The interaction between fixed charges is carried out through an electric field called electrostatic.

The electrostatic field is considered completely specified if its voltage is known anywhere.

Tension electrostatic field - VFV, which is a power characteristic electric field and numerically equal to force acting on a single, positive spot charge, placed at this field, and aimed just as the force attached to the positive point charge:

Units of measurement of tension [E] \u003d B / M \u003d N / CL.

, (2.2)

where Q. - a charge that creates the field,

r. - Distance from charge to a given point.

If the field is specified by the charge system, then it is fair for it superposition principle (overlay) electrostatic fields: the tension of the resulting field created by the charge system is equal to the geometric sum of the fields of fields created at this point by each of the charges separately:

. (2.3)

From formula (2.1) follows:

,

E. the disease is brought in the field of tension charge q. > 0, then the direction of force acting on it and tensions coincide, and if q. < 0, then power directed to the side opposite (Fig. 9).

Graphically electrostatic field are depicted using power lines.

Power lines of tension - Lines tangents to which at each point coincide with the direction of the vector . Voltage lines never intersect.

G.

the elimination of the lines is proportional to the voltage vector module, the direction of the lines is taken from positive charge to negative (Fig. 10, a).

The simplest electrostatic field is about dnorodno field - field, anywhere whose vector the same and in the module and in the direction (Fig. 10, b). Uniform is a field of a uniformly charged plane, two planes (Fig. 10, B).

1

. Determine the direction of the intensity of the field of two point charges at a point of A, equidistant from them ( | q. 1 | = |q. 2 |).

2

. In which drawings the tension of the resulting field at point A is directed vertically up, if the distance between charges is equally ( | q 1. | = | q 2. | )?

3

. In which drawings charged dusting mass m. maybe being equilibrium?

4. The field was created by two charged infinitely long planes (| σ 1 | = |σ 2 |). In fact, the tension of the electrostatic field at the point is zero? Perform constructions and explain.

5



. The field was created by two charged endlessly long hollow cylinders. Determine the directions of the tension of the resulting field at points A, B, C.

6. The field is created evenly charged concentric and intersecting spheres. Determine the direction of tension of the result field at points A, B, C.

2.2. Stream vector stream. Ostrogradsky Theorem - Gauss

The line of tension of the electrostatic field is carried out so that their density through a single perpendicular platform is proportional to the vector module .

Then for the elementary platform

through which tension lines are used, you can enter such a characteristic as the strength of the voltage vector of the electrostatic field

- SFV, which characterizes the intensity of the electrostatic field and numerically equal to the scalar product of vectors and

:

where α - angle between positive normal to the site and tension vector (Fig. 11).

For an arbitrary closed surface S. Stroy vector stream through this surface

Flow measurement units [F] \u003d B ∙ m.

Depending on the corner α , the flow can be:

maximum ( F. = max), if a α = 0;

positive ( F. \u003e 0) if 0< α < 90º;

equal to zero ( F. \u003d 0) if α \u003d 90º;

negative ( F. < 0), если 90º < α < 180º.

P rights to count the stream of the vector coming out of the surface, positive, and incoming negative (Fig. 12, a). If the closed surface does not cover the charge, then the flow through it is equal to 0, since the number of voltage lines included in the surface is equal to the number of lines overlooking it . (Fig. 12, b).

Ostrogradsky theorem - Gauss determines F. E through any closed surface and is used to calculate the intensity of the electrostatic field in the case of a large number of charges with symmetry.

Ostrogradsky Theorem - Gauss : The strength of the tension of the electrostatic field through an arbitrary closed surface is equal to the ratio of the algebraic amount of charges covered by this surface, to ε ε 0 :

. (2.4)

If the charged body is in vacuo or in the air, the dielectric permeability of which ε \u003d 1, then in further conclusions we omit it.

The method of calculating fields using the Ostrogradsky - Gauss theorem is provided in section 2.2.2.

The tasks of this paragraph are devoted to finding the tension of the electrostatic field, and the calculated methods used depend on how the charges that create the field are distributed.

