Find the interaction energy of two uniformly charged balls. Electric field energy

One of the most interesting and useful discoveries in mechanics is the law of conservation of energy. Knowing the formulas for the kinetic and potential energies of a mechanical system, we are able to detect the relationship between the states of the system at two different points in time, without going into the details of what happens between these moments. We want to determine now the energy of electrostatic systems. In electricity, the conservation of energy will prove equally useful in discovering many curious facts.

The law by which energy changes during electrostatic interaction is very simple; in fact, we have already discussed it. Let there be charges q 1 And q2, separated by a gap r 12 . This system has some energy because it took some work to bring the charges closer together. We counted the work done when two charges approached from a great distance; it is equal to

We know from the principle of superposition that if there are many charges, then the total force acting on any of the charges is equal to the sum of the forces acting on the part of all other charges. It follows that the total energy of a system of several charges is the sum of terms expressing the interaction of each pair of charges separately. If q¡ And qj- some two of the charges, and the distance between them rij(Fig. 8.1), then the energy of this particular pair is equal to

Total electrostatic energy U is the sum of the energies of all possible pairs of charges:

If the distribution is given by the charge density ρ, then the sum in (8.3) must, of course, be replaced by an integral.

We will talk about energy here from two points of view. First - application concepts of energy to electrostatic problems; second - different ways estimates energy values. Sometimes it is easier to calculate the work done in some case than to estimate the value of the sum in (8.3) or the value of the corresponding integral. For the sample, we calculate the energy required to collect a uniformly charged ball from the charges. The energy here is nothing but the work that is expended on collecting charges from infinity.

Imagine that we are constructing a sphere by layering spherical layers of infinitely small thickness one after the other. At each stage of the process, we collect a small amount of electricity and place it in a thin layer from r to r+dr. We continue this process until we reach the given radius A(Fig. 8.2). If Q r — is the charge of the ball at the moment when the ball is brought to radius r, then the work required to deliver the charge to the ball dQ, is equal to

If the charge density inside the ball is ρ, then the charge Q r equals

and the charge dQ equals

Example 2

Determine the electrical energy of the interaction of a charged ring with a dipole located on its axis, as shown in Fig.4. Known distances a, l, charges Q, q and ring radius R.

Solution.

When solving the problem, one should take into account all the energies of pair interactions of the charges of one body (ring) with the charges of another body (dipole). Interaction energy of a point charge q with charge Q distributed over the ring is determined by the sum

where is the charge of an infinitely small ring fragment, - the distance from this fragment to the charge q. Since everyone is the same and equal, then

Similarly, we find the interaction energy of a point charge - q with charged ring:

Summing up W 1 and W 2 , we obtain for the energy of interaction of the ring with the dipole:

Electrical energy of charged conductors

Example 3

Determine the work of electric forces when the radius of a uniformly charged sphere is reduced by a factor of 2. sphere charge q, its initial radius R.

Solution.

The electrical energy of a solitary conductor is determined by the formula, where q is the charge of the conductor, j is its potential. Considering that the potential of a uniformly charged sphere of radius R equal to , find its electrical energy:

After halving the radius of the sphere, its energy becomes equal to

The electrical forces do work

Example 4

Two metal spheres whose radii are r and 2 r, and the corresponding charges are 2 q And - q, located in vacuum at a great distance from each other. How many times will the electrical energy of the system decrease if the balls are connected by a thin wire?

Solution.

After connecting the balls with a thin wire, their potentials become the same

and the steady charges of the balls Q 1 and Q 2 are obtained as a result of the flow of charge from one ball to another. In this case, the total charge of the balls remains constant:

From these equations we find

The energy of the balls before connecting them with a wire is equal to

and after connecting

Substituting in the last expression the values Q 1 and Q 2 , we obtain after simple transformations

Example 5

Merged into one ball N\u003d 8 identical balls of mercury, the charge of each of which q. Assuming that in the initial state the mercury balls were at a great distance from each other, determine how many times the electrical energy of the system increased.

Solution.

When mercury balls merge, their total charge and volume are preserved:

Where Q- the charge of the ball, R is its radius, r is the radius of each small mercury ball. Total electrical energy N of solitary balls is equal to

The electrical energy of the ball obtained as a result of the merger

After algebraic transformations, we get

= 4.

