A dipole is a system consisting of two charges equal in magnitude and opposite in sign. The vector I drawn from a negative to a positive charge is called a dipole arm.

Electric dipole moment

Where – dipole charge.

The electric dipole moment of a molecule is usually expressed in atomic scale units - debye (D) = 3.33∙10 -30 C∙m.

A dipole is called point if the distance r from the center of the dipole to the point at which the action of the dipole is considered is much greater than the dipole arm .

Field strength point dipole:

a) on the dipole axis

, or
;

b) perpendicular to the dipole axis

, or
;

c) in general case

, or
,

Where
─ the angle between the radius vector r and the electric dipole moment r (Fig. 2.1).

Dipole field potential

.

Potential energy dipoles in an electrostatic field

Mechanical moment acting on a dipole with an electric dipole moment , placed in a uniform electric field with intensity ,

or
,

Where
– angle between the direction of the vectors And .

Force F acting on a dipole in a non-uniform electrostatic field with axial (along the axes) symmetry,

,

Where ─ quantity characterizing the degree of inhomogeneity of the electrostatic field along the x axis; – angle between vectors And .

Examples of problem solving

Example 1. Dipole with electric moment

. Electric torque vector makes an angle
with the direction of the field lines. Define JobA external forces, perfect when the dipole is rotated through an angle
.

R decision. From the starting position (Fig. 2.2, A) the dipole can be rotated by an angle
, rotating it clockwise to the angle (Fig. 2.2, b), or counterclockwise to the corner (Fig. 2.2, V).

In the first case, the dipole will rotate under the influence of field forces. Consequently, the work of external forces is negative. In the second case, the rotation can be made only under the influence of external forces and the work of external forces is positive.

The work done when turning the dipole can be calculated in two ways: 1) directly integrating the expression for the elementary work; 2) using the relationship between work and the change in potential energy of a dipole in an electric field.

a B C

1st method. Elementary work when turning the dipole at an angle
:

and full work when turning an angle from before
:

.

After performing the integration, we get

Work done by external forces when rotating the dipole clockwise

counterclock-wise

2nd method. Work A of external forces is associated with a change in potential energy
ratio

,

Where
─ potential energies of the system in the initial and final states, respectively. Since the potential energy of a dipole in an electric field is expressed by the formula
,That

which coincides with formula (2.1) obtained by the first method.

Example 2. Three point charge ,
,
, form an electrically neutral system, and
. The charges are located at the vertices of an equilateral triangle. Determine maximum tension values
and potential
field created by this system of charges at a distance
from the center of a triangle whose side length is
.

Solution. A neutral system consisting of three point charges can be represented as a dipole. Indeed, the “center of gravity” of the charges And
lies in the middle of the straight line connecting these charges (Fig. 2.3). At this point the charge can be considered concentrated
. And since the charge system is neutral (
), That

Since the distance between the charges Q 3 and Q is much less than the distance r (Fig. 2.4), the system of these two charges can be considered a dipole with an electric moment
,Where
─ dipole arm. Electric dipole moment

.

The same result can be obtained in another way. Let us imagine a system of three charges as two dipoles with electric moments (Fig. 2.5) equal in magnitude:
;
. Electric torque of the charge system find it as a vector sum And , And
.As it follows from Fig. 2.5, we have
.Because

,That

,

which coincides with the previously found value.

Tension and potential dipole fields are expressed by the formulas

;
,

G de
─ angle between radius vector and electric dipole moment (Fig. 2.1).

Tension and potential will have maximum values ​​at
= 0, therefore,

;
.

Because
,That

;
.

The calculations give the following values:

;
.

Tasks

201. Calculate the electric moment p of a dipole if its charge
,
. (Answer: 50 nC∙m).

202. Distance between charges
And
dipole is 12 cm. Find the tension E and potential field created by a dipole at a point distant by
both from the first and from the second charge. (Answer:
;
).

203. Dipole with electric moment
formed by two point charges
And
. Find tension E and potential electric field at point A (Fig. 2.6), located at a distance
from the center of the dipole. (Answer:
;
).

204. Electric moment of a dipole
field created at point A (Fig. 2.6), located at a distance
from the center of the dipole. (Answer:
;
).

205. Determine tension E and potential
on distance

with the electric torque vector. (Answer:
;
).

206. Dipole with electric moment
rotates uniformly at frequency
relative to an axis passing through the center of the dipole and perpendicular to its arm. Point C is at a distance
from the center of the dipole and lies in the plane of rotation of the dipole. Derive the law of potential change as a function of time at point C. Accept that at starting moment time potential at point C
. Build a dependency graph
. (Answer:
;
;
).