The main types of tasks of this section:

the field is formed by one or more point charges (Section 2.2.1);

the field is created charged: infinitely long cylinder (thread), infinite plane, sphere, ball (Section 2.2.2);

the field was created by a charged body of a simple form, which is not infinitely cylinder (thread), an infinite plane, sphere, a ball (Section 2.2.3).

Vologda State Technical University

Department of Physics

Electrostatic and constant current

Part 3.

Compiled by: L.A.Kuzina, Cand. Fiz.-Mat.Nuk, Associate Professor

Vologda

2011

1. The law of Coulomb. Field tension. Theorem Gaussa

The law of the coulon.

- determination of the field strength;

,

- principle of superposition;

- dielectric permeability of the dielectric;

- point of point charge field;

,

,

- bulk, surface, linear charge density;

- plane field strength;

- capacitor field strength;

- thread field thread (cylinder with r.>R., R. - cylinder radius);

- Vector electric displacement;

,

- stream of tension vector;

,

- the flow of the electrical displacement vector;

,

- Theorem Gauss.

^ Examples of solving problems

Task 1.

Determine the intensity of the field created by the charge, evenly distributed by a thin straight rod with a linear density of 200 nkl / m, at a point lying on a perpendicular restored in the middle of the rod, at a distance of 40 cm from its middle. Rod length 60 cm.

R messing

We break the rod on infinitely small elements dL \u003d DY.; y. - coordinate of this element. Charge element dQ \u003d.τ dY. You can read point. Field Tension Created dQ. At a point a at a distance r. From charge, equal:

, (1)

where

; (2)

α - the angle between the perpendicular to the rod and the radius vector r. The element of the rod conducted from point A. The direction of the tension vector, see Fig.1. As

T.

T.

. (3)

We find projections dE. on coordinate axes:

;

, (4)

Finally, the projections of full tensions on the axis are calculated by integration:

;

, (5)

Moreover, integration is performed along the entire length of the rod. It uses the principle of superposition in the projections on the axis. Complete tension is calculated by the Pythagora theorem:

. (6)

Taking into account (1) - (4) we get from (5):

Permanent quantity

We carry out the integral sign and simple integration limits: the angle α varies from (-α 0) to α 0, where

. Next, a primitive function from

- this is

and from -

. Then

,

Finally get for tensions:

,

.

Answer: E.\u003d 5.4. 10 3 V / m.

Task 2.

On two coaxial infinite cylinders with radii 5 cm and 10 cm, charges with linear densities of charge τ 1 \u003d 100 NK / M and τ 2 \u003d 50 NL / M, respectively, are uniformly distributed. The space between the cylinders is filled with paraffin with dielectric permeability 2. Find the electric field strength at points removed from the axis of cylinders at a distance of 3 cm, 9 cm, 15 cm.

The symmetry of the problem allows you to use the Gauss theorem: the stream of the intensity of the electrostatic field through any closed surface is equal to the amount of free charges covered by this surface divided by (εε 0):

. (1)

Z. oh α - angle between the vector And the normal to the surface at this point. Take the Gaussian surface in the form of a cylinder, coaxial data, the height of which is equal h., but radius r.. Vector The electrostatic field strength can only be directed perpendicular to the side surface of the cylinder, parallel to the bases, (see Fig. 2), then on the left side (1) it is necessary to take into account only the contribution through the side surface of the cylinder (for the base α \u003d 90 0, cosα \u003d 0) , and for the side surface α \u003d 0, cosα \u003d 1. In addition, due to symmetry, the value of tension at any point of the side surface of Gaussian cylinder is equally, and the value E. You can endure the integral sign. Then

, (2)

Where

- Square of the side surface of Gaussian cylinder.

Now we calculate the right side (1). At the same time you need to consider 3 cases:

1) R 1
2) R 1
q \u003d τ. 1 h.. (3)

From (1) - (3) We will get:

From where. Here is replacement

.

3) R 2 Q \u003d ( τ 1 +τ 2 )h., then

,

Answer: E. 1 =0; E. 2 \u003d 10 4 V / m; E. 3 \u003d 6. 10 3 V / m.

2. Energy interaction of point charges. Potential

- energy of interaction of point charges;

- determination of the potential;

- Potential field of point charge;

,

- principle of superposition;

- the potential energy of the point charges;

Work fields to move charge;

,

,

- The connection of tension and potential.