Example 6

metal ball radius R= 1 mm and charge q\u003d 0.1 nC of a long distance is slowly approached to an uncharged conductor and stopped when the potential of the ball becomes equal to j \u003d 450 V. What work should be done for this?

Solution.

Where q 1 and q 2 - charges of conductors, j 1 and j 2 - their potentials. Since the conductor is not charged according to the condition of the problem, then

Where q 1 and j 1 charge and potential of the ball. When the ball and the uncharged conductor are at a great distance from each other,

and electrical energy of the system

In the final state of the system, when the potential of the ball became equal to j, the electrical energy of the system:

The work of external forces is equal to the increment of electrical energy:

= -0.0225 μJ.

Note that the electric field in the final state of the system is created by charges induced on the conductor, as well as by charges nonuniformly distributed over the surface of the metal ball. Calculating this field with a known geometry of the conductor and a given position of the metal ball is very difficult. We did not need to do this, since the problem does not specify the geometric configuration of the system, but the potential of the ball in the final state.

Example 7

The system consists of two concentric thin metal shells with radii R 1 and R 2 (and corresponding charges q 1 and q 2. Find electrical energy W systems. Consider also the special case where .

Solution.

The electrical energy of a system of two charged conductors is determined by the formula

To solve the problem, it is necessary to find the potentials of the inner (j 1) and outer (j 2) spheres. This is not difficult to do (see the corresponding section of the tutorial):

, .

Substituting these expressions into the formula for energy, we obtain

At , the energy is

Own electric energy and interaction energy

Example 8

Two conducting spheres whose charges q And - q, radii R 1 and R 2 are located in vacuum at a large distance from each other. Sphere of larger radius R 2 consists of two hemispheres. The hemispheres are separated, brought to the sphere of radius R 1 , and reconnect, thus forming a spherical capacitor. Determine the work of electric forces in this composition of the capacitor.

Solution.

The electric energy of two charged spheres distant from each other is equal to

The electrical energy of the resulting spherical capacitor:

The potential of the inner sphere is the potential of the outer sphere. Hence,

The work of electric forces with this composition of the capacitor:

Note that the electrical energy of a spherical capacitor W 2 is equal to the work of external forces in charging the capacitor. In this case, electrical forces do work. This work is done not only when the charged plates approach each other, but also when a charge is applied to each of the plates. That's why A EL differs from the work found above A, perfected by electric forces only when the plates approach each other.

Example 9

point charge q= 1.5 μC is located in the center of a spherical shell, over the surface of which the charge is uniformly distributed Q= 5 μC. Find the work of electric forces during the expansion of the shell - an increase in its radius from R 1 = 50 mm up to R 2 = 100 mm.

Solution.

Interaction energy of a point charge q with charges located on a spherical shell of radius R is equal to

The self-electric energy of the shell (the energy of interaction of the charges of the shell with each other) is equal to:

The work of electric forces during the expansion of the shell:

After transformations, we get

1.8 J

Another way to solve

We represent a point charge as a uniformly charged sphere of small radius r and charge q. The total electrical energy of the system is

Radius Sphere Potential r,

Radius Sphere Potential R. As the outer sphere expands, the electrical forces do work

After substitutions and transformations, we get the answer.

Example 10

What part of the electrical energy of a charged conducting sphere located in vacuum is contained within an imaginary sphere concentric with the ball, the radius of which is n times the radius of the sphere?

Solution.

Volumetric energy density of the electric field

defines electrical energy localized in an infinitesimal volume ( E is the modulus of the electric field strength vector in this volume, e is the permittivity). To calculate the total electrical energy of a charged conducting ball, let us mentally divide the entire space into infinitely thin spherical layers concentric with the charged ball. Consider one of these layers of radius r and thickness dr(see fig.5). Its volume is

and the electrical energy concentrated in the layer

tension E field of a charged conducting ball depends, as is well known, on the distance r to the center of the ball. Inside the ball , therefore, when calculating the energy, it is sufficient to consider only those spherical layers, the radius r which exceeds the radius of the ball R.

At field strength

permittivity and hence

Where q is the charge of the ball.

The total electric energy of a charged ball is determined by the integral

and the energy concentrated inside an imaginary sphere of radius nR, is equal to

Hence,


Fig.5	Fig.6	Fig.7

Example 11.

