The formula for finding tension. Voltage Formula

Definition

Tension vector is the power characteristic of the electric field. At some point in the field, the intensity is equal to the force with which the field acts on a unit positive charge placed at the specified point, while the direction of the force and the intensity are the same. The mathematical definition of tension is written as follows:

where is the force with which the electric field acts on a stationary, “trial”, point charge q, which is placed at the considered point of the field. At the same time, it is considered that the “trial” charge is small enough that it does not distort the field under study.

If the field is electrostatic, then its intensity does not depend on time.

If the electric field is uniform, then its strength is the same at all points in the field.

Graphically, electric fields can be represented using lines of force. Lines of force (tension lines) are lines whose tangents at each point coincide with the direction of the intensity vector at this point of the field.

The principle of superposition of electric field strengths

If the field is created by several electric fields, then the strength of the resulting field is equal to the vector sum of the strengths of the individual fields:

Let us assume that the field is created by a system of point charges and their distribution is continuous, then the resulting intensity is found as:

integration in expression (3) is carried out over the entire area of ​​charge distribution.

Field strength in a dielectric

The field strength in the dielectric is equal to the vector sum of the field strengths created by free charges and bound (polarization charges):

In the event that the substance that surrounds the free charges is a homogeneous and isotropic dielectric, then the intensity is equal to:

where is the relative permittivity of the substance at the studied point of the field. Expression (5) means that for a given charge distribution, the strength of the electrostatic field in a homogeneous isotropic dielectric is less than in vacuum by a factor of.

Field strength of a point charge

The field strength of a point charge q is:

where F / m (SI system) - electrical constant.

Relationship between tension and potential

In the general case, the electric field strength is related to the potential as:

where is the scalar potential and is the vector potential.

For stationary fields, expression (7) is transformed into the formula:

Electric field strength units

The basic unit of measurement of electric field strength in the SI system is: [E]=V/m(N/C)

Examples of problem solving

Example

Exercise. What is the modulus of the electric field strength vector at a point defined by the radius vector (in meters) if the electric field creates a positive point charge (q=1C) that lies in the XOY plane and its position specifies the radius vector , (in meters)?

Decision. The voltage modulus of the electrostatic field, which creates a point charge, is determined by the formula:

r is the distance from the charge that creates the field to the point where we are looking for the field.

From formula (1.2) it follows that the modulus is equal to:

Substitute in (1.1) the initial data and the resulting distance r, we have:

Answer.

Example

Exercise. Write down an expression for the field strength at a point, which is determined by the radius - vector, if the field is created by a charge that is distributed over the volume V with density.

ELECTRICAL BIAS

Basic formulas

 Electric field strength

E=F/Q,

where F is the force acting on a point positive charge Q placed at the 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 an inhomogeneous field,

Or
,

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

 The intensity of the electric field created by a point charge Q on distance r from the charge

.

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

a) inside the sphere (r<.R)

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

;

c) outside the sphere (r>R)

.

 The principle of superposition (superposition) of electric fields, according to which the intensity E of 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 strengths E 1 and E 2 strength vector modulus

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

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

, where  is the linear charge density.

The 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 intensity of the field created by an infinite uniformly charged plane,

where  is the surface charge density.

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

.

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

.

The above formula is valid for 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 displacement D associated with tension E electric field ratio

D= 0 E.

This relation is valid only for isotropic dielectrics.

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

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

;

b) in the case of an inhomogeneous field and an arbitrary surface

,

where D n - vector projection D to the direction of the normal to the surface element, the area of ​​which is equal to d S.

 Ostrogradsky-Gauss theorem. Electric displacement vector flux through any closed surface enclosing charges Q 1 ,Q 2 , ...,Q n ,

,

where P- the number of charges (with its own sign) enclosed 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. The circulation is expressed by the closed-loop integral
, where E l - the projection of the intensity vector E at a given point of the contour on 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 \u003d 15 cm from the first and at a distance r 2 =10 cm from the second charges.

Decision. 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 strengths of the electric field created in vacuum by the first and second charges are, respectively, equal to

(1)

Vector E 1 (Fig. 14.1) is 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 , as Q 2 <0.

Vector modulus E find by the law of cosines:

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

.

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

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

.

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

Example 2 The electric field is created by two parallel infinite charged planes with surface charge densities  1 \u003d 0.4 μC / m 2 and  2 \u003d 0.1 μC / m 2. Determine the strength of the electric field created by these charged planes.

R
solution. According to the principle of superposition, the fields created by each charged plane individually are superimposed on each other, with each charged plane creating an electric field regardless of the presence of another charged plane (Fig. 14.2).

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

;
.

