller axis.
In practical calculations of the rotor, the following formulas are used most often:
Since, in accordance with equation (6), the circulation of the tension vector is zero, we obtain:
Condition (8) must be satisfied for any surface S that rests on the contour L. This is only possible if the integrand is:
and for each point of the field.
By analogy with the impeller in Fig. 2 imagine an electric “impeller”. At the ends of such an “impeller” there are charges q of equal magnitude. The system is placed in a uniform field with intensity E. In those places where $rot\overrightarrow(E)\ne 0$ such a “device” will rotate with acceleration, which depends on the projection of the rotor onto the impeller axis. In the case of an electrostatic field, such a “device” would not rotate at any axis orientation. Since the distinctive feature of the electrostatic field is that it is irrotational. Equation (9) represents the circulation theorem in differential form.
Example 1
Assignment: In Fig. 3 shows the electrostatic field. What can you tell about the characteristics of this field from the figure?
About this field we can say that the existence of such an electrostatic field is impossible. If you select the outline (it is shown as a dotted line). For such a circuit, the circulation of the tension vector is:
\[\oint\limits_L(\overrightarrow(E)d\overrightarrow(s)\ne 0)\left(1.1\right),\]
which contradicts the circulation theorem for the electrostatic field. The field strength is determined by the density of the field lines, it is not the same in different parts of the field, as a result, the work along a closed loop will differ from zero, therefore, the circulation of the strength vector is not equal to zero.
Example 2
Assignment: Based on the circulation theorem, show that the tangential components of the electrostatic field strength vector do not change when passing through the dielectric interface.
Let's consider the boundary between two dielectrics with dielectric constants $(\varepsilon )_2\ and\ (\varepsilon )_1$ (Fig. 4). Let us select a small rectangular contour on this boundary with parameters a - length, b - width. The X axis passes through the midpoints of sides b.
For the electrostatic field, the circulation theorem is satisfied, which is expressed by the equation:
\[\oint\limits_L(\overrightarrow(E)d\overrightarrow(s)=0\ \left(2.1\right).)\]
For small circuit sizes, the circulation of the tension vector and in accordance with the indicated direction of traversal of the circuit, the integral in formula (2.1) can be represented as:
\[\oint\limits_L(\overrightarrow(E)d\overrightarrow(s)=E_(1x)a-E_(2x)a+\left\langle E_b\right\rangle 2b=0\ \left(2.2\right) ,)\]
where $\left\langle E_b\right\rangle $ is the average value of $\overrightarrow(E)$ in areas perpendicular to the interface.
From (2.2) it follows that:
\[((E)_(2x)-E_(1x))a=\left\langle E_b\right\rangle 2b\ (2.3).\]
If $b\to 0$, then we get that:
Expression (2.4) is satisfied with an arbitrary choice of the X axis, which lies at the dielectric interface. If we imagine the tension vector in the form of two components (tangential $E_(\tau )\ $ and normal $E_n$):
\[\overrightarrow(E_1)=\overrightarrow(E_(1n))+\overrightarrow(E_(1\tau )),\overrightarrow(E_2)=\overrightarrow(E_(2n))+\overrightarrow(E_(2\ tau ))\ \left(2.5\right).\]
In this case, from (2.4) we write:
where $E_(\tau i)$ is the projection of the intensity vector onto the unit unit $\tau $ directed along the dielectric interface.
When a charge moves along an arbitrary closed path L, the work done by the electrostatic field forces is zero. Since the final position of the charge is equal to the initial position r 1 =r 2, then (the circle near the integral sign indicates that the integration is carried out along a closed path). Since and , then 
. From here we get . Reducing both sides of the equality by q 0, we obtain or, where E l=Ecosa - projection of vector E onto the direction of elementary displacement. The integral is called circulation of the tension vector. Thus, circulation of the electrostatic field strength vector along any closed loop is zero
. This conclusion is a condition field potentiality.
Potential charge energy.