Determine the electrical energy of a system consisting of a charged conductive ball and an uncharged conductive spherical layer concentric with it (Fig. 6). Inner and outer layer radii a And b, ball radius , charge q, the system is in a vacuum.

Chapter 8

ELECTROSTATIC ENERGY

§1. Electrostatic energy of charges. uniform ball

§2. Capacitor energy. Forces acting on charged conductors

§3. Electrostatic energy of an ionic crystal

§4. Electrostatic energy of the nucleus

§5.Energy in an electrostatic field

§6. Energy of a point charge

Repeat: ch. 4 (Issue 1) "Energy Conservation"; ch. 13 and 14 (issue 1) "Work and potential energy"

§ 1. Electrostatic energy of charges. uniform ball

The law by which energy changes during electrostatic interaction is very simple; in fact, we have already discussed it. Let there be charges q 1 and q 2 , separated by a gap r 12 . This system has some energy because it took some work to bring the charges closer together. We counted the work done when two charges approached from a great distance; it is equal to

We know from the principle of superposition that if there are many charges, then the total force acting on any of the charges is equal to the sum of the forces acting on the part of all other charges. It follows that the total energy of a system of several charges is the sum of terms expressing the interaction of each pair of charges separately. If q i And q j - - some two of the charges, and the distance between them is r ij(fig. 8.1),

Fig. 8.1. The electrostatic energy of a system of particles is the sum of the electrostatic energies of each pair.

then the energy of this pair is equal to

Total electrostatic energy U is the sum of the energies of all possible pairs of charges:

If the distribution is given by the charge density r, then the sum in (8.3) must, of course, be replaced by an integral.

Imagine that we are constructing a sphere by layering spherical layers of infinitely small thickness one after the other. At each stage of the process, we collect a small amount of electricity and place it in a thin layer from r to r + dr. We continue this process until we reach the given radius A(Fig. 8.2). If Q r is the charge of the ball at the moment when the ball is brought to radius r, then the work required to deliver the charge to the ball dQ, is equal to

Fig. 8.2. The energy of a uniformly charged ball can be calculated by imagining that it was molded by sequentially layering spherical layers on top of each other.

If the charge density inside the ball is r, then the charge Q r equals

Equation (8.4) becomes

The total energy required to accumulate a full ball of charges is equal to the integral over dU from r=0 to r=a, i.e.

and if we want to express the result in terms of the full charge Q ball, then

The energy is proportional to the square of the total charge and inversely proportional to the radius. (8.7) can also be represented as follows: the average value (1/r ij) over all pairs of points inside the ball is 6/5 au.

§ 2. Capacitor energy. Forces acting on charged conductors

Consider now the energy required to charge the capacitor. If the charge Q was removed from one plate of the capacitor and transferred to another, then a potential difference arises between the plates, equal to

Where WITH - capacitor capacitance. How much work is required to charge the capacitor? Acting in exactly the same way as we did with the ball, imagine that the capacitor is already charged by transferring charge from one plate to another in small portions dQ. Work required to transfer charge dQ, is equal to

Taking V from (8.8), we write

Or, integrating from Q=0 to final charge Q, we get

This energy can also be written as

Recalling that the capacitance of a conducting sphere (with respect to infinity) is

we immediately obtain from equation (8.9) the energy of the charged sphere

This expression, of course, also applies to the energy of the subtle spherical layer fully charged Q; it turns out 5 / 6 energy uniformly charged ball [equation (8.7)].

Let's see how the concept of electrostatic energy is applied. Let's consider two questions. What is the force acting between the capacitor plates? What rotational (torque) moment around a certain axis does a charged conductor experience in the presence of another conductor with an opposite charge? It is easy to answer such questions using our expression (8.9) for the electrostatic energy of a capacitor and the principle of virtual work (see Issue 1, Chapters 4, 13 and 14).

We apply this method to determine the force acting between two plates of a flat capacitor. If we imagine that the gap between the plates has expanded by a small amount Dz, then the mechanical work performed from the outside in order to push the plates apart would be equal to

Where F- force acting between the plates. This work must be equal to the change in the electrostatic energy of the capacitor, unless the charge of the capacitor has changed.

According to equation (8.9), the energy of the capacitor was originally equal to

The change in energy (if we don't allow the amount of charge to change) is then

Equating (8.12) and (8.13), we get

which can also be written as

It is clear that this force here arises from the attraction of charges on the plates; we see, however, that we have nothing to worry about how they are distributed there; the only thing we need is to take into account the capacitance WITH.

