Along with Coulomb's law, another description of the interaction is possible electric charges.

Long-range and short-range. Coulomb's law, similar to the law universal gravity, interprets the interaction of charges as “action at a distance”, or “long-range action”. Indeed, the Coulomb force depends only on the size of the charges and the distance between them. Coulomb was convinced that the intermediate medium, i.e., the “void” between charges, did not take any part in the interaction.

This point of view was undoubtedly inspired by the impressive successes of Newton's theory of gravity, which was brilliantly confirmed by astronomical observations. However, Newton himself wrote: “It is not clear how inanimate inert matter, without the mediation of something else that is immaterial, could act on another body without mutual contact.” Nevertheless, the concept of long-range action, based on the idea of ​​the instantaneous action of one body on another at a distance without the participation of any intermediate medium, dominated the scientific worldview for a long time.

The idea of ​​a field as a material medium through which any interaction of spatially distant bodies is carried out was introduced into physics in the 30s of the 19th century by the great English naturalist M. Faraday, who believed that “matter is present everywhere, and there is no intermediate space not occupied

by her." Faraday developed a consistent concept electromagnetic field, based on the idea of ​​a finite speed of interaction propagation. A complete theory of the electromagnetic field, expressed in a strict mathematical form, was subsequently developed by another great English physicist, J. Maxwell.

By modern ideas electric charges give the space around them special physical properties- create an electric field. The main property of the field is that a charged particle located in this field is acted upon by a certain force, i.e., the interaction of electric charges is carried out through the fields they create. The field created by stationary charges does not change with time and is called electrostatic. To study a field you need to find it physical characteristics. Two such characteristics are considered - strength and energy.

Tension electric field. To experimentally study the electric field, you need to place a test charge in it. In practice, this will be some kind of charged body, which, firstly, must have small enough dimensions so that one can judge the properties of the field at a certain point in space, and, secondly, its electric charge must be small enough so that one can neglect the influence of this charge on the distribution of charges creating the field under study.

A test charge placed in an electric field is acted upon by a force that depends both on the field and on the test charge itself. This force is greater, the larger the test charge. By measuring the forces acting on different test charges placed at the same point, one can verify that the ratio of the force to the test charge no longer depends on the size of the charge. This means that this relationship characterizes the field itself. The force characteristic of the electric field is the intensity E - a vector quantity equal at each point to the ratio of the force acting on the test charge placed at this point to the charge

In other words, field strength E is measured by the force acting on a unit positive test charge. In general, the field strength is different at different points. A field in which the intensity at all points is the same both in magnitude and direction is called homogeneous.

Knowing the electric field strength, you can find the force acting on any charge placed in this point. In accordance with (1), the expression for this force has the form

How to find the field strength at any point?

The electric field strength created by a point charge can be calculated using Coulomb's law. We will consider a point charge as a source of electric field. This charge acts on a test charge located at a distance from it with a force whose modulus is equal to

Therefore, in accordance with (1), dividing this expression by we obtain the modulus E of the field strength at the point where the test charge is located, i.e. at a distance from the charge

Thus, the field strength of a point charge decreases with distance in inverse proportion to the square of the distance or, as they say, according to the inverse square law. Such a field is called Coulomb. When approaching a point charge creating a field, the field strength of the point charge increases indefinitely: from (4) it follows that when

The coefficient k in formula (4) depends on the choice of system of units. In SGSE k = 1, and in SI . Accordingly, formula (4) is written in one of two forms:

The unit of tension in the SGSE does not have a special name, but in the SI it is called “volt per meter”

Due to the isotropy of space, i.e., the equivalence of all directions, the electric field of a solitary point charge is spherically symmetrical. This circumstance is manifested in formula (4) in the fact that the modulus of the field strength depends only on the distance to the charge creating the field. The intensity vector E has a radial direction: it is directed from the field-creating charge if it is a positive charge (Fig. 6a, a), and towards the field-creating charge if this charge is negative (Fig. 6b).

The expression for the field strength of a point charge can be written in vector form. It is convenient to place the origin of coordinates at the point where the charge creating the field is located. Then the field strength at any point characterized by the radius vector is given by the expression

This can be verified by comparing definition (1) of the field strength vector with formula (2) § 1, or starting from

directly from formula (4) and taking into account the considerations formulated above about the direction of the vector E.

Superposition principle. How to find the strength of the electric field created by an arbitrary distribution of charges?

Experience shows that electric fields satisfy the superposition principle. The field strength created by several charges is equal to vector sum field strengths created by each charge separately:

The principle of superposition actually means that the presence of other electric charges has no effect on the field created by a given charge. This property, when individual sources act independently and their actions simply add up, is inherent in the so-called linear systems, and this very property of physical systems is called linearity. The origin of this name is due to the fact that such systems are described linear equations(equations of the first degree).

Let us emphasize that the validity of the superposition principle for the electric field is not a logical necessity or something taken for granted. This principle is a generalization of experimental facts.