207. Dipole with electric moment

relative to an axis passing through the center of the dipole and perpendicular to its arm. Determine the average potential energy
charge
located at a distance
and lying in the plane of rotation, a time equal to a half-cycle (from
before
). At the initial moment of time, count
. (Answer:).

208. Two dipoles with electric moments
And
are at a distance
from each other. Find the force of their interaction if the axes of the dipoles lie on the same straight line. (Answer:
).

209. Two dipoles with electric moments
And
are at a distance
from each other, so that the axes of the dipoles lie on the same straight line. Calculate the mutual potential energy of the dipoles corresponding to their stable equilibrium. (Answer:
).

210. Dipole with electric moment
attached to an elastic thread (Fig. 2.7). When an electric field of intensity was created in the space where the dipole is located
, perpendicular to the arm of the dipole and the thread, the dipole rotated at an angle
. Determine the moment of force M that causes the thread to twist by 1 rad. (Answer:
).

211. Dipole with electric moment
attached to an elastic thread (Fig. 2.7). When an electric field intensity was created in the space where the dipole is located
, perpendicular to the arm of the dipole and the thread, the dipole has rotated at a small angle
. Determine the moment of force M that causes the thread to twist by 1 rad. (Answer: ).

212. Dipole with electric moment
is in a uniform electric field of intensity
. The electric torque vector makes an angle
with field lines. What is the potential energy P of the field? Count
, when the vector of the electric moment of the dipole is perpendicular to the field lines. (Answer: ).

213. Dipole with electric moment
freely established in a uniform electric field of strength

. (Answer: ).

214. Dipole with electric moment



. (Answer: ).

215. Perpendicular to the arm of a dipole with an electric moment
a uniform electric field of intensity is excited
. Under the influence of field forces, the dipole begins to rotate about an axis passing through its center. Find the angular velocity
dipole at the moment it passes the equilibrium position. The moment of inertia of the dipole about an axis perpendicular to the arm and passing through its center. (Answer:
;
).

216. Dipole with electric moment
freely established in a uniform electric field of intensity
. The dipole was turned to a small angle and left to its own devices. Determine the natural frequency of dipole oscillations in an electric field. Moment of inertia of a dipole about an axis passing through its center
. (Answer:
).

217. Dipole with electric moment
is in a non-uniform electric field. The degree of field inhomogeneity is characterized by the value
, taken in the direction of the dipole axis. Calculate the force F acting on the dipole in this direction. (Answer: ).

218. Dipole with electric moment
settled along power line in the field of a point charge
on distance
From him. Determine the value for this point
, characterizing the degree of field inhomogeneity in the direction of the field line and the force F acting on the dipole. (Answer:
;
).

219. Dipole with electric moment
established along a line of force in a field created by an infinite straight thread charged by an infinite straight thread charged with linear density
on distance
from her. Determine the value at this point
, characterizing the degree of field inhomogeneity in the direction of the field line and the force F acting on the dipole. (Answer:
;
).

220. Dipole with electric moment
formed by two point charges
And
. Find tension E and potential electric field at point B (Fig. 2.6), located at a distance
from the center of the dipole. (Answer:
;
).

221. Electric moment of a dipole
. Determine tension E and potential field created at point B (Fig. 3.6), located at a distance
from the center of the dipole. (Answer:
;
).

222. Determine tension E and potential field created by a dipole with electric moment
on distance
from the center of the dipole, in a direction constituting an angle
with the electric torque vector. (Answer:
;
).

223. Dipole with electric moment
rotates uniformly at angular velocity
relative to an axis passing through the center of the dipole and perpendicular to its arm. Determine the average potential energy
charge
located at a distance
and lying in the plane of rotation, over time
.At the initial moment of time, count
. (Answer:
).

224. Dipole with electric moment
freely established in a uniform electric field of strength
. Calculate the work A required to rotate the dipole through an angle
. (Answer:
).

225. Dipole with electric moment
freely established in a uniform electric field of intensity
. Determine the change in potential energy
dipole when rotated by an angle
. (Answer: ).

226. The HF molecule has an electric moment
. Internuclear distance
. Find the charge such a dipole and explain why the found value differs significantly from the value of the elementary charge
. (Answer:
).

227. Point charge
is at a distance

. Determine the potential energy P and the force F of their interaction in the case when the point charge is located on the dipole axis. (Answer:
;
).

228. Point charge
is at a distance
from a point dipole with electric moment
. Determine the potential energy P and the force F of their interaction in the case when the point charge is perpendicular to the dipole axis. (Answer:
;
).