In a potential field, bodies have potential energy and the work of conservative forces is done due to the loss of potential energy.
Therefore work A 12 can be represented as the difference in potential charge energies q 0 at the initial and final points of the charge field q :
Potential charge energy q 0 located in the charge field q on distance r equal to
Assuming that when the charge is removed to infinity, the potential energy goes to zero, we get: const = 0 .
For namesake charges potential energy of their interaction ( repulsion) positive, For different names charges potential energy from interaction ( attraction) negative.
If the field is created by the system n point charges, then the potential energy of the charge q 0 located in this field is equal to the sum of its potential energies created by each of the charges separately:
Electrostatic field potential.
The ratio does not depend on the test charge q0 and is, energy characteristic of the field, called potential :
Potential ϕ at any point in the electrostatic field is scalar physical quantity, determined by the potential energy of a unit positive charge placed at this point.
1.7 Relationship between tension and potential.
Relationship between potential and electrostatic field strength. Equipotential surfaces.
As previously shown, the work of electrostatic field forces when moving a charge q 0 can be written on the one hand as 
, on the other hand, as a decrease in potential energy, i.e. . Here dr is the projection of elementary displacement d l charge to the direction of the field line, 
- there is a small potential difference between two closely located field points. Let's equate the right-hand sides of the equalities and reduce by q 0 . We get the ratios 
, . From here.
The last relationship represents the connection between the main characteristics of the electrostatic field E and j. Here is the rate of change of potential in the direction of the field line. The minus sign indicates that the vector is directed in the direction of decreasing potential. Because the 
, we can write the projections of the vector onto the coordinate axes: 
. It follows that . The expression in parentheses is called the gradient of the scalar j and is denoted as gradj.
The electrostatic field strength is equal to the potential gradient taken with the opposite sign.
To graphically depict the distribution of the electrostatic field potential, use equipotential surfaces - surfaces, the potential of all points of which is the same. Field potential of a single point charge. Equipotential surfaces in this case are concentric spheres with the center at the point where the charge q is located (Fig. 1.13). An infinite number of equipotential surfaces can be drawn, but it is customary to draw them with a density proportional to the value of E.
1.8 Electrical capacity, flat capacitor.
Electrical capacity.
Let's consider solitary guide - a conductor remote from other bodies and charges. From experience it follows that different conductors, being equally charged, have different potentials.
Physical quantity C, equal to the conductor charge ratio q to its potential ϕ , called electrical capacity this conductor.
The electrical capacity of an isolated conductor is numerically equal to the charge that must be imparted to this conductor in order to change its potential by one.
It depends on the shape and size of the conductor and on the dielectric properties of the environment. The capacitances of geometrically similar conductors are proportional to their linear dimensions.
Example: Consider a solitary ball of radius R located in a homogeneous medium with dielectric constant e. Previously, it was found that the potential of the ball is equal to 
. Then the capacity of the ball 
, i.e. depends only on its radius.
Unit of electrical capacity-farad (F): 1F is the capacitance of such an isolated conductor, the potential of which changes by 1V when a charge of 1C is imparted to it. A sphere with a radius has a capacity of 1F R= 9 ⋅10 6 km. The earth's capacitance is 0.7 mF.
The interaction of stationary charges is realized through an electrostatic field. The electrostatic field is described using the intensity vector ($\overline(E)$), which is defined as the force ($\overline(F)$) acting on a unit positive charge located at the field point under consideration:
\[\overline(E)=\frac(\overline(F))(q)\left(1\right).\]
Electrostatic forces are conservative, which means that their work along a closed path ($L$) is zero:
where $\overline(r)$ is the displacement.
The integral in formula (2) is called the circulation of the electrostatic field strength vector. The circulation of the vector $\overline(E)$ is the work that Coulomb forces can do by moving a positive charge equal to one along the contour.
Considering that $q\ne 0$, we get:
\[\oint\nolimits_L(\overline(E)d\overline(r)=)0\ \left(3\right).\]
The theorem on the circulation of the electrostatic field strength vector says that the circulation of $\overline(E)$ along a closed loop is zero.