Examples of solving problems

Task 3.

Determine the potential of the field created by the charge, evenly distributed by a thin straight rod with a linear density of 200 nkl / m, at a point lying on a perpendicular restored in one of the ends of the rod, at a distance of 40 cm from it. Rod length 30 cm.

Decision

R azobe Stem on infinitely small elements dL \u003d DY.; y. - coordinate of this element (Fig. 3). Charge element dQ \u003d.τ dY. You can read point. The potential of the field created by charge dQ. At a point a at a distance r. From charge, equal to:

, (1)

Where
^

. (2)

On the principle of superposition full potential

. (3)

Integration is conducted along the entire length of the rod. Then

Here, Constanta

put out the integral sign and used that the primitive function for the function

is an

what m eZHNA MAKE DIFFERENCE:

By formula (4), calculate the potential:

Answer: φ \u003d 1250 V.

Task 4.

On two coaxial infinite cylinders with radii 5 cm and 10 cm, charges with linear densities of charge τ 1 \u003d 100 NK / M and τ 2 \u003d 50 NL / M, respectively, are uniformly distributed. The space between the cylinders is filled with paraffin with dielectric constant 2. Find the potential difference of cylinders.

R messing

We use the results obtained in Task 2: the electrostatic field strength between the cylinders, R. 1 RR 2, calculated by the Gauss Theorem, is equal to:

. (1)

According to the communication formula between tensions and potential

, (2)

Where the integral is more convenient to take on the power line of the field, so, since the direction of tension coincides with the direction of the radius-vector and element of the length of the integration circuit

, α \u003d 0. Substitting (1) in (2), we get:

,

.

Answer: Δ. φ =624 IN.

3. Polarization of dielectrics. Dipole

- electric dipole moment;

- moment of force acting on the dipole in the electric field;

,

- polarity (polarization vector) dielectric;

where - dielectric susceptibility of dielectric;

Electric displacement motor.

Examples of solving problems

Task 5.

The intensity of the field of the air capacitor, charged and disconnected from the source is equal to E. 0. The dielectric plate with dielectric permeability ε was placed in parallel to the condenser in parallel to the plates. To find surface density associated charges on the edges of the dielectric, express it through the surface density of free charges on the condenser plates; find the field strength in the dielectric, as well as the intensity of the field created only by the associated charges; The value of the electrical displacement and polarity of the dielectric.

The field strength in the dielectric decreases compared to the voltage in vacuo in ε times, therefore

. (1)

The total (complete) field in the dielectric folds from the field of free charges and related (induced) :

, but and sent the opposite (see Fig.4), therefore E \u003d E. 0 -E ',

N. the edge of the associated charges can be expressed through the surface density of the associated charges (condenser field strength):

, (3)

Then taking into account (2):

. (4)

Similarly, the field strength of only free charges

, then from (4):

. (5)

Vector electrical offset, so

. (6)

Next, since

and vectors , I. Directed the same, then:

You can check (7): by definition, the polarization vector is equal to the total dipole moment of the volume of the volume of the substance:

, (8)

And the dipole moment of the dielectric plate is equal to the work related chargeLocalized on one of the faces

, on the shoulder of the dipole - the thickness of the plate d., then

, (9)

Since the volume of the plate Δ V.=S. . d.. From (4) and (9) we get (7).

Answer:

;

;

;

;

;

.

^

4. Conductors. Capacitors. Conductor capacity; condenser.

Charged particle in electric field

,

- Determination of the capacity of the conductor, the capacitor;

- Bowl.

- Communication between field strength and voltage on the condenser.

- Capacity of a flat capacitor;

- general container with parallel condenser connection;

- Total capacity with a consecutive condenser connection

Energy acquired by a particle in an electric field.

^ Examples of solving problems

Task 6.

Two identical flat air capacitors with a capacity of 100 PF each are connected to the battery sequentially. Determine how much the battery capacity changes if the space between the plates of one of the capacitors is filled with paraffin with dielectric constant 2.