The principle of superposition allows one to calculate the field strength created by any distribution of stationary electric charges. In the case of several point charges, the recipe for calculating the resulting intensity is obvious. Any non-point charge can be mentally broken down into such small parts that each of them can be considered as a point charge. The electric field strength at an arbitrary point is found as

the vector sum of the intensities created by these “point” charges. The corresponding calculations are greatly simplified in cases where there is a certain symmetry in the distribution of the charges creating the field.

Lines of tension. Visual graphic image Electric fields are produced by tension lines or lines of force.

Rice. 7. Field strength lines of positive and negative point charges

These electric field lines are drawn in such a way that at each point the tangent to the line coincides in direction with the intensity vector at this point. In other words, anywhere the tension vector is directed tangentially to power line passing through this point. Lines of force are assigned a direction: they come from positive charges or come from infinity. They either end at negative charges or go to infinity. In the figures, this direction is indicated by arrows on the power line.

A line of force can be drawn through any point in the electric field.

The lines are drawn more densely in places where the field strength is greater, and less often where it is less. Thus, the density of the field lines gives an idea of ​​the intensity modulus.

Rice. 8. Field strength lines of opposite identical charges

In Fig. Figure 7 shows the field lines of solitary positive and negative point charges. From symmetry it is obvious that these are radial straight lines, distributed with equal density in all directions.

More complex look has a pattern of field lines created by two charges of opposite signs. Such a field is obviously

has axial symmetry: the whole picture remains unchanged when rotated through any angle around an axis passing through the charges. When the charge moduli are the same, the pattern of lines is also symmetrical relative to the plane passing perpendicular to the segment connecting them through its middle (Fig. 8). In this case, the lines of force come out of the positive charge and they all end in the negative, although in Fig. 8 it is impossible to show how lines that go far from the charges are closed.

ELECTRICAL DISPLACEMENT

Basic formulas

 Electric field strength

E=F/Q,

Where F- force acting on a point positive charge Q, placed at a given point in the field.

 Force acting on a point charge Q, placed in an electric field,

F=QE.

E electric field:

a) through an arbitrary surface S, placed in a non-uniform field,

Or
,

where  is the angle between the tension vector E and normal n to a surface element; d S- area of ​​the surface element; E n- projection of the tension vector onto the normal;

b) through a flat surface placed in a uniform electric field,

F E =ES cos.

 Tension vector flow E through a closed surface

,

where integration is carried out over the entire surface.

 Ostrogradsky-Gauss theorem. Tension vector flow E through any closed surface enclosing charges Q l , Q 2 , . . ., Q n ,

,

Where - algebraic sum of charges enclosed inside a closed surface; p - number of charges.

 Electric field strength created by a point charge Q at a distance r from charge,

.

The electric field strength created by a metal sphere with a radius R, charge-carrying Q, at a distance r from the center of the sphere:

a) inside the sphere (r<.R)

b) on the surface of the sphere (r=R)

;

c) outside the sphere (r>R)

.

 The principle of superposition (imposition) of electric fields, according to which the intensity E the resulting field created by two (or more) point charges is equal to the vector (geometric) sum of the strengths of the added fields:

E=E 1 +E 2 +...+E n .

In the case of two electric fields with intensities E 1 And E 2 voltage vector module

where  is the angle between the vectors E 1 And E 2 .

 The field strength created by an infinitely long uniformly charged thread (or cylinder) at a distance r from its axis,

, where  is the linear charge density.

Linear charge density is a value equal to the ratio of the charge distributed along the thread to the length of the thread (cylinder):

 The field strength created by an infinite uniformly charged plane is

where  is the surface charge density.

Surface charge density is a value equal to the ratio of the charge distributed over the surface to the area of ​​this surface:

.

 The field strength created by two parallel infinite uniformly and oppositely charged planes, with the same absolute charge surface density (field of a flat capacitor)

.

The above formula is valid for calculating the field strength between the plates of a flat capacitor (in the middle part of it) only if the distance between the plates is much less than the linear dimensions of the capacitor plates.

 Electrical displacement D associated with tension E electric field relation

D= 0 E.

This relationship is valid only for isotropic dielectrics.

 The flux of the electric displacement vector is expressed similarly to the flux of the electric field strength vector:

a) in the case of a uniform field, flow through a flat surface

;

b) in the case of a non-uniform field and an arbitrary surface

,

Where D n - vector projection D to the direction of the normal to a surface element whose area is d S.

 Ostrogradsky-Gauss theorem. Flow of electric displacement vector through any closed surface enclosing charges Q 1 ,Q 2 , ...,Q n ,

,

Where n-the number of charges (with their own sign) contained inside a closed surface.

 Circulation of the electric field strength vector is a value numerically equal to the work of moving a single point positive charge along a closed loop. Circulation is expressed by a closed loop integral
, Where E l - projection of the tension vector E at a given point of the contour onto the direction of the tangent to the contour at the same point.