It is easy to see how to generalize this idea to free-form conductors and other force components. Let's replace in the equation (8.14) F the component that interests us, and Dz - by a small shift in the corresponding direction. Or if we have an electrode mounted on some axis, and we want to know the torque t, then we write the virtual work in the form

where Dq is a small angular turn. Of course, now D(1/C) must be a change 1/C, corresponding to a rotation by Dq.

Fig. 8.3. What is the torque acting on the variable capacitor?

In this way we can determine the torque acting on the moving plates of the variable capacitor shown in Fig. 8.3.

Let us return to the particular case of a flat capacitor; we can take the capacitance formula derived in Chap. 6:

Where A- the area of each lining. If the gap increases by Dz, then

From (8.14) it then follows that the force of attraction between the two plates is equal to

Let's take a closer look at equation (8.17) and see if we can tell how this force is generated. If the charge on one of the plates we write in the form

then (8.17) can be rewritten as follows:

Or since the field between the plates is

One could immediately guess that the force acting on one of the plates would be equal to the charge Q this plate, multiplied by the field acting on the charge. But what is surprising is the 1/2 multiplier. The fact is that E 0 - this is not the field which acts on charges. If we imagine that the charge on the surface of the plate occupies some thin layer (Fig. 8.4), then the field will change from zero at the inner boundary of the layer to E 0 in the space outside the plates. The average field acting on surface charges is E 0 /2. That's why in (8.18) there is a factor 1 / 2 .

You should note that in calculating the virtual work, we assumed that the charge on the capacitor was constant, that the capacitor was not electrically connected to other objects, and that the total charge could not change.

Fig. 8.4. The field near the surface of the conductor varies from zero to E 0 =s/e 0 , when the surface charge layer is crossed. 1 - conductive plate; 2 - surface charge layer.

Now let us assume that during virtual displacements the capacitor is maintained at a constant potential difference. Then we would have to take

and instead of (8.15) we would have

which results in a force equal in magnitude to that obtained in equation (8.15) (because V = Q/C), but with the opposite sign!

Of course, the force acting between the plates of a capacitor does not change sign when we disconnect the capacitor from the source of electricity. In addition, we know that two plates with opposite electric charges must attract. The principle of virtual work in the second case was applied incorrectly, we did not take into account the virtual work produced by the source charging the capacitor. This means that in order to keep the potential at a constant value V, when the capacitance changes, the source of electricity must supply the capacitor with a VDC charge. But this charge comes at a potential V, so the work done by the electrical system to keep the charge constant is V 2 DC. Mechanical work.FDz plus this electrical work V 2 DC together leads to a change in the total energy of the capacitor by 1/2 V 2 DC. Therefore, mechanical work, as before, accounts for F D z=- 1 / 2 V 2 DC.

§ 3. Electrostatic energy of an ionic crystal

Let us now consider the application of the concept of electrostatic energy in atomic physics. We cannot easily measure the forces acting between atoms, but often we are interested in the difference in energies of two arrangements of atoms (for example, the energy of chemical changes). Since atomic forces are basically electrical forces, chemical energy in its main part is simply electrostatic energy.

Consider, for example, the electrostatic energy of an ionic lattice. An ionic crystal such as NaCl is made up of positive and negative ions, which can be thought of as hard spheres. They are electrically attracted until they touch; then the repulsive force comes into play, which increases rapidly if we try to bring them closer together.

For the initial approximation, imagine a set of hard spheres representing atoms in a salt crystal. The structure of such a lattice was determined using X-ray diffraction. This lattice is cubic - something like a three-dimensional chessboard. Its cross section is shown in Fig. 8.5. The gap between the ions is 2.81 E (or 2.81 10 -8 cm).

If our understanding of the system is correct, we should be able to test it by asking the following question: how much energy is needed to scatter these ions, i.e., completely divide the crystal into ions? This energy must be equal to the heat of vaporization of the salt plus the energy required to dissociate the molecules into ions. The total energy of separation of NaCl into ions, as follows from experience, is 7.92 ev per molecule.

Fig. 8.5. Cross section of a salt crystal on a scale of a few atoms.

in two perpendicular To the plane of the cross-sectional pattern will be the same staggered arrangement of ions Na And Cl (see issue 1, fig. 1.7).