Decision

General capacity with condenser condensation FROM 1 I. FROM 2 You can: Find out of the formula:

. Therefore, the overall capacity of the battery consisting of two identical capacitors capacity FROM 0 (before filling one of the capacitors of paraffin) is equal to:

. After filling in paraffin one of the capacitors, its capacity

, and before filling was equal

, that is, the tank has increased in ε time:

. Find a new total battery capacity:

. Thus, changing the capacity of the battery is:

. Substitute numerical values:

.

Answer:

.

5. Energy electrostatic field. Field energy density

- energy charged conductor;

- energy of a charged condenser;

- communication between conservative power and potential energy;

- determination of the bulk energy density of the field;

- bulk energy density of the electrostatic field.
^

Examples of solving problems

Task 7.

E. the ledctic field is created charged ( Q.\u003d 0.2 μCl) with a metal sphere with a radius of 5 cm. What is the energy of the field concluded in the spherical layer bounded by the sphere and the concentric with it a spherical surface, which is 3 times the radius of the sphere?

Decision:

The energy of the field concluded in the spherical layer will be found through the bulk energy density equal to the definition

, (1)

And for the energy of the electrostatic field

. (2)

The tension of the electrostatic field created by a solitary metal charged sphere is beyond this area (when r.>R. 0) The same as the tension of the point charge field located in the center of the sphere:

, (3)

From (1) - (3) it follows that the energy concluded in any small volume dVequal to:

. (4)

Since the field is spherically symmetrically, as dV You should take a thin ball layer, the concentric in this area, with the inner radius r., external radius ( r.+dr.), Then within this layer, the value of tension can be considered the same and equal to (3). The volume of the layer can be found by moving the area of \u200b\u200bthe sphere on its thickness, as the layer is thin:

. (5)

Finally, we find the desired energy, configuring (4) by volume, that is, within R. 0 RR:

,

Answer: W.\u003d 2.4 MJ.

6. Electricity. Laws of Ohm and Kirchhoff

- determination of current;

- the charge pasted through the conductor cross section;

- determination of current density;

- current density with directional motion of charged particles;

- Ohm law in local form;

- Communication of specific electrical conductivity and specific resistance;

- resistance of the conductor;

- general resistance with a sequential connection;

- general resistance with parallel compound;

- Ohm law for uniform plot chains;

- voltage on an inhomogeneous section of the chain;

- determination of the electromotive force;

- Ohm law for a closed chain;

- the first Kirchhoff rule (for the node);

- The second rule of Kirchhoff (for a closed contour).

- The dependence of metal resistance on temperature.

^ Examples of solving problems

Z. adagha 8.

Determine the strength of the current in the resistance R 3 and the voltage at the ends of this resistance (Fig. 10). E 1 \u003d 1 V, E 2 \u003d 5 V, R 1 \u003d 1 Ohm, R 2 \u003d 2 Ohm, R 3 \u003d 3 ohms.

Decision:

To solve the problem, we use Kirchhoff rules. First of all, we choose the directions of currents in all the branches of the chain (in this task three) and the protruding current designations (see Fig. 11). In the chain, two nodes (B and E), therefore, one equation should be recorded according to the first rule (permanently than the number of nodes):

- Algebraic amount of currents converging in the node is zero. We write this rule for the node B:

I 1 -i 2 + i 3 \u003d 0, (1)

Moreover, the currents coming to the node take a positive sign, leaving - with negative.

According to the second rule of Kirchhoff, we write down the two remaining equations (the total equations are as much as the currents):

- Algebraic amount of stress drops on resistances in any closed loop is equal to an algebraic amount electromotive power. Here you also need to follow the rules of signs: if the direction of circuit around the contour in this area is opposite to the current direction, then the voltage drop takes a negative sign; If EMF is passing from the plus to minus, then take it with a negative sign.


- determination of the specific thermal power of the current;

- Law of Joule-Lenza in local form.

^ Examples of solving problems

Task 9.

The power of the current in the conductor resistance 12 ohm evenly decreases from the maximum value to zero for 10 s. What amount of heat is highlighted in this conductor during the specified period of time, if the charge has passed 50 kl on the conductor?