In the case of an electrostatic field, the circulation of the intensity vector is zero:

.

Examples of problem solving

P
example 1. The electric field is created by two point charges: Q 1 =30 nC and Q 2 = –10 nC. Distance d between charges is 20 cm. Determine the electric field strength at a point located at a distance r 1 =15 cm from the first and at a distance r 2 =10 cm from the second charges.

Solution. According to the principle of superposition of electric fields, each charge creates a field regardless of the presence of other charges in space. Therefore tension E electric field at the desired point can be found as the vector sum of the strengths E 1 And E 2 fields created by each charge separately: E=E 1 +E 2 .

The electric field strengths created in vacuum by the first and second charges are respectively equal to

(1)

Vector E 1 (Fig. 14.1) directed along the field line from the charge Q 1 , since the charge Q 1 >0; vector E 2 also directed along the line of force, but towards the charge Q 2 , because Q 2 <0.

Vector module E we find using the cosine theorem:

where angle  can be found from a triangle with sides r 1 , r 2 And d:

.

In this case, in order to avoid cumbersome entries, we calculate the value of cos separately. Using this formula we find

Substituting expressions E 1 And E 2 and using formulas (1) into equality (2) and taking out the common factor 1/(4 0 ) for the sign of the root, we get

.

Substituting the values ​​of  ,  0 , Q 1 , Q 2 , r 1 -, r 2 and  into the last formula and after performing calculations, we find

Example 2. The electric field is created by two parallel infinite charged planes with surface charge densities  1 =0.4 µC/m 2 and  2 =0.1 µC/m2. Determine the strength of the electric field created by these charged planes.

R
decision. According to the principle of superposition, the fields produced by each individual charged plane are superimposed on each other, with each charged plane producing an electric field regardless of the presence of the other charged plane (Figure 14.2).

The strengths of uniform electric fields created by the first and second planes are respectively equal to:

;
.

The planes divide the entire space into three regions: I, II and III. As can be seen from the figure, in the first and third regions the electric field lines of both fields are directed in the same direction and, therefore, the strengths of the total fields E (I) And E(III) in the first and third areas are equal to each other and equal to the sum of the field strengths created by the first and second planes: E (I) = E(III) = E 1 +E 2 , or

E (I) = E (III) =
.

In the second region (between the planes), the electric field lines are directed in opposite directions and, therefore, the field strength E (II) equal to the difference in field strengths created by the first and second planes: E (II) =|E 1 -E 2 | , or

.

Substituting the data and performing calculations, we get

E (I) =E (III) =28,3 kV/m=17 kV/m.

The distribution of field lines of the total field is shown in Fig. 14.3.

Example 3. There is a charge on the plates of a flat air capacitor Q=10 nC. Square S each plate of the capacitor is 100 cm 2 Determine the force F, with which the plates are attracted. The field between the plates is considered uniform.

Solution. Charge Q one plate is in the field created by the charge of the other plate of the capacitor. Consequently, a force acts on the first charge (Fig. 14.4)

F=E 1 Q,(1)

Where E 1 - the field strength created by the charge of one plate. But
where  is the surface charge density of the plate.

Formula (1) taking into account the expression for E 1 will take the form

F=Q 2 /(2 0 S).

Substituting the values ​​of the quantities Q,  0 And S into this formula and performing calculations, we get

F=565 µN.

Example 4. An electric field is created by an infinite plane charged with surface density  = 400 nC/m 2 , and an endless straight thread charged with a linear density =100 nC/m. At a distance r=10 cm from the thread there is a point charge Q=10 nC. Determine the force acting on the charge and its direction if the charge and the thread lie in the same plane parallel to the charged plane.

Solution. The force acting on a charge placed in a field is

F=EQ, (1)

Where E - Q.

Let's determine the tension E field created, according to the conditions of the problem, by an infinite charged plane and an infinite charged thread. The field created by an infinite charged plane is uniform, and its strength at any point is

. (2)

The field created by an infinite charged line is non-uniform. Its intensity depends on the distance and is determined by the formula

. (3)

According to the principle of superposition of electric fields, the field strength at the point where the charge is located Q, is equal to the vector sum of the intensities E 1 And E 2 (Fig. 14.5): E=E 1 +E 2 . Since vectors E 1 And E 2 mutually perpendicular, then

.

Substituting expressions E 1 And E 2 using formulas (2) and (3) into this equality, we obtain

,

or
.

Now let's find the strength F, acting on the charge, substituting the expression E into formula (1):

. (4)

Substituting the values ​​of the quantities Q,  0 , , ,  and r into formula (4) and making calculations, we find

F=289 µN.

Direction of force F, acting on a positive charge Q, coincides with the direction of the tension vector E fields. The direction of the vector E is given by the angle  to the charged plane. From Fig. 14.5 it follows that

, where
.