229. Two dipoles (Fig. 2.8) with electric moments
are at a distance
apart from each other (
─ dipole arm). Determine the potential energy P of the interaction of dipoles. (Answer:
).

230. Two identically oriented dipoles (Fig. 2.9) with electric moments
are at a distance
apart from each other (
─ dipole arm). Determine the potential energy P and the force F of the interaction of dipoles. (Answer:
;
).

Let us consider the field of the simplest system of point charges. The simplest system point charges is electric dipole. An electric dipole is a collection of two point charges equal in magnitude but opposite in sign –q And +q, shifted relative to each other by some distance. Let be the radius vector drawn from the negative charge to the positive one. Vector

is called the electric moment of the dipole or dipole moment, and the vector is called the dipole arm. If the length is negligible compared to the distance from the dipole to the observation point, then the dipole is called a point dipole.

Let's calculate the electric field of an electric point dipole. Since the dipole is a point one, it makes no difference, within the limits of calculation accuracy, from which point of the dipole the distance is measured r to the observation point. Let the observation point A lies on the continuation of the dipole axis (Fig. 1.13). In accordance with the principle of superposition for the intensity vector, the electric field strength at this point will be equal to

,

it was assumed that , .

IN vector form

where and are the field strengths excited by point charges –q and + q. From Fig. 1.14 it is clear that the vector is antiparallel to the vector and its modulus for a point dipole is determined by the expression

,

It is taken into account here that under the assumptions made.

In vector form, the last expression will be rewritten as follows

Doesn't have to be perpendicular JSC passed through the center of a point dipole. In the accepted approximation, the resulting formula remains true even when beyond the point ABOUT any dipole point is accepted.

The general case is reduced to the analyzed special cases (Fig. 1.15). Let's lower it from the charge + q perpendicular CD to the surveillance line VA. Let's put it on point D two point charges + q And –q. This will not change the fields. But the resulting set of four charges can be considered as a set of two dipoles with dipole moments and . We can replace the dipole with the geometric sum of dipoles and . Now applying to dipoles the previously obtained formulas for the intensity on the extension of the dipole axis and on the perpendicular restored to the dipole axis, in accordance with the principle of superposition we obtain:

.

Considering that , we get:

,

used here is that .

Thus, characteristic of the electric field of a dipole is that it decreases in all directions in proportion to , that is, faster than the field of a point charge.

Let us now consider the forces acting on a dipole in an electric field. In a uniform field charges + q And –q will be under the influence of forces equal in magnitude and opposite in direction (Fig. 1.16). The moment of this pair of forces will be:

The moment tends to rotate the dipole axis to the equilibrium position, that is, in the direction of the vector. There are two equilibrium states of a dipole: when the dipole is parallel to the electric field and when it is antiparallel to it. The first position will be stable, but the second will not, since in the first case, with a small deviation of the dipole from the equilibrium position, a moment of a pair of forces will arise, tending to return it to its original position; in the second case, the resulting moment takes the dipole even further from the equilibrium position.

Gauss's theorem

As mentioned above, it was agreed to draw the lines of force with such density that the number of lines piercing a unit of surface perpendicular to the lines of the site would be equal to the modulus of the vector. Then, from the pattern of tension lines, one can judge not only the direction, but also the magnitude of the vector at various points in space.

Let us consider the field lines of a stationary positive point charge. They are radial lines extending from the charge and ending at infinity. Let's carry out N such lines. Then at a distance r from the charge, the number of lines of force intersecting a unit surface of a sphere of radius r, will be equal. This value is proportional to the field strength of a point charge at a distance r. Number N you can always choose such that the equality holds

where . Since the lines of force are continuous, the same number of lines of force intersect a closed surface of any shape enclosing the charge q. Depending on the sign of the charge, the lines of force either enter this closed surface or go outside. If the number of outgoing lines is considered positive and the number of incoming lines negative, then we can omit the modulus sign and write:

Tension vector flow. Let us place an elementary pad with area . The area should be so small that the electric field strength at all its points can be considered the same. Let's draw a normal to the site (Fig. 1.17). The direction of this normal is chosen arbitrarily. The normal makes an angle with the vector. The flow of the electric field strength vector through a selected surface is the product of the surface area and the projection of the electric field strength vector onto the normal to the area:

where is the projection of the vector onto the normal to the site.

Since the number of field lines piercing a single area is equal to the modulus of the intensity vector in the vicinity of the selected area, the flow of the intensity vector through the surface is proportional to the number of field lines crossing this surface. Therefore, in the general case, the flow of the field strength vector through the area can be visually interpreted as the quantity equal to the number lines of force penetrating this area:

Note that the choice of the direction of the normal is conditional; it can be directed in the other direction. Consequently, the flow is an algebraic quantity: the sign of the flow depends not only on the configuration of the field, but also on the relative orientation of the normal vector and the intensity vector. If these two vectors form sharp corner, the flux is positive, if blunt, it is negative. In the case of a closed surface, it is customary to take the normal outside the area covered by this surface, that is, to choose the outer normal.