In differential form, the circulation theorem is written as:
This type of notation as (4) is convenient to use to check the potentiality of a vector field. The potential field is irrotational.
As a consequence of the circulation theorem $\overline(E)$: the work done when moving a charge from one point in the field to another does not depend on the shape of the trajectory.
From the circulation theorem it follows that the lines of the electrostatic field are not closed; they begin on positive and end on negative charges.
Theorem on the circulation of the magnetic field strength vector
Physical quantity ($\overline(H)$), which is a characteristic of the magnetic field, is equal to:
\[\overline(H)=\frac(\overline(B))((\mu )_0)-(\overline(P))_m(5)\]
called magnetic field strength. $\overline(B)$ - vector of magnetic field induction; $(\mu )_0$ - magnetic constant; $(\overline(P))_m$ is the magnetization vector.
The circulation of the magnetic field strength vector is equal to the algebraic sum of the conduction currents that are covered by the closed loop along which the circulation is considered:
\[\oint\limits_L(\overline(H)d\overline(r)=\sum(I_m)\left(6\right).)\]
If the direction of bypassing the circuit is associated with the direction of the current by the rule of the right screw, then the current in the sum (5) has a plus sign.
The circulation of the intensity vector is generally different from zero, which means that the magnetic field is a vortex field, it is not potential.
The theorem on the circulation of the magnetic field strength vector is proven based on the Biot-Savart-Laplace law and the principle of superposition.
The circulation theorem for the vector $\overline(H)$ plays a role similar to the role of the Gauss theorem for the electric field strength vector. If there is symmetry in the distribution of currents, then using the circulation theorem $\overline(H),$ the magnetic field strength itself is found.
Examples of problems with solutions
Example 1
Exercise. Determine whether the electric field given by the equation is potential: $\overline(E)\left(x,y\right)=A\left(2xy\ \overline(i)+\left(x^2-y^2 \right)\overline(j)\right).$
Solution. From the circulation theorem, which is written in differential form:
it follows that if the field vortex is zero, then the field is potential. Using the rotor definition:
\=\frac(\partial E_y)(\partial x)\overline(k)-\frac(\partial E_x)(\partial y)\overline(k)\left(1.3\right).\]
The partial derivatives of $\overline(E)$ are:
\[\frac(\partial E_y)(\partial x)=A\cdot 2x;;\ \frac(\partial E_x)(\partial y)=A\cdot 2x\ \left(1.4\right).\]
Substituting (1.4) into (1.3), we obtain that
\=0.\]
Answer. The field is potential.
Example 2
Exercise. What is the circulation of the magnetic field strength vector for a closed loop $L$ (Fig. 1), if $I_1=5\ A;;\ I_2=2\ A;;\ I_3=10\ A;;\ I_4=1\ A? $
Solution. The basis for solving the problem is the theorem on the circulation of the magnetic field strength vector:
\[\oint\limits_L(\overline(H)d\overline(r)=\sum(I_m)\left(2.1\right).)\]
Circuit $L$ covers three currents, therefore:
\[\oint\limits_L(\overline(H)d\overline(r)=I_1-I_2+I_3.)\]
Let's calculate the circulation:
\[\oint\limits_L(\overline(H)d\overline(r)=5-2+10=13\ (A.)\]
Answer.$\oint\limits_L(\overline(H)d\overline(r)=13A\ .)$
If in the electrostatic field of a point charge Q from point 1 exactly 2 another point charge moves along an arbitrary trajectory (Fig. 132) Q 0 , then the force applied to the charge does work. Work of force F on elementary displacement d l equal to
Since d/cos=d r, That
Work when moving a charge Q 0 from point 1 exactly 2
(83.1)
does not depend on the trajectory of movement, but is determined only by the positions of the initial 1 and final 2 points. Therefore, the electrostatic field of a point charge is potential, and electrostatic forces - conservative(see § 12).