Using the conversion factor

and the Avogadro number (the number of molecules in a gram-molecule)

we can represent the energy of evaporation in the form

The favorite unit of energy used by physical chemists is the kilocalorie, equal to 4190 j; so 1 ev per molecule is the same as 23 kcal/mol. The chemist would therefore say that the dissociation energy of NaCl is

Can we theoretically obtain this chemical energy by calculating how much work it takes to gut a crystal? According to our theory, it is equal to the sum of the potential energies of all pairs of ions. The easiest way to get an idea of this energy is to choose one ion and calculate its potential energy with respect to all other ions. This will give doubled energy per ion, because the energy belongs couples charges. If we need the energy associated with one of some ions, then we must take half the sum. But what we really need is energy per molecule containing two ions, so that the sum we calculate will directly give us the energy per molecule.

The energy of an ion with respect to its nearest neighbor is -e 2 /a, where e 2 =q 2 e/4pe 0 , and A- the gap between the centers of the ions. (We are considering monovalent ions.) This energy is -5.12 ev; we can already see that the answer is of the right order of magnitude. But we have yet to calculate an infinite series of terms.

Let's start by adding the energies of all the ions lying in a straight line. Counting the ion marked in FIG. 8.5 with the sign Na, our distinguished ion, we first consider those ions that lie on the same horizontal line with it. There are two chlorine ions closest to it with negative charges, each at a distance i from Na. Then there are two positive ions at distances of 2a, and so on. Denoting this sum of energies U 1 , write

The series converges slowly, so it is difficult to evaluate it numerically,

but it is known that it is equal to ln2. Means,

Now let's move on to the nearest line adjoining from above. The nearest ion is negative and is at a distance A. Then there are two positive ones at distances Ts2a. The next pair is at a distance of C5a, the next is at C10a, etc. For the entire line, a series is obtained

Such lines four: above, below, front and back. Then there are the four lines that are closest diagonally, and so on and so forth.

If you patiently do the calculations for all the lines and then add everything up, you will see that the total is as follows:

This number is slightly larger than what was obtained in (8.20) for the first line. Given that e 2 /a=- 5,12 ev, we will get

Our answer is about 10% more than the experimentally observed energy. It shows that our idea that the entire lattice is held together by electric Coulomb forces is fundamentally correct. We first obtained a specific property of macroscopic matter from our knowledge of atomic physics. Over time, we will achieve much more. The field of science that tries to understand the behavior of large masses of matter in terms of the laws of atomic behavior is called solid state physics.

But what about the error in our calculations? Why aren't they completely correct? We did not take into account the repulsion between ions at close distances. They're not completely rigid spheres, so they flatten out a bit as they get closer. But they are not very soft and flatten just a little bit. Nevertheless, some energy is spent on this deformation, and now, when the ions fly apart, this energy is released. The energy actually needed to pull all the ions apart is slightly less than what we calculated; repulsion helps overcome electrostatic attraction.

Is it possible to somehow estimate the share of this repulsion? Yes, if we know the law of the repulsive force. We are not yet able to analyze the details of the repulsion mechanism, but we can get some idea of its characteristics from macroscopic measurements. measuring compressibility crystal as a whole, one can get a quantitative idea of the law of repulsion between ions, and hence - of its contribution to energy. In this way, it was found that this contribution should be 1/9.4 of the contribution from electrostatic attraction and, naturally, have the opposite sign. If we subtract this contribution from the purely electrostatic energy, we get the number 7.99 for the dissociation energy per molecule ev. This is much closer to the observed result of 7.92 ev, but still not in perfect agreement. There is one more thing that we didn't take into account: we didn't make any assumptions about the kinetic energy of the crystal's vibrations. If we make a correction for this effect, then a very good agreement with the experimental value immediately arises. Hence, our ideas are correct: the main contribution to the energy of a crystal such as NaCl is electrostatic.

§ 4. Electrostatic energy of the nucleus

Let us now turn to another example of electrostatic energy in atomic physics - to the electrostatic energy of the atomic nucleus. Before dealing with this question, we must consider some of the properties of those basic forces (called nuclear forces) that hold protons and neutrons together in the nucleus. For the first time after the discovery of nuclei - and protons with the neutrons that make them up - it was hoped that the law of the strong, non-electric part of the force acting, for example, between one proton and another, would have some simple form, similar to, say, the law of inverse squares in electricity. If it were possible to determine this law of forces and, in addition, the forces acting between a proton and a neutron and between a neutron and a neutron, then it would be possible to theoretically describe the entire behavior of these particles in nuclei. Therefore, a great program began to unfold to study the scattering of protons in the hope of finding the law of the forces acting between them; but after thirty years of effort, nothing simple has come up. A considerable body of knowledge has accumulated about the forces acting between proton and proton, but it has been discovered that these forces are as complex as one can imagine.