Substituting the values ​​of , r,  and  into this expression and calculating, we get

Example 5. Point charge Q=25 nC is in the zero created by a straight infinite cylinder of radius R= 1 cm, uniformly charged with surface density =2 µC/m 2. Determine the force acting on a charge placed from the cylinder axis at a distance r=10 cm.

Solution. Force acting on the charge Q, located in the field,

F=QE,(1)

Where E - field strength at the point where the charge is located Q.

As is known, the field strength of an infinitely long uniformly charged cylinder

E=/(2 0 r), (2)

where  is the linear charge density.

Let us express the linear density  through the surface density . To do this, select a cylinder element with length l and express the charge on it Q 1 in two ways:

Q 1 =  S= 2  Rl and Q 1 = l.

Equating the right-hand sides of these equalities, we obtain  l=2 Rl . After reduction by l let's find =2 R . Taking this into account, formula (2) will take the form E=R /( 0 r). Substituting this expression E into formula (1), we find the required force:

F=Q R/( 0 r).(3)

Because R And r are included in the formula in the form of a ratio, then they can be expressed in any, but only identical units.

Having performed calculations using formula (3), we find

F=2510 -9 210 -6 10 -2 /(8.8510 -12 1010 -2)H==56510 -6 H=565 µH.

Direction of force F coincides with the direction of the tension vector E, and the latter, due to symmetry (the cylinder is infinitely long), is directed perpendicular to the cylinder.

Example 6. The electric field is created by a thin infinitely long thread, uniformly charged with a linear density =30 nC/m. At a distance A= 20 cm from the thread there is a flat round area with a radius r=1 cm. Determine the flow of the tension vector through this area if its plane makes an angle  = 30° with the tension line passing through the middle of the area.

Solution. The field created infinitely uniformly by a charged thread is inhomogeneous. The flux of the tension vector in this case is expressed by the integral

, (1)

Where E n - vector projection E to normal n to the surface of the site dS. Integration is performed over the entire surface of the site, which is penetrated by tension lines.

P
projection E n the tension vector is equal, as can be seen from Fig. 14.6,

E n =E cos,

where  is the angle between the direction of the vector and the normal n. Taking this into account, formula (1) will take the form

.

Since the surface dimensions of the pad are small compared to the distance to the thread (r<E very little. changes in magnitude and direction within the site, which allows you to replace the values ​​under the integral sign E and cos by their average values<E> and and move them beyond the integral sign:

Performing integration and substitution<E> and their approximate values E A and cos  A , calculated for the midpoint of the site, we obtain

F E =E A cos  A S= r 2 E A cos A . (2)

Tension E A calculated by the formula E A=/(2 0 a). From

rice. 14.6 follows cos  A=cos(/2 -  )=sin.

Given the expression E A and cos  A equality (2.) will take the form

.

Substituting the data into the last formula and performing calculations, we find

F E=424 mV.m.

Example 7 . Two concentric conducting spheres with radii R 1 =6 cm and R 2 = 10 cm carry charges accordingly Q 1 =l nC and Q 2 = –0.5 nC. Find tension E fields at points spaced from the center of the spheres at distances r 1 =5 cm, r 2 =9 cm r 3 =15cm. Build a graph E(r).

R
decision. Note that the points at which it is necessary to find the electric field strength lie in three regions (Fig. 14.7): region I ( r<R 1 ), region II ( R 1 <r 2 <R 2 ), region III ( r 3 >R 2 ).

1. To determine the tension E 1 in region I we draw a spherical surface S 1 radius r 1 and use the Ostrogradsky-Gauss theorem. Since there are no charges inside region I, then according to the indicated theorem we obtain the equality

, (1)

Where E n- normal component of the electric field strength.

For symmetry reasons, the normal component E n must be equal to the tension itself and constant for all points of the sphere, i.e. En=E 1 = const. Therefore, it can be taken out of the integral sign. Equality (1) will take the form

.

Since the area of ​​the sphere is not zero, then

E 1 =0,

i.e. the field strength at all points satisfying the condition r 1 <.R 1 , will be equal to zero.

2. In region II we draw a spherical surface with a radius r 2 . Since inside this surface there is a charge Q 1 , then for it, according to the Ostrogradsky-Gauss theorem, we can write the equality

. (2)

Because E n =E 2 =const, then from the symmetry conditions it follows

, or ES 2 =Q 1 / 0 ,

E 2 =Q 1 /( 0 S 2 ).

Substituting here the expression for the area of ​​a sphere, we get

E 2 =Q/(4
). (3)

3. In region III we draw a spherical surface with a radius r 3 . This surface covers the total charge Q 1 +Q 2 . Consequently, for it the equation written on the basis of the Ostrogradsky-Gauss theorem will have the form

.