R measure:

We write the law on which the current power in the explorer changes over time. Current decreases evenly, that is, according to the linear law, from the maximum value I. 0, then:

I.=I. 0 –kt., (1)

Where

- speed descending Current:

Detlaf, A.A. Course of physics: studies. Handbook for universities / A.A. Detlaf, V.M. Yavorsky. - M.: Horsis.Shk., 1989.- 608 p.

Course of physics: studies. For universities: 2 t. T. 1 / Ed. V.N.Lozovsky. - St. Petersburg: Lan, 2000. - 576 p.

Trofimova, T.I. Course of Physics / T.I. Trofimova.-M.: Higher. Shk., 1999.-542 p.

Requirements for registration and general guidelines .............. ......

1. Cool law. Field tension. Gaussian Theorem ..................... .. ....

2. Energy of interaction of point charges. Potential…………..………

3. Polarization of dielectrics. Dipole….………………..………………..…..

4. Conductors. Capacitors. Conductor capacity; condenser. The charged particle in the electric field .............................. .. ...... ..

5. Energy electrostatic field. Energy density field ...... .. ...... ..

6. Electric current. Laws of Oma and Kirchhoff ........................... .. ...... ..

7. Law of Joule-Lenza ............ .. ................................................... ..

8. Current in fluid and gas. Thermoelectronic emission ..................... ..........

Bibliographic list ...................................................... .. ... ..

ELECTRIC FIELD.

Electric field strength.

1.Electric field.

How do two charged bodies interact in some distance from each other?

Great English Physicist Michael Faradays in the 1930s 19th. suggested that any charged body creates an electric field around him in an all over it. It is through this field that interacts, i.e. The field created by one charge acts on another charge, and vice versa. Subsequently, this hypothesis was confirmed by James Maxwell.

Electric field can be considered as some space, in of each point of which force acts on the charged body.

The electric field is a special kind of matter surrounding the charged bodies by means of the interaction of charges..

In other words, electric field generate charges I. act On charges.

Electric field continuously in space.

Since the field specifies on charges, then characterize it precisely on the charge. In electric field a trial charge is made.

Trial charge It is called a point charge, small in size (so as not to distort its field the field under study) and a positive sign (so agreed).

If in the same point of the electric field (in the figure it is created point charge Q) Submit the varied test charges Q, it turns out that the force acting on these charges is proportional to the magnitude of these charges. It means that the ratio of force acting on the charge, which was made to this point of the electric field, to the magnitude of this charge always has the same meaning that does not depend on the value of the test charge.

Therefore, this ratio was taken for the characteristics of the electric field at this point and called the electric field strength.

The electric field strength at this point is a vector physical quantity, the module of which is equal to the ratio of the force acting on the test charge made at this point to the magnitude of this charge. Direction of voltage vector coincides with the direction of force.

The determination of tension can be formulated differently.

The voltage of the electric field at a given point is a vector physical value, the module of which is numerically equal to the force acting on a single test charge at this point point, and the direction coincides with the direction of force.

Unit of voltage measurement [ E.] \u003d 1N / CL.

The field strength module created by a point charge Q at a distance of R from it, by definition, is the ratio of force acting on the pricnial charge Q to the magnitude of this charge. Note that the force acting on the charged charge is the strength of the Coulomb interaction of two charges - q and q located at a distance of r from each other.

Thus, the field strength module created by a point charge Q at a distance of R is equal to

The direction of the voltage vector coincides, as already mentioned, with the direction of force acting on the pricnial charge Q.

The figure shows a graph of the intensity of the point charge field of the distance from the distance.

Outside the ball, the module and the direction of tension is defined in the same way as in the case of a point charge, but under R here is meant the distance from the center of the ball to the point in which the tension is calculated, i.e. r.= R.+ h.where R is a ball radius, and H is the distance from the surface of the ball to the point.

On the surface of the ball tension

Inside the ball tension is zero E \u003d 0..

A graph of the dependence of the battery field of the charged ball (sphere) from the distance is shown in the figure.

5. Superposition.

If the field is created by several charged bodies, then it helps to calculate its tension at some point superposition principlewhose essence is as follows.

Suppose the field is created by two point charges - positive Q 1 and negative Q 2. It is required to find tensions at a point at a distance of R 1 and R 2 from the first and second charges, respectively.