By "as complex as possible" we mean that the forces depend on all the quantities on which they could depend.

First, force is not a simple function of the distance between protons. At greater distances there is attraction, at smaller distances there is repulsion.

Fig. 8.6. The strength of the interaction of two protons depends on all conceivable parameters.

Distance dependence is some complex function, still not well known. Secondly, the force depends on the orientation of the proton spin. Protons have spin, and two interacting protons can spin either in the same direction or in opposite directions. And the force when the spins are parallel is different from what happens when the spins are antiparallel (Fig. 8.6, A And b). The difference is great; it cannot be neglected.

Thirdly, the force changes markedly, depending on how parallel whether there is no gap between the protons to their spins (Fig. 8.6, c and d) or whether it is perpendicular(Fig. 8.6, A And b).

Fourth, the force, as in magnetism, depends (and even much more strongly) on the speed of the protons. And this speed dependence of the force is by no means a relativistic effect; it is large even when the speeds are much less than the speed of light. Moreover, this part of the force depends, besides the magnitude of the speed, on other things. Say, when a proton is moving close to another proton, the force varies depending on whether the orbital motion in direction coincides with the spin rotation (Fig. 8.6, e) or these two directions are opposite (Fig. 8.6, e). This is what is called the "spin-orbital" part of the force.

No less complex are the forces of interaction of a proton with a neutron and a neutron with a neutron. To this day, we do not know the mechanism that determines these forces, we do not know any simple way to understand them.

However, in one important respect, nuclear forces are still easier, what could be. Nuclear the forces acting between two neutrons coincide with the forces acting between a proton and a neutron, and with the forces acting between two protons! If in some system in which there are nuclei, we replace the neutron with a proton (and vice versa), then nuclear interactions won't change! The "fundamental reason" for this equality is not known to us, but it is a manifestation of an important principle that can be extended to the laws of interaction of other strongly interacting particles, such as n-mesons and "strange" particles.

This fact is beautifully illustrated by the arrangement of energy levels in similar nuclei.

Fig. 8.7. Energy levels of nuclei B 11 and C 11 (energy in MeV). Ground state C 11 1.982 MeV higher than the same state B 11 .

Consider a nucleus such as B 11 (boron-eleven), consisting of five protons and six neutrons. In the nucleus, these eleven particles interact with each other, performing some kind of intricate dance. But there is a combination of all possible interactions that has the lowest possible energy; this is the normal state of the nucleus and is called main. If the nucleus is perturbed (say, by hitting it with a high-energy proton or some other particle), then it can go into any number of other configurations, called excited states, each of which will have its own characteristic energy, which is higher than the energy of the ground state. In nuclear physics research, say carried out with a Van de Graaff generator, the energies and other properties of these excited states are determined experimentally. The energies of the fifteen lowest known excited states of B 11 are shown in the one-dimensional diagram in the left half of FIG. 8.7. The horizontal bar at the bottom represents the ground state. The first excited state has an energy of 2.14 mev higher than the main, the next - by 4.46 mev higher than basic, etc. Researchers are trying to find an explanation for this rather confusing pattern of energy levels; as yet, however, there is no complete general theory of such nuclear energy levels.

If in B 11 one of the neutrons is replaced by a proton, the nucleus of the carbon isotope C 11 will be obtained. The energies of the sixteen lowest excited states of the C 11 nucleus were also measured; they are shown in Fig. 8.7 right. (Dashed lines indicate levels for which experimental information is questionable.)

Looking at FIG. 8.7, we notice a striking similarity between the energy level patterns of both nuclei. The first excited states are approximately 2 mev above main. Then there is a wide slot with a width of 2.3 mev, separating the second excited state from the first, then a small jump by 0.5 mev up to the third level. Then again a big jump from the fourth to the fifth level, but between the fifth and sixth a narrow gap of 0.1 Mev. And so on. At about the tenth level, the correspondence seems to disappear, but it can still be found by labeling the levels with other characteristics, say their angular momentum, and the way in which they lose their excess energy.