From formula (83.1) it follows that the work done when moving an electric charge in an external electrostatic field along any closed path L, is equal to zero, i.e.
If we take a single point positive charge as the charge transferred in an electrostatic field, then the elementary work of field forces on the path d l equal to E d l=E l dl, Where E l =E cos - vector projection E to the direction of elementary movement. Then formula (83.2) can be written in the form
(83.3)
Integral 
called circulation of the tension vector. Consequently, the circulation of the electrostatic field strength vector along any closed loop is zero. A force field with property (83.3) is called potential. From the circulation of a vector going to zero E it follows that the electric field strength lines cannot be closed; they begin and end on charges (positive or negative, respectively) or go to infinity.
Formula (83.3) is valid only for the electrostatic field. Subsequently, it will be shown that for the field of moving charges, condition (83.3) is not satisfied (for it, the circulation of the intensity vector is nonzero).
§ 84. Electrostatic field potential
A body located in a potential field of forces (and the electrostatic field is potential) has potential energy, due to which work is done by the field forces (see § 12). As is known (see (12.2)), the work of conservative forces is accomplished due to a decrease in potential energy. Therefore, the work (83.1) of the electrostatic field forces can be represented as the difference in potential energies possessed by a point charge Q 0 at the initial and final points of the charge field Q:
(84.1)
whence it follows that the potential energy of the charge qq in the charge field Q equal to
It, as in mechanics, is determined ambiguously, but to within an arbitrary constant WITH. If we assume that when the charge is removed to infinity ( r) potential energy goes to zero ( U=0), That WITH=0 and potential charge energy Q 0 , charge located in the field Q at a distance r from it, is equal to
(84.2)
For charges of the same name Q 0 Q> 0 and the potential energy of their interaction (repulsion) is positive, for unlike charges Q 0 Q<0 и потенциальная энергия их взаимодействия (притяжения) отрицательна.
If the field is created by the system n point charges Q 1 , Q 2 , ..., Q n, then the work of electrostatic forces performed on the charge Q 0 is equal to the algebraic sum of the work of forces caused by each of the charges separately. Therefore potential energy U charge Q 0 , located in this field is equal to the sum of potential energies U i , each of the charges:
(84.3)
From formulas (84.2) and (84.3) it follows that the relation U/ Q 0 does not depend on Q 0 and is therefore energy characteristics of the electrostatic field, called potential:
Potential at any point in the electrostatic field there is a physical quantity determined by the potential energy of a unit positive charge placed at this point.
From formulas (84.4) and (84.2) it follows that the potential of the field created by a point charge Q, is equal
The work done by the cells of the electrostatic field when moving a charge Q 0 from point 1 exactly 2 (see (84.1), (84.4), (84.5)), can be represented as
i.e., equal to the product of the moved charge and the potential difference at the starting and ending points. Potential difference two points 1 And 2 in an electrostatic field is determined by the work done by field forces when moving a unit positive charge from a point 1 exactly 2 .
Work done by field forces when moving a charge Q 0 from point 1 exactly 2 can also be written in the form
(84.7)
Equating (84.6) and (84.7), we arrive at the expression for the potential difference:
(84.8)
where integration can be performed along any line connecting the start and end points, since the work of the electrostatic field forces does not depend on the trajectory of movement.
If you move the charge Q 0 from an arbitrary point beyond the field, i.e. to infinity, where, by condition, the potential is zero, then the work of the electrostatic field forces, according to (84.6), A = Q 0 , where
Thus, potential- a physical quantity determined by the work of moving a single positive charge when it is removed from a given point in the field to infinity. This work is numerically equal to the work done by external forces (against the forces of the electrostatic field) to move a unit positive charge from infinity to a given point in the field.
From expression (84.4) it follows that the unit of potential is volt(B): 1 V is the potential of a point in the field at which a charge of 1 C has a potential energy of 1 J (1 V = 1 J/C). Taking into account the dimension of the volt, it can be shown that the unit of electrostatic field strength introduced in § 79 is indeed equal to 1 V/m: 1 N/C=1 Nm/(Cm)=1 J/(Cm)=1 V /m.