The voltage modules created at this point each charge separately and. To determine the direction of voltage vectors mentally I will make a test charge Q (positive) to this point. The direction of voltage vectors is supreming with the direction of forces acting on the test charge from charges of chargesq 1 and Q 2. These vectors we put on the rule of the parallelogram. The resulting embedding vector is and there is a vector of electric field strength at this point.

The principle of superposition can be formulated as follows.

The intensity of the field created by the charge system is equal to the vector sum of the fields of fields created at this point by each charge separately.

6. Relia tension.

Graphically electric field is depicted using lines of tension.

Voltage lines are built in such a way that at each point the direction of tangent to the voltage line coincides with the direction of the voltage vector at this point.

Thus, knowing how the voltage line passes through any point of the field and spending the tangent to it, you can determine the direction of the voltage vector at this point.

Electric field strength lines have the following properties.

1. In the same field, they do not intersect anywhere.

2. The decline begin on positive charges or come from infinity, and end in negative charges or go into infinity, i.e. are not closed.

3. Nowhere in space are not interrupted.

4. The flow (density) of the lines is proportional to the size of the field strength in this area.

Examples of the graphic image of electric fields of point positive and negative charges and two different point charges.

7.Ohnoye field.

The electric field is called homogeneous if the vector of tension has in each point same meaning and direction.

Graphically, such a field is depicted by parallel lines of tension, separated from each other at the same distance.

Examples of homogeneous fields are fields created by endless charged planes.

If these two planes are brought apart and applied the principle of superposition, it turns out that the voltage lines between the planes are directed in one direction and, therefore, the field strength increases, and on the right and to the left of the voltage line planes are directed in different directions and, therefore, the field strength decreases.

If the charges of the planes in the module are the same, the intensity of the field on the right and to the left of the planes will be equal to zero.

1. Electric field . This concept introduced Michael Faraday in the middle of the XIX century. Starting with Faraday, physics began to consider the electric field as a special form of matter capable of carrying the action of one charge to another.

The concept of the gravitational field appeared later, magnetic field, nuclear field, etc. All fields carry the action with the final speed. All fields are quantized, i.e. The interaction is carried out with the help of appropriate particles.

Quantha electromagnetic field Do not have the masses of peace, but have energy and pulse (momentum momentum).

The source of the electromagnetic field is electrical charges. To measure and describe the field created by a certain charge Q., I need another charge q.which could be made in different points of the charge field Q.. This auxiliary charge q. Call trial. It is assumed that the test charge is always positive, and its value is much smaller than charge Q..

2. Electric field strengthcall physical quantityequal to the ratio of force acting on the trial charge q., To the magnitude of this test charge. . (3.1)

Faraday suggested graphically to depict electrical fields with continuous power lines or voltage lines, at each point of which the vector of force or tension is aimed at the tangent of them. All lines begin on positive and end on negative charges. If the field is created by a secluded charge Q., the lines go to infinity. Than more charge Q., the more lines come out of it or ends on it.

3. Field of point charge. Let the electric field be created by a secluded point charge Q.. To measure it at some point, you need to make a trial charge at this point q.. Field strength for charge q. By the law of Kulon (3.2)

Here - a radius of a vector spent from a point charge Q. In the point of the field where the trial charge is q.(Fig. 6 - a).

Charge Tension Q., equal (3.3)

The field of a secluded point charge has a central symmetry. Figure 6 - B shows the lines of the Positive Dot Charge Field Q.lying in the plane passing through the charge Q.. Lines are directed from the center to the periphery. Field lines negative charge Directed from the periphery to the center (Fig. 6-c).

The advantage of graphical interpretation of the field is not only the ability to evaluate the configuration of the line direction of the vector E. , but also the opportunity to evaluate its magnitude, as the lines are proportional to tensions E. .

Graphic image quantitative characteristics of the electric field are possible due to the fact that the field E. Point charge decreases proportionally 1 ç r 2, and at any distance from charge r. The density of the lines, that is, their number per unit of surface area perpendicular power lines decreases in proportion to 1 ç r 2.

4. Superposition of electric fields. To answer the question; What is equal to the tension of the field created by several different point charges located in different places, you need to know how the fields are folded.