From here, using the provisions applied in the first two cases, we find

Let us make sure that the right-hand sides of equalities (3) and (4) give the unit of electric field strength;

Let us express all quantities in SI units ( Q 1 =10 -9 C, Q 2 = –0.510 -9 C, r 1 =0.09 m, r 2 =15m , l/(4 0 )=910 9 m/F) and perform the calculations:

4. Let's build a graph E(r).IN region I ( r 1 1 ) tension E=0. In area II (R 1 r<.R 2 ) tension E 2 (r) varies according to the law l/r 2 . At the point r=R 1 tension E 2 (R 1 )=Q 1 /(4 0 R )=2500 V/m. At a point r=R 1 (r strives for R 1 left) E 2 (R 2 )=Q 1 /(4 0 R )=900V/m. In area III ( r>R 2 )E 3 (r) changes according to the law 1/ r 2 , and at the point r=R 2 (r strives for R 2 right) E 3 (R 2 ) =(Q 1 –|Q 2 |)/(4 0 R )=450 V/m. So the function E(r) at points r=R 1 And r=R 2 suffers a break. Dependency graph E(r) shown in Fig. 14.8.

Tasks

Field strength of point charges

14.1. Determine tension E electric field created by a point charge Q=10 nC at a distance r=10 cm from it. Dielectric - oil.

14.2. Distance d between two point charges Q 1 =+8 nC and Q 2 = –5.3 nC equals 40 cm. Calculate the tension E fields at a point lying in the middle between the charges. What is the voltage if the second charge is positive?

14.3. Q 1 =10 nC and Q 2 = –20 nC located at a distance d=20 cm from each other. Determine tension E fields at a point distant from the first charge by r 1 =30 cm and from the second to r 2 =50 cm.

14.4. Distance d between two point positive charges Q 1 =9Q And Q 2 =Q is equal to 8 cm. At what distance r from the first charge is the point at which the tension E is the field of charges equal to zero? Where would this point be if the second charge were negative?

14.5. Two point charges Q 1 =2Q And Q 2 = –Q are at a distance d from each other. Find the position of the point on the line passing through these charges, the tension E fields in which it is equal to zero,

14.6. Electric field created by two point charges Q 1 =40 nC and Q 2 = –10 nC located at a distance d=10 cm apart. Determine tension E fields at a point distant from the first charge by r 1 =12 cm and from the second to r 2 =6 cm.

Field strength of a charge distributed over a ring and sphere

14.7. Thin ring with radius R=8 cm carries a charge uniformly distributed with a linear density =10 nC/m. What is the tension E electric field at a point equidistant from all points of the ring at a distance r=10 cm?

14.8. The hemisphere carries a charge uniformly distributed with a surface density  = 1.nC/m 2. Find tension E electric field at the geometric center of the hemisphere.

14.9. On a metal sphere with radius R=10 cm is the charge Q=l nCl. Determine tension E electric field at the following points: 1) at a distance r 1 =8 cm from the center of the sphere; 2) on its surface; 3) at a distance r 2 =15 cm from the center of the sphere. Build a dependency graph E from r.

14.10. Two concentric metal charged spheres with radii R 1 =6cm and R 2 =10 cm carry charges accordingly Q 1 =1 nC and Q 2 = – 0.5 nC. Find tension E fields in dots. distances from the center of the spheres r 1 =5 cm, r 2 =9 cm, r 3 =15 cm. Build a dependence graph E(r).

Charged line field strength

14.11. A very long, thin, straight wire carries a charge evenly distributed along its entire length. Calculate the linear charge density  if the voltage E fields in the distance A=0.5 m from the wire opposite its middle is equal to 200 V/m.

14.12. Distance d between two long thin wires located parallel to each other is 16 cm. The wires are uniformly charged with opposite charges with linear density ||=^150. µC/m. What is the tension E fields at a point distant by r=10 cm from both the first and second wire?

14.13. Straight metal rod with diameter d=5 cm long l=4 m carries a charge uniformly distributed over its surface Q=500 nC. Determine tension E fields at a point located opposite the middle of the rod at a distance A=1 cm from its surface.

14.14. An infinitely long thin-walled metal tube with a radius R= 2 cm carries a charge uniformly distributed over the surface ( = 1 nC/m 2). Determine tension E fields at points spaced from the tube axis at distances r 1 =l cm, r 2 =3 cm. Build a dependence graph E(r).

Objective of the lesson: give the concept of electric field strength and its definition at any point in the field.

Lesson objectives:

• formation of the concept of electric field strength; give the concept of tension lines and a graphical representation of the electric field;
• teach students to apply the formula E=kq/r 2 in solving simple problems of calculating tension.

An electric field is a special form of matter, the existence of which can only be judged by its action. It has been experimentally proven that there are two types of charges around which there are electric fields characterized by lines of force.

When depicting the field graphically, it should be remembered that the electric field strength lines:

1. do not intersect with each other anywhere;
2. have a beginning on a positive charge (or at infinity) and an end on a negative charge (or at infinity), i.e. they are open lines;
3. between charges are not interrupted anywhere.