From formulas (84.3) and (84.4) it follows that if the field is created by several charges, then the field potential of the system of charges is equal to algebraic the sum of the field potentials of all these charges:
Let's take an arbitrary contour (G) and an arbitrary surface S in a non-uniform electrostatic field (Fig. 3.7, a, b).
Then circulation of a vector along an arbitrary contour (Г) is called an integral of the form:
and the flow of the FE vector through an arbitrary surface S is the following expression
The vectors and included in these formulas are defined as follows. In modulus they are equal to the elementary length dl of the contour (G) and the area dS of the elementary area of the surface S. The direction of the vector coincides with the direction of traversing the contour (G), and the vector is directed along the normal vector to the area dS (Fig. 3.7).
In the case of an electrostatic field, the circulation of a vector along an arbitrary closed contour (G) is equal to the ratio of the work Akkrug of the field forces to move a point charge q along this contour to the magnitude of the charge and, in accordance with formula (3.20), will be equal to zero
It is known from theory that if for an arbitrary vector field the circulation of the vector along an arbitrary closed contour (G) is equal to zero, then this field is potential. Hence, the electrostatic field is potential and the electric charges in it have potential energy.
If we take into account that the density of lines determines the magnitude of the vector at a given point in the field, then the flux of the vector will be numerically equal to the number N of lines piercing the surface S.
Figure 3.8 shows examples of calculating the flow through various surfaces S (Figure 3.8, a, b, c, surface S is flat; Figure 3.8, d S is a closed surface). In the latter case, the flux through the closed surface is zero, since the number of lines entering () and leaving () from it is the same, but they are taken with opposite signs ( +>0, -<0).
For a vector we can formulate Gauss's theorem, which determines the flow of a vector through an arbitrary closed surface.
Gauss's theorem in the absence of a dielectric (vacuum) is formulated as follows: the flux of a vector through an arbitrary closed surface is equal to the algebraic sum of the free charges covered by that surface divided by .
This theorem is a consequence of Coulomb's law and the principle of superposition of electrostatic fields.
Let us show the validity of the theorem for the case of a point charge field. Let the closed surface be a sphere of radius R, in the center of which there is a point positive charge q (Fig. 3.9, a).
The obtained result will not change if instead of a sphere we choose an arbitrary closed surface (Fig. 3.9, b), since the vector flux is numerically equal to the number of lines piercing the surface, and the number of such lines in cases a and b is the same.
The same reasoning using the principle of superposition of electrostatic fields can be given in the case of several charges falling inside a closed surface, which confirms Gauss’s theorem.
Gaussian tower for vector in the presence of a dielectric. In this case, in addition to free charges, it is necessary to take into account bound charges that appear on opposite faces of the dielectric when it is polarized in the external electric (for more details, see the section on dielectrics). Therefore, Gauss’s theorem for a vector in the presence of a dielectric will be written as follows:
where the right side of the formula includes the algebraic sum of free and bound charges covered by the surface S.
From formula (3.28) it follows physical meaning of Gauss's theorem for the vector : The sources of the electrostatic field vector are free and bound charges.
In the particular case of a symmetrical arrangement of charges and a dielectric, in the presence of axial or spherical symmetry or in the case of an isotropic homogeneous dielectric, the relative dielectric permittivity of the medium remains a constant value, independent of the point considered inside the dielectric, and therefore the presence of a dielectric can be taken into account in formula (3.28) without only by introducing bound charges , but also the parameter , which is more convenient for practical calculations. So, we can write (see paragraph 3.1.12.6, formula (3.68))
Then the Gauss theorem for the vector in this case will be written as follows
where is the relative dielectric constant of the medium in which the surface S is located.
Note that formula (3.29) is used when solving problems in this section, as well as for most cases encountered in practice.