Experience shows that the strength of the interaction of any two charges does not depend on the availability of other charges. This means that the force acting from the charge system q. 1 …q. n on a trial charge q, It is equal to the geometric amount of forces from each of them, and the tension of the total electric field is equal to the geometric sum of the field strengths created by each charge separately.

(The law of addition of electric fields) (3.4)

The ability of electric fields to develop without mutual distortion is called the principle of superposition This is an objective property of linear power fields, known to us from mechanics. Thanks to the superposition of electric fields, there is a possibility of calculating the fields of the point charges and extended charged macotel.

Example 3.1. Field of two point charges. Let two charges q. 1 I. q. 2 are at a distance of 2 but Friend from each other. It is necessary to find the tension of the total field.

Both charges on the axis OX. Cartesian coordinate system at points with coordinates q. 1 (-a.,0,0), q. 2 (+a.0.0). Since the axis OX. is the axis of symmetry of the charge system, the two-dimensional solution of the task in the plane XoY. It is exhaustive.

Field strength anywhere BUT With coordinates x., y. equal to the sum of tensions (Fig. 17 - a) :. (3.5)

Here - the charge field q. 1, - Charge field q. 2 , E. 1 X. and E. 2 X. - Projections of vectors and on the axis OX., E. 1 Y. and E. 2 Y. - Projections of vectors on the axis Oy., and - single orthops of these axes.

In this way, you can calculate the field arbitrarily more point charges. Just add to formula (3.5) projections of the fields of the following charges: E. 3 X., E. 3 Y., E. 4 X., E. 4 Y. etc. If the charge system is not flat or does not have the symmetry axis, the task must be solved in three-dimensional space.

Example 3.2. Field of charged thread. Let on the length of the thread length l. There is a charge with linear density t., [t.] \u003d CL ç m. (numerically t. - This is a charge per 1 m length of the thread). It is necessary to find the field strength around the thread.

To calculate the electric field created by an extended charged body, this charge is broken mentally on sufficiently small elements that can be considered point charges. The fields created by these elements are summed up (integrated).

The field of straight charged thread has axial symmetry, so it is enough to calculate the field in the thread plane. Position the segment of the charged thread along the straight Oy.. The ends of the thread are at points with coordinates (0, y. 1) and (0, y. 2) (Fig. 8).

Take on the thread an infinitely small cut dY. on distance y. from the beginning of the coordinates. This cut dY. carries charge dQ.= tDY and can be considered point.

Field dE.created by this point charge at the point BUTEqually: . (3.9)

And its components on the axes:

, (3.10)

. (3.11)

Here . (3.12)

Integrate expressions (3.10) and (3.11) is easier around the corner j.. Since, then , and. Then .

Integrating from the corner j. 1, under which "visible" bottom end of the segment, to the corner j. 2 (upper end), we get:, (3.13)

Computing components E X. and E Y. at different coordinate values x. 0 , y. 0 Points BUT, We get the vector of tension at any point of space. (3.15)

Near an endless uniformly charged thread with a linear charge density t.power lines perpendicular to it and field strength decreases inversely distance R. to the thread: E. = tç.2pr..

6. Work on moving electric charge . We calculate the work performed by the field when moving a test positive charge q. In the field of point positive charge Q.. We believe that the charge q.moves endlessly slow from the point but To point b. For some kind of arbitrary trajectory (Fig. 9).

Work on an infinitely small section of the path is equal to where - the force acting from the side of the charge Q. on a trial charge q.. The work is equal to the decrease potential energy systems, dA= -dW.. The work of the final movement from but before b. There is integration.

. (3.17)

As T. . (3.18)

If you take a point b. on infinity, then work when moving charge q. From the point but On infinity is equal to its potential energy at the point but: . (3.19)

When r. ® ¥, Coulomb force drawn to zero. Therefore, the work of the discharge of charges Q. and q. With distance r A. Infinity determines the total potential energy of the system of two charges Q. and q.located at a distance r A. Friend from each other.