Fig.1

Positive charge lines:

Fig.2

Negative charge lines:

Fig.3

Field lines of interacting charges of the same name:

Fig.4

Field lines of unlike interacting charges:

Fig.5

The strength characteristic of the electric field is intensity, which is denoted by the letter E and has units of measurement or. Tension is a vector quantity, as it is determined by the ratio of the Coulomb force to the value of a unit positive charge

As a result of transforming the formula of Coulomb's law and the intensity formula, we have the dependence of the field strength on the distance at which it is determined relative to a given charge

Where: k– proportionality coefficient, the value of which depends on the choice of units of electric charge.

In the SI system N m 2 / Cl 2,

where ε 0 is the electrical constant equal to 8.85·10 -12 C 2 /N m 2 ;

q – electric charge (C);

r is the distance from the charge to the point at which the voltage is determined.

The direction of the tension vector coincides with the direction of the Coulomb force.

An electric field whose strength is the same at all points in space is called uniform. In a limited region of space, the electric field can be considered approximately uniform if the field strength within this region varies slightly.

The total field strength of several interacting charges will be equal to the geometric sum of the strength vectors, which is the principle of field superposition:

Let's consider several cases of determining tension.

1. Let two opposite charges interact. Let's place a point positive charge between them, then at this point there will be two voltage vectors directed in the same direction:

According to the principle of field superposition, the total field strength at a given point is equal to the geometric sum of the strength vectors E 31 and E 32.

The tension at a given point is determined by the formula:

E = kq 1 /x 2 + kq 2 /(r – x) 2

where: r – distance between the first and second charge;

x is the distance between the first and point charge.

Fig.6

2. Consider the case when it is necessary to find the voltage at a point distant at a distance a from the second charge. If we take into account that the field of the first charge is greater than the field of the second charge, then the intensity at a given point of the field is equal to the geometric difference in intensity E 31 and E 32.

The formula for tension at a given point is:

E = kq1/(r + a) 2 – kq 2 /a 2

Where: r – distance between interacting charges;

a is the distance between the second and point charge.

Fig.7

3. Let's consider an example when it is necessary to determine the field strength at a certain distance from both the first and second charge, in this case at a distance r from the first and at a distance b from the second charge. Since like charges repel, and unlike charges attract, we have two tension vectors emanating from one point, then to add them we can use the method; the opposite angle of the parallelogram will be the total tension vector. We find the algebraic sum of vectors from the Pythagorean theorem:

E = (E 31 2 + E 32 2) 1/2

Hence:

E = ((kq 1 /r 2) 2 + (kq 2 /b 2) 2) 1/2

Fig.8

Based on this work, it follows that the intensity at any point in the field can be determined by knowing the magnitude of the interacting charges, the distance from each charge to a given point and the electrical constant.

4. Reinforcing the topic.

Test work.

Option #1.

1. Continue the phrase: “electrostatics is...

2. Continue the phrase: an electric field is….

3. How are the field lines of intensity of this charge directed?

4. Determine the signs of the charges:

Homework tasks:

1. Two charges q 1 = +3·10 -7 C and q 2 = −2·10 -7 C are in a vacuum at a distance of 0.2 m from each other. Determine the field strength at point C, located on the line connecting the charges, at a distance of 0.05 m to the right of the charge q 2.

2. At a certain point in the field, a charge of 5·10 -9 C is acted upon by a force of 3·10 -4 N. Find the field strength at this point and determine the magnitude of the charge creating the field if the point is 0.1 m away from it.

Coulomb's Law:

where F is the force of interaction between two point charges q 1 and q 2; r – distance between charges;  - dielectric constant of the medium;  0 - electrical constant

.

Law of conservation of charge:

,

Where – algebraic sum of charges included in an isolated system; n – number of charges.

Electrostatic field strength and potential:

;
, or
,

Where – force acting on a point positive charge q 0 placed at a given point in the field; P – potential charge energy; A ∞ is the work spent on moving the charge q 0 from a given point in the field to infinity.

Tension vector flow electric field:

a) through an arbitrary surface S placed in a non-uniform field:

, or
,

where  is the angle between the tension vector and normal to the surface element; dS – area of ​​the surface element; E n – projection of the tension vector onto the normal;

b) through a flat surface placed in a uniform electric field:

.

Tension vector flow through a closed surface -

(integration is carried out over the entire surface).

Ostrogradsky-Gauss theorem. The flow of the intensity vector through any closed surface covering charges q1, q2, …, qn, –

,

Where – algebraic sum of charges contained inside a closed surface; n – number of charges.

The strength of the electrostatic field created by a point charge q at a distance r from the charge is –

.

The electric field strength created by a sphere having a radius R and carrying a charge q at a distance r from the center of the sphere is:

inside the sphere (r R) E=0;

on the surface of the sphere (r=R)
;

outside the sphere (r  R)
.