If the field is inhomogeneous and the surface is arbitrary, then the flow is defined as follows. The entire surface must be divided into small elements with area , calculate the stress fluxes through each of these elements, and then sum the fluxes through all elements:

Thus, field strength characterizes the electric field at a point in space. The intensity flow does not depend on the value of the field strength at a given point, but on the distribution of the field over the surface of a particular area.

Electric field lines can only begin on positive charges and end on negative ones. They cannot begin or end in space. Therefore, if there is no electric charge inside some closed volume, then full number lines entering and exiting a given volume must be equal to zero. If more lines leave the volume than enter it, then there is a positive charge inside the volume; if there are more lines coming in than coming out, then there must be a negative charge inside. When the total charge inside the volume is equal to zero or when there is no electric charge in it, the field lines penetrate through it, and full flow equal to zero.

These simple considerations do not depend on how the electric charge is distributed within the volume. It can be located in the center of the volume or near the surface that bounds the volume. The volume may contain several positive and negative charges, distributed within the volume in any way. Only the total charge determines the total number of incoming or outgoing voltage lines.

As can be seen from (1.4) and (1.5), the flow of the electric field strength vector through an arbitrary closed surface enclosing the charge q, equal to . If inside the surface there is n charges, then, according to the principle of field superposition, the total flux will be the sum of the fluxes of field strengths of all charges and will be equal to , where in this case we mean algebraic sum all charges covered by a closed surface.

Gauss's theorem. Gauss was the first to discover the simple fact that the flow of the electric field strength vector through an arbitrary closed surface must be associated with the total charge located inside this volume.

To understand how dielectrics behave in a field at the microscopic level, we first need to explain how an electrically neutral system can respond to an external electric field. The simplest case - complete absence charges - we are not interested. We know for sure that the dielectric contains electric charges- composed of atoms, molecules, ions crystal lattice etc. Therefore, we will consider the electrically neutral system that is the next simplest in design - two point charges equal in magnitude and opposite in sign + q And - q, located at a distance l from each other. Such a system is called electric dipole.

Rice. 3.6. Electric dipole

Electric field strength lines and equipotential surfaces electric dipole look like this (Fig. 3.7, 3.8, 3.9)

Rice. 3.7. Electric field strength lines of an electric dipole

Rice. 3.8. Equipotential surfaces of an electric dipole

Rice. 3.9. Electric field lines and equipotential surfaces

The main characteristic of a dipole is. Let's introduce the vector l, directed away from the negative charge (– q) to positive (+ q), then the vector R , called electric dipole moment or simply dipole moment, is defined as

Let's consider the behavior of a “hard” dipole - that is, whose distance does not change - in an external field E (Fig. 3.10).

Rice. 3.10. Forces acting on an electric dipole placed in an external field

Let the direction of the dipole moment be with the vector E corner . The positive charge of the dipole is acted upon by a force coinciding in direction with E and equal F 1 = +q E , and for negative - oppositely directed and equal F 2 = –q E . The torque of this pair of forces is equal to

Because ql = R, That M = pE sin or in vector notation

(Recall that the symbol

means vector product vectors A And b .) Thus, with a constant dipole moment of the molecule (), the mechanical moment acting on it is proportional to the tension E external electric field and depends on the angle between the vectors R And E .

Under the influence of moment of force M The dipole rotates and work is done

which goes to increase its potential energy. From here we get potential energy of a dipole in an electric field

if we put const = 0.

It can be seen from the figure that the external electric field tends to rotate the dipole in such a way that the vector of its electric moment R coincided in direction with the vector E . In this case, and, therefore, M = 0. On the other hand, when the potential energy of the dipole in the external field takes minimum value, which corresponds to the position sustainable balance. When the dipole deviates from this position, a mechanical moment again arises, which returns the dipole to its original position. Another equilibrium position when the dipole moment is directed against the field is unstable. Potential energy in this case takes maximum value and with small deviations from this position, the resulting forces do not return the dipole back, but deflect it even more.

In Fig. 3.11 shows an experiment illustrating the occurrence of the moment electrical forces, acting on a dielectric in an electric field. On an elongated dielectric sample located at a certain angle to the power lines electrostatic field, there is a moment of force that tends to turn this sample along the field. Dielectric rod suspended from the middle inside flat capacitor, turns perpendicular to its plates after feeding them high voltage from an electrostatic machine. The appearance of torque is due to the interaction of the polarized rod with the electric field of the capacitor.