Planes divide all space into three regions: I, II and III. As can be seen from the figure, in the first and third regions, the electric lines of force of both fields are directed in the same direction and, consequently, the strengths of the total fields E (I) and E(III) in the first and third regions 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 lines of force of the fields are directed in opposite directions and, therefore, the field strength E (II) is equal to the difference in the field strengths created by the first and second planes: E (II) =|E 1 -E 2 | , or

.

Substituting the data and doing the calculations, we get

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

The picture of the distribution of force lines of the total field is shown in fig. 14.3.

Example 3. On the plates of a flat air capacitor there is a charge Q=10 nC. Square S each plate of the capacitor is equal to 100 cm 2 Determine the force F, with which the plates are attracted. The field between the plates is assumed to be uniform.

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

F=E 1 Q,(1)

where E 1 - the strength of the field 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 quantities Q,  0 and S into this formula and doing the calculations, we get

F=565 µN.

Example 4 The electric field is created by an infinite plane charged with a surface density  = 400 nC/m 2 , and an infinite straight thread charged with a linear density =100 nC/m. On distance r\u003d 10 cm from the thread there is a point charge Q=10 nC. Determine the force acting on the charge, its direction, if the charge and the thread lie in the same plane parallel to the charged plane.

Decision. The force acting on a charge placed in a field

F=EQ, (1)

where E - Q.

Let's define tension E field created, according to the condition 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 intensity at any point

. (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 Q, is equal to the vector sum of the intensities E 1 and E 2 (Fig. 14.5): E=E 1 +E 2 . Since the vectors E 1 and E 2 mutually perpendicular, then

.

Substituting Expressions E 1 and E 2 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 quantities Q,  0 , , ,  and r into formula (4) and making calculations, we find

F=289 µN.

Force direction F, acting on a positive charge Q, coincides with the direction of the intensity vector E fields. Direction same 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\u003d 25 nC is in the field created by a straight infinite cylinder with a radius R= 1 cm, uniformly charged with surface density =2 μC/m 2 . Determine the force acting on a charge placed at a distance from the axis of the cylinder r=10 cm.

Decision. Force acting on a 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  in terms of the surface density . To do this, select a cylinder element with length l and express the charge on it Q 1 two ways:

Q 1 =  S= 2  Rl and Q 1 = l.

Equating the right parts of these equalities, we get  l=2 Rl . After shortening to l find =2 R . With this in mind, formula (2) takes the form E=R /( 0 r). Substituting this expression E into formula (1), we find the desired force:

F=Q R/( 0 r).(3)

As R and r are included in the formula as a ratio, then they can be expressed in any, but only the same units.

After performing calculations using formula (3), we find

F\u003d 2510 -9 210 -6 10 -2 / (8.8510 -12 1010 -2)H==56510 -6 H=565μH.

Force direction 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. On distance a\u003d 20 cm from the thread there is a flat round area with a radius r\u003d 1 cm. Determine the flow of the tension vector through this area if its plane makes an angle  \u003d 30 ° with the line of tension passing through the middle of the area.

Decision. The field created infinitely uniformly by a charged filament is inhomogeneous. The intensity vector flux 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 pierced by lines of tension.

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

E P =E cos,

where  is the angle between the direction of the vector and the normal n. With this in mind, formula (1) takes the form

.

Since the dimensions of the area surface are small compared to the distance to the thread (r<E very little. varies in absolute value and direction within the site, which allows you to replace the values ​​under the integral sign E and cos their average values<E> and and take them out of the integral sign:

By integrating and replacing<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.) takes 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 respectively Q 1 =l nC and Q 2 = -0.5 nC. Find tension E fields at points separated from the center of the spheres at distances r 1 =5 cm, r 2 =9 cm r 3 =15cm. Build Graph E(r).

R
solution. Note that the points at which you want to find the electric field strength lie in three areas (Fig. 14.7): area 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 draw a spherical surface S 1 radius r 1 and use the Ostrogradsky-Gauss theorem. Since there are no charges inside the region I, then, according to the indicated theorem, we obtain the equality

, (1)

where E n is the normal component of the electric field strength.

For reasons of symmetry, 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) takes the form

.

Since the area of ​​a 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 there is a charge inside this surface Q 1 , then for it, according to the Ostrogradsky-Gauss theorem, we can write the equality

. (2)

As E n =E 2 =const, then the symmetry conditions imply

, or ES 2 =Q 1 / 0 ,

E 2 =Q 1 /( 0 S 2 ).

Substituting here the expression for the area of ​​the 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 . Therefore, for it, the equation written on the basis of the Ostrogradsky-Gauss theorem will have the form

.

Hence, using the provisions applied in the first two cases, we find

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

We express all quantities in SI units ( Q 1 \u003d 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).AT area 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 the point r=R 1 (r tends to R 1 left) E 2 (R 2 )=Q 1 /(4 0 R )=900V/m. In region III ( r>R 2 )E 3 (r) varies according to the law 1/ r 2 , and at the point r=R 2 (r tends to R 2 on 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. Define tension E electric field generated by a point charge Q=10 nC at distance r\u003d 10 cm from it. Dielectric - oil.