The principle of superposition (imposition) of electrostatic fields, according to which the intensity the resulting field created by two (or more) point charges is equal to the vector (geometric) sum of the strengths of the added fields, expressed by the formula

In the case of two electric fields with intensities And the absolute value of the tension vector is

where  is the angle between the vectors And .

The field strength created by an infinitely long and uniformly charged thread (or cylinder) at a distance r from its axis is –

,

where  is the linear charge density.

The linear charge density is a value equal to its ratio to the length of the thread (cylinder):

.

The field strength created by an infinite uniformly charged plane is –

,

where  is the surface charge density.

Surface charge density is a value equal to the ratio of the charge distributed over the surface to its area:

.

The field strength created by two infinite and parallel planes, charged uniformly and differently, with the same absolute value of the surface charge density  (field of a flat capacitor) –

.

The above formula is valid when calculating the field strength between the plates of a flat capacitor (in its middle part) only if the distance between the plates is much less than the linear dimensions of the capacitor plates.

Electrical bias associated with tension electric field relation

,

which is valid only for isotropic dielectrics.

The electric field potential is a quantity equal to the ratio of potential energy and a point positive charge placed at a given point in the field:

.

In other words, the electric field potential is a value equal to the ratio of the work done by the field forces to move a point positive charge from a given point in the field to infinity to the magnitude of this charge:

.

The electric field potential at infinity is conventionally assumed to be zero.

Electric field potential created by a point charge q on

distance r from the charge, –

.

The electric field potential created by a metal sphere having a radius R and carrying a charge q at a distance r from the center of the sphere is:

inside the sphere (r  R)
;

on the surface of the sphere (r = R)
;

outside the sphere (r  R)
.

In all formulas given for the potential of a charged sphere,  is the dielectric constant of a homogeneous infinite dielectric surrounding the sphere.

The potential of the electric field generated by a system of n point charges at a given point in accordance with the principle of superposition of electric fields is equal to the algebraic sum of potentials
, created by individual point charges
:

.

Energy W of interaction of a system of point charges
is determined by the work that this system can do when they move relative to each other to infinity, and is expressed by the formula

,

Where - field potential created by all (n-1) charges (except for the i-th) at the point where the charge is located .

The potential is related to the electric field strength by the relation

.

In the case of an electric field with spherical symmetry, this relationship is expressed by the formula

,

or in scalar form

.

In the case of a uniform field, i.e. field, the strength of which at each point is the same both in absolute value and in direction -

,

where  1 and  2 are the potentials of points of two equipotential surfaces; d is the distance between these surfaces along the electric field line.

The work done by the electric field when moving a point charge q from one point of the field, having a potential  1, to another, having a potential  2, is equal to

, or
,

where E – vector projection on the direction of movement;
- movement.

In the case of a homogeneous field, the last formula takes the form

,

Where – displacement; - angle between vector directions and movement .

A dipole is a system of two point (equal in absolute value and opposite in sign) charges located at some distance from each other.

Electric torque dipole is a vector directed from a negative charge to a positive charge, equal to the product of the charge to vector , drawn from a negative charge to a positive one, and called the dipole arm, i.e.

.

A dipole is called a point dipole if its arm much less than the distance from the center of the dipole to the point at which we are interested in the action of the dipole (  r), see fig. 1.

Field strength of a point dipole:

,

where p is the electric moment of the dipole; r is the absolute value of the radius vector drawn from the center of the dipole to the point where the field strength is of interest to us;  - angle between the radius vector and shoulder dipole.

Field strength of a point dipole at a point lying on the dipole axis

(=0), found by the formula

;

at a point lying perpendicular to the dipole arm, reconstructed from its middle
, – according to the formula

.

The field potential of a point dipole at a point lying on the dipole axis (=0) is

,

and at a point lying perpendicular to the dipole arm, reconstructed from its middle
, –

The tension and potential of a non-point dipole are determined in the same way as for a system of charges.

The mechanical moment acting on a dipole with an electric moment p, placed in a uniform electric field with intensity E, is

, or
,

where  is the angle between the directions of the vectors And .

Electrical capacity of an isolated conductor or capacitor –

,

where q is the charge imparted to the conductor;  - the change in potential caused by this charge.

The electrical capacity of a solitary conducting sphere of radius R located in an infinite medium with dielectric constant , –

.

If the sphere is hollow and filled with a dielectric, then its electrical capacity does not change.

Electrical capacity of a flat capacitor:

,

where S is the area of ​​each capacitor plate; d – distance between plates;  is the dielectric constant of the dielectric filling the space between the plates.

The electrical capacity of a flat capacitor filled with n layers of dielectric with thickness d i and dielectric constant  i each (layered capacitor) is

.

The electrical capacity of a spherical capacitor (two concentric spheres of radius R 1 and R 2, the space between which is filled with a dielectric with dielectric constant ) is found as follows:

.