Holding the index " but"We obtain a general formula for the potential energy of a system of two charges in vacuum at a distance r.: . (3.20)

7. Potential of the electrostatic field. If split energy W.trial charge (3.19) on the magnitude of its charge, then we get another energy characteristic Fields - potential j.at a point at a distance r. From point charge: , (3.21)

The absolute values \u200b\u200bof the potential and potential energy of the interaction of charges in theory are taken equal to zero at infinity. You can only measure the potential difference or the difference of potential energies of two states of charge system - two charge configurations in space. Therefore, as in the mechanics, when calculating the potential energy of bodies in the gravitational field, the potential in electrostatics is taken or an extremely solid and extended body with permanent potential in terms of experience.

The difference of potentials of two points is usually called voltage And denote U. = j. 1 - j. 2. Charging Travel Field q. between points with the potential difference between them U. equal to the work A. = qU.(3/22)

The potential of the electric field is a scalar, at each point of the field it is determined by one number, while the field strength is the vector, at each point of the field it is determined by three numbers, with its own projections on the axis: E x \u003d.¶ jç.¶ x., E y \u003d.¶ jç.¶ y., E Z \u003d.¶ jç.¶ z.. (3.23)

The field strength does not depend on the choice of the level of zero potential, because Determined only by the speed of its change in space.

The work of the movement of charge in the electric field does not depend on the path of movement (3.18). This suggests that the electrostatic field has a field of conservative forces. The circulation of the voltage vector of the electrostatic field along the closed contour is zero. . (3.24)

The set of field points with the same potential continuously and forms equipotential surface (from Latin aequi. - Equally). When the charge is moving along the equipotential surface, the operation is zero ,. (3.25)

Hence it follows that vector line at any point of the orthogonal equipotential surface at this point..

Capacity and Voltage Unit with SI - Volt, 1 V \u003d 1 J ç CL. Two points of the field have the difference in the potentials of 1 B, if the charge between them is moving between them 1 J.

8. Potential of the charge system field At some point but Regarding the point of the field b. Determined by the work of moving per unit of positive charge from the point but exactly b..

The potential of the charge system field is equal to the algebraic sum of the potentials of the fields of each of the charges separately. This statement expresses the principle of superposition of electric fields for potential.

Example 3.3. Potential of the dipole field. Electric dipole is a system of two equal in size and opposite on the sign of point charges located at some distance from each other. Electrical characteristic The dipole is its electric moment, where the vector is directed from the negative charge to the positive. In Example 3.1, the electrical field strength of the system of two point charges is calculated, the special case of which is the dipole.

I calculate the potential of the dipole field, using the polar coordinates. The electric moment of the dipole will send along the polar axis (Fig. 10).

The potential of the total field at an arbitrary point BUT Regarding the infinity on the principle of superposition there is the sum of the potentials of the folded charge fields q. - I. q. + .

, (3.27)

Because q. - = -q. +, T. . (3.28)

If point BUT removed for a long distance from the dipole r.>>l. T. r. 1 -r. 2 " l.cos. a., r. 1 r. 2 " r. 2, I. . (2.29)

Compared to a point charge, the dipole field decreases with a distance faster, the potential is proportional to 1 ç r 2, but not 1 ç r, like a point charge.

Tensions can be calculated through the potential by the formula . In the polar coordinates, the differential operator ñ (it is called the operator Nabel) has the form: . (3.30)

Here is a single ort, directed along the polar radius - a single ort, directed perpendicular to the radius in the direction of an increase in the polar angle a.. Calculate the components of the vector.

From here. (3.33)

The dipole field has axial symmetry relative to the line passing through charges (Fig. 11). The solid lines depict the line of tension of the electric field, the bar - the cross sections of the surfaces of equal potential - equipotential surfaces. On the axis of the dipole its electric moment because a.\u003d 0, and . (3.34)

9. Electric field theory. Due to the fact that the law of Kulon is very similar to the law of World Newton, to the description electrical phenomena It turned out to be possible to apply the developed earlier mathematical theory of gravity. Moreover, the electrical interaction turned out to be richer gravitational, because there is not only attraction, but also the repulsion of tel. Therefore, a variety of specific applications of phenomenological theories of the electric field are much larger.

During the XIX century D. Poisson, J. Green, M. Faraday, K. Gauss, W. Thomson developed a strict mathematical theory of the electric field. And summarized it in general Theory Electromagnetic field J. Maxwell.