14.2. Distance d between two point charges Q 1 =+8 nC and Q 2 \u003d -5.3 nC is equal to 40 cm. Calculate the intensity E field at a point midway between the charges. What is the intensity if the second charge is positive?

14.3. Q 1 =10 nC and Q 2 = –20 nC, located at a distance d=20 cm apart. Define tension E field at a point remote from the first charge by r 1 \u003d 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 \u003d Q is equal to 8 cm. At what distance r from the first charge is the point at which the intensity E charge field is 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 a point on a straight line passing through these charges, the intensity E fields in which 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. Define tension E field at a point remote from the first charge by r 1 \u003d 12 cm and from the second to r 2 =6 cm.

The field strength of the charge distributed over the ring and the sphere

14.7. A thin ring with a radius R\u003d 8 cm carries a charge uniformly distributed with a linear density  \u003d 10 nC/m. What is the tension E electric field at a point equidistant from all points of the ring at a distance r\u003d 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 a radius R\u003d 10 cm is a charge Q=l nC. Define 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. Plot dependency graph E from r.

14.10. Two concentric metallic charged spheres with radii R 1 =6cm and R 2 \u003d 10 cm carry charges, respectively Q 1 =1 nC and Q 2 = – 0.5 nC. Find tension E dot fields. spaced from the center of the spheres at distances r 1 =5 cm, r 2 =9 cm, r 3 \u003d 15 cm. Plot dependency 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 intensity E fields in the distance a\u003d 0.5 m from the wire against its middle is 200 V / m.

14.12. Distance d between two long thin wires parallel to each other is 16 cm. The wires are uniformly charged with opposite charges with a linear density ||=^150. µC/m. What is the tension E fields at a point remote on r\u003d 10 cm from both the first and second wire?

14.13. Straight metal rod diameter d=5cm and long l\u003d 4 m carries a charge uniformly distributed over its surface Q=500 nC. Define tension E field at a point 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 radius R\u003d 2 cm carries a charge evenly distributed over the surface ( \u003d 1 nC / m 2). Define tension E fields at points separated from the axis of the tube at distances r 1 \u003d l cm, r 2 \u003d 3 cm. Plot dependency E(r).

The purpose 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 \u003d kq / r 2 in solving simple problems for 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 proved that there are two types of charges around which there are electric fields characterized by lines of force.

Graphically depicting the field, 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 of force:

Fig.2

Negative charge lines of force:

Fig.3

Force lines of like interacting charges:

Fig.4

Force lines of opposite interacting charges:

Fig.5

The power characteristic of the electric field is the intensity, which is denoted by the letter E and has units of measurement or. The 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 the transformation of the Coulomb law formula and the strength formula, we have the dependence of the field strength on the distance at which it is determined relative to a given charge

where: k– coefficient of proportionality, 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 an electrical constant equal to 8.85 10 -12 C 2 /N m 2;

q is the electric charge (C);

r is the distance from the charge to the point where the intensity 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 homogeneous. In a limited region of space, an electric field can be considered approximately uniform if the field strength within this region changes insignificantly.

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

Consider several cases of determining tension.

1. Let two opposite charges interact. We place a point positive charge between them, then at this point two intensity vectors will act, directed in the same direction:

According to the principle of superposition of fields, 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 \u003d kq 1 / x 2 + kq 2 / (r - x) 2

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

x is the distance between the first and the point charge.

Fig.6

2. Consider the case when it is necessary to find the intensity at a point remote 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 between the intensity E 31 and E 32 .

The formula for tension at a given point is:

E \u003d kq1 / (r + a) 2 - kq 2 / a 2

Where: r is the distance between interacting charges;

a is the distance between the second and the point charge.

Fig.7

3. Consider an example when it is necessary to determine the field strength at some distance from both the first and the second charge, in this case at a distance r from the first and at a distance b from the second charge. Since charges of the same name repel and unlike charges attract, we have two tension vectors emanating from one point, then for their addition you can apply the method to the opposite corner of the parallelogram will be the total tension vector. We find the algebraic sum of vectors from the Pythagorean theorem:

E \u003d (E 31 2 + E 32 2) 1/2

Hence:

E \u003d ((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 of 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. Fixing the topic.

Verification work.

Option number 1.

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

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

3. How are the lines of force of this charge directed?

4. Determine the signs of the charges:

Home tasks:

1. Two charges q 1 = +3 10 -7 C and q 2 = −2 10 -7 C are in 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 some point of the field, a force of 3 10 -4 N acts on a charge of 5 10 -9 C. Find the field strength at this point and determine the magnitude of the charge that creates the field if the point is 0.1 m away from it.