The electrical capacity of series-connected capacitors is:

in general -

,

where n is the number of capacitors;

in the case of two capacitors -

;

.

The electrical capacity of parallel-connected capacitors is determined as follows:

in general -

С=С 1 +С 2 +…+С n;

in the case of two capacitors -

C=C 1 +C 2;

in the case of n identical capacitors with electrical capacity C 1 each –

The energy of a charged conductor is expressed in terms of charge q, potential  and electrical capacity C of the conductor as follows:

.

Energy of a charged capacitor –

,

where q is the charge of the capacitor; C – electrical capacity of the capacitor; U is the potential difference across its plates.

Objective of the lesson: give the concept of electric field strength and its definition at any point in the field.

Lesson objectives:

• formation of the concept of electric field strength; give the concept of tension lines and a graphical representation of the electric field;
• teach students to apply the formula E=kq/r 2 in solving simple problems of calculating tension.

An electric field is a special form of matter, the existence of which can only be judged by its action. It has been experimentally proven that there are two types of charges around which there are electric fields characterized by lines of force.

When depicting the field graphically, it should be remembered that the electric field strength lines:

1. do not intersect with each other anywhere;
2. have a beginning on a positive charge (or at infinity) and an end on a negative charge (or at infinity), i.e. they are open lines;
3. between charges are not interrupted anywhere.

Fig.1

Positive charge lines:

Fig.2

Negative charge lines:

Fig.3

Field lines of interacting charges of the same name:

Fig.4

Field lines of unlike interacting charges:

Fig.5

The strength characteristic of the electric field is intensity, which is denoted by the letter E and has units of measurement or. Tension is a vector quantity, as it is determined by the ratio of the Coulomb force to the value of a unit positive charge

As a result of transforming the formula of Coulomb's law and the intensity formula, we have the dependence of the field strength on the distance at which it is determined relative to a given charge

Where: k– proportionality coefficient, the value of which depends on the choice of units of electric charge.

In the SI system N m 2 / Cl 2,

where ε 0 is the electrical constant equal to 8.85·10 -12 C 2 /N m 2 ;

q – electric charge (C);

r is the distance from the charge to the point at which the voltage is determined.

The direction of the tension vector coincides with the direction of the Coulomb force.

An electric field whose strength is the same at all points in space is called uniform. In a limited region of space, the electric field can be considered approximately uniform if the field strength within this region varies slightly.

The total field strength of several interacting charges will be equal to the geometric sum of the strength vectors, which is the principle of field superposition:

Let's consider several cases of determining tension.

1. Let two opposite charges interact. Let's place a point positive charge between them, then at this point there will be two voltage vectors directed in the same direction:

According to the principle of field superposition, the total field strength at a given point is equal to the geometric sum of the strength vectors E 31 and E 32.

The tension at a given point is determined by the formula:

E = kq 1 /x 2 + kq 2 /(r – x) 2

where: r – distance between the first and second charge;

x is the distance between the first and point charge.

Fig.6

2. Consider the case when it is necessary to find the voltage at a point distant at a distance a from the second charge. If we take into account that the field of the first charge is greater than the field of the second charge, then the intensity at a given point of the field is equal to the geometric difference in intensity E 31 and E 32.

The formula for tension at a given point is:

E = kq1/(r + a) 2 – kq 2 /a 2

Where: r – distance between interacting charges;

a is the distance between the second and point charge.

Fig.7

3. Let's consider an example when it is necessary to determine the field strength at a certain distance from both the first and second charge, in this case at a distance r from the first and at a distance b from the second charge. Since like charges repel, and unlike charges attract, we have two tension vectors emanating from one point, then to add them we can use the method; the opposite angle of the parallelogram will be the total tension vector. We find the algebraic sum of vectors from the Pythagorean theorem:

E = (E 31 2 + E 32 2) 1/2

Hence:

E = ((kq 1 /r 2) 2 + (kq 2 /b 2) 2) 1/2

Fig.8

Based on this work, it follows that the intensity at any point in the field can be determined by knowing the magnitude of the interacting charges, the distance from each charge to a given point and the electrical constant.

4. Reinforcing the topic.

Test work.

Option #1.

1. Continue the phrase: “electrostatics is...

2. Continue the phrase: an electric field is….

3. How are the field lines of intensity of this charge directed?

4. Determine the signs of the charges:

Homework tasks:

1. Two charges q 1 = +3·10 -7 C and q 2 = −2·10 -7 C are in a vacuum at a distance of 0.2 m from each other. Determine the field strength at point C, located on the line connecting the charges, at a distance of 0.05 m to the right of the charge q 2.

2. At a certain point in the field, a charge of 5·10 -9 C is acted upon by a force of 3·10 -4 N. Find the field strength at this point and determine the magnitude of the charge creating the field if the point is 0.1 m away from it.