Rice. 3.11. Moment of electrical forces acting on a dielectric in an electric field

When inhomogeneous field the dipole under consideration will also be acted upon by a resultant force F equal, trying to move him. We will look here special case. Let's direct the x axis along the field E . Let the dipole under the influence of the field have already rotated along the field line, so that the negative charge is located at the point with the coordinate x, and the positive charge is located at the point with coordinate X +l. Let us imagine that the magnitude of the field strength depends on the coordinate X. Then the resultant force F equals equal

The same result can be obtained from general ratio

where the energy P is defined in (3.8). If E increases with growth x, That

and the projection of the resultant force is positive. This means that it tends to pull the dipole into the region where the field strength is greater. This explains the well-known effect when neutral pieces of paper are attracted to an electrified comb. In a flat capacitor with uniform field they would remain motionless.

Let us consider several experiments illustrating the occurrence of a force acting on a dielectric placed in a non-uniform electric field.

In Fig. Figure 3.12 shows the retraction of the dielectric into the space between the plates of a flat capacitor. In a non-uniform electrostatic field, forces act on the dielectric, pulling it into a region of a stronger field.

Rice. 3.12. Drawing a liquid dielectric into a parallel-plate capacitor

This is demonstrated using a transparent vessel in which a flat capacitor is placed and a certain amount of liquid dielectric - kerosene - is poured (Fig. 3.13). The capacitor is connected to a high-voltage power source - an electrostatic machine. When it operates at the lower edge of the capacitor, in the region of a non-uniform field, a force acts on the kerosene, drawing it into the space between the plates. Therefore, the kerosene level inside the condenser is set higher than outside. After the field is turned off, the level of kerosene between the plates drops to the level in the vessel.

Rice. 3.13. Drawing kerosene into the space between the plates of a flat-plate capacitor

In real substances, dipoles formed by only two charges are rarely encountered. Usually we deal with more complex systems. But the concept of an electric dipole moment is also applicable to systems with many charges. In this case, the dipole moment is defined as

where , is the amount of charge with number i and a radius vector defining its location, respectively. In case of two charges we come to the same expression

Let our system of charges be electrically neutral. It contains positive charges, the magnitudes and locations of which we will denote by the index “+”. We will provide the index “–” absolute values negative charges and their radius vectors. Then expression (3.10) can be written as

In (3.11), in the first term the summation is carried out over all positive charges, and in the second - over all negative charges of the system.

Expressions (3.13) are similar to formulas for the center of mass in mechanics, and therefore we called them centers of positive and negative charges, respectively. With these notations and taking into account relation (3.12), we write electric dipole moment(3.11) charge systems as

Where l -vector drawn from the center of negative charges to the center positive charges. The point of our exercise is to demonstrate that any electrically neutral system of charges can be represented as some kind of equivalent dipole.

Example 3. B arbitrary point C (Fig. 2.1.7).

Rice. 2.1.7. Finding the E dipole at an arbitrary point

at .

From the above examples it is clear that the electric field strength of the system of charges is equal to geometric sum field strengths of each charge separately ( superposition principle).

2.1.6. Interaction of two dipoles

Let us consider the interaction of dipoles located along the same axis. Let us denote the distance between the centers of the dipoles as r; let this distance be much greater than the dipole arm:

(Fig. 2.1.8).

Rice. 2.1.8. Interaction of dipoles located along the same axis

The interaction force consists of four components - two repulsive forces between like charges and two attractive forces between unlike charges:

It is not difficult to generalize this expression for the case of interaction of dipoles with different electric moments and:

So if dipole moments If two dipoles are located along the same straight line and have the same direction, then they attract, and the force of attraction is proportional to the product of the electric moments of the dipoles and inversely proportional to the fourth power of the distance between them. Consequently, the dipole interaction decreases with distance much faster than the interaction between point charges.

Show yourself what will happen - attraction or repulsion - between dipoles whose moments are located on the same straight line and directed in opposite directions.

Let's calculate the interaction force between dipoles located as shown in Figure 2.1.9.

Rice. 2.1.9. Calculation of the interaction strength between dipoles

Resultant force

Assuming, as above, that we therefore have

Calculate on your own what the force will be equal to when the dipole moments are antiparallelly oriented.

Comparing expressions (2.1.18) and (2.1.19), we are convinced that, unlike central forces(gravitational and Coulomb), the strength of interaction between dipoles depends not only on the distance between them, but also on their mutual orientation. Nuclear forces have similar properties.