The purpose 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 \u003d kq / r 2 in solving simple problems for 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 proved that there are two types of charges around which there are electric fields characterized by lines of force.

Graphically depicting the field, 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 of force:

Fig.2

Negative charge lines of force:

Fig.3

Force lines of like interacting charges:

Fig.4

Force lines of opposite interacting charges:

Fig.5

The power characteristic of the electric field is the intensity, which is denoted by the letter E and has units of measurement or. The 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 the transformation of the Coulomb law formula and the strength formula, we have the dependence of the field strength on the distance at which it is determined relative to a given charge

where: k– coefficient of proportionality, 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 an electrical constant equal to 8.85 10 -12 C 2 /N m 2;

q is the electric charge (C);

r is the distance from the charge to the point where the intensity 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 homogeneous. In a limited region of space, an electric field can be considered approximately uniform if the field strength within this region changes insignificantly.

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

Consider several cases of determining tension.

1. Let two opposite charges interact. We place a point positive charge between them, then at this point two intensity vectors will act, directed in the same direction:

According to the principle of superposition of fields, 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 \u003d kq 1 / x 2 + kq 2 / (r - x) 2

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

x is the distance between the first and the point charge.

Fig.6

2. Consider the case when it is necessary to find the intensity at a point remote 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 between the intensity E 31 and E 32 .

The formula for tension at a given point is:

E \u003d kq1 / (r + a) 2 - kq 2 / a 2

Where: r is the distance between interacting charges;

a is the distance between the second and the point charge.

Fig.7

3. Consider an example when it is necessary to determine the field strength at some distance from both the first and the second charge, in this case at a distance r from the first and at a distance b from the second charge. Since charges of the same name repel and unlike charges attract, we have two tension vectors emanating from one point, then for their addition you can apply the method to the opposite corner of the parallelogram will be the total tension vector. We find the algebraic sum of vectors from the Pythagorean theorem:

E \u003d (E 31 2 + E 32 2) 1/2

Hence:

E \u003d ((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 of 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. Fixing the topic.

Verification work.

Option number 1.

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

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

3. How are the lines of force of this charge directed?

4. Determine the signs of the charges:

Home tasks:

1. Two charges q 1 = +3 10 -7 C and q 2 = −2 10 -7 C are in 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 some point of the field, a force of 3 10 -4 N acts on a charge of 5 10 -9 C. Find the field strength at this point and determine the magnitude of the charge that creates the field if the point is 0.1 m away from it.

As you know, electrical voltage must have its own measure, which initially corresponds to the value that is calculated to power a particular electrical device. Exceeding or reducing the value of this supply voltage negatively affects electrical equipment, up to its complete failure. What is tension? This is the difference in electrical potential. That is, if, for ease of understanding, it is compared with water, then this will approximately correspond to pressure. According to the scientific, electrical voltage is a physical quantity that shows what work the current does in a given area when a unit charge moves through this area.

The most common formula for voltage is the one in which there are three basic electrical quantities, namely the voltage itself, current and resistance. Well, this formula is known as Ohm's law (finding the electrical voltage, potential difference).

This formula sounds as follows - the electrical voltage is equal to the product of the current strength and resistance. Let me remind you that in electrical engineering for various physical quantities there are their own units of measurement. The unit of voltage measurement is "Volt" (in honor of the scientist Alessandro Volta, who discovered this phenomenon). The unit of measurement for current is "Ampere", and resistance is "Ohm". As a result, we have - an electrical voltage of 1 volt will be equal to 1 ampere times 1 ohm.

In addition, the second most used voltage formula is the one in which this same voltage can be found knowing the electrical power and current strength.

This formula sounds as follows - the electrical voltage is equal to the ratio of power to current strength (to find the voltage, you need to divide the power by the current). The power itself is found by multiplying the current by the voltage. Well, to find the current strength, you need to divide the power by the voltage. Everything is extremely simple. The unit of electrical power is "Watt". So 1 volt is equal to 1 watt divided by 1 amp.

Well, now I will give a more scientific formula for electrical voltage, which contains "work" and "charges".

This formula shows the ratio of the work done to move the electric charge. In practice, this formula is unlikely to be needed. The most common will be the one that contains current, resistance and power (that is, the first two formulas). But, I want to warn you that it will be true only for the case of active resistances. That is, when calculations are made for an electrical circuit that has resistance in the form of conventional resistors, heaters (with a nichrome spiral), incandescent bulbs, and so on, then the above formula will work. In the case of using reactance (the presence of inductance or capacitance in the circuit), a different voltage formula will be needed, which also takes into account the voltage frequency, inductance, capacitance.

P.S. The formula of Ohm's law is fundamental, and it is from it that you can always find one unknown quantity out of two known ones (current, voltage, resistance). In practice, Ohm's law will be applied very often, so it is simply necessary for every electrician and electronics to know it by heart.