The impressive similarity of the picture of the energy levels of the B 11 and C 11 nuclei is by no means a mere coincidence. It hides some physical law behind it. Indeed, it shows that even in the difficult conditions of the nucleus, replacing a neutron with a proton will change little. This can only mean that the neutron-neutron and proton-proton forces should be almost the same. Only then could we expect the nuclear configurations of five protons and six neutrons to coincide with the "five neutrons - six protons" combination.

Note that the properties of these nuclei tell us nothing about the neutron-proton forces; the number of neutron-proton combinations in both nuclei is the same. But if we compare two other nuclei, such as C 14 with its six protons and eight neutrons, and N 14, in which there are seven of them, then we will reveal the same correspondence in energy levels. It can be concluded that rr-, n-n- And R-n-forces coincide with each other in all details. An unexpected principle arose in the laws of nuclear forces. Although the forces acting between each pair of nuclear particles are very confused, the interaction forces for any of the three conceivable pairs are the same.

However, there are also some slight differences. There is no exact correspondence between the levels; in addition, the ground state of C 11 has an absolute energy (mass), which is 1.982 mev above the ground state B 11 . All other levels are also higher in absolute energy value by the same number. So the forces are not exactly equal. But we already know very well that complete, the magnitude of the forces is not exactly the same; between two protons electrical forces, because each of them is positively charged, and there are no such forces between neutrons. Perhaps the difference between B 11 and C 11 is explained by the fact that in these two cases the electrical interactions of protons are different? Or perhaps the remaining minimal level difference is caused by electrical effects? Since the nuclear forces are so strong compared to the electrical ones, the electrical effects could only slightly perturb the level energies.

In order to test this notion, or rather, to find out what consequences it will lead to, we first consider the difference in the energies of the ground states of the two nuclei. To keep the model quite simple, let us assume that the nuclei are balls of radius r (to be determined) containing Z protons. If we consider the nucleus to be a ball with a uniformly distributed charge, then we can expect that the electrostatic energy [from equation (8.7)] will be equal to

Where q e - elementary charge of the proton. Due to the fact that Z is equal to five for B 11, and six for C 11, the electrostatic energies will differ.

But with such a small number of protons, equation (8.22) is not entirely correct. If we calculate the electrical energy of interaction of all pairs of protons, considered as points approximately uniformly distributed over the ball, we will see that the value of Z 2 in (8.22) will have to be replaced by Z(Z- 1), so that the energy will be equal to

If the radius of the nucleus r is known, we can use expression (8.23) to determine the difference in the electrostatic energies of the B 11 and C 11 nuclei. But let's do the opposite: from the observed difference in energies, we calculate the radius, assuming that the entire existing difference in origin is electrostatic. In general, this is not entirely true. Energy difference 1.982 mev two basic states B 11 and C 11 includes rest energies, i.e. energies tc 2 all particles. Moving from B 11 to C 11, we replace the neutron with a proton, the mass of which is slightly smaller. So part of the energy difference is the difference in the rest masses of the neutron and proton, which is 0.784 Mev. The difference to be compared with the electrostatic energy is thus greater than 1.982 Mev; it is equal to

Substituting this energy into (8.23), for the radius B 11 or C 11 we get

Does this number make any sense? To check this, let's compare it with other definitions of the radii of these nuclei.

For example, one can determine the radius of the nucleus differently by observing how it scatters fast particles. These measurements revealed that density matter in all nuclei is approximately the same, i.e., their volumes are proportional to the number of particles contained in them. If through A denote the number of protons and neutrons in the nucleus (a number very closely proportional to its mass), it turns out that the radius of the nucleus is given by

From these measurements, we get that the radius of the nucleus B 11 (or C 1 1) should be approximately equal to

Comparing this with expression (8.24), we will see that our assumptions about the electrostatic origin of the difference in energies B 11 and C 11 are not so wrong; the discrepancy hardly reaches 15% (and this is not so bad for the first calculation according to the theory of the nucleus!).

The reason for the discrepancy is likely to be the following. According to our current understanding of nuclei, an even number of nuclear particles (in the case of B11, five neutrons with five protons) forms a sort of shell; when another particle is added to this shell, instead of being absorbed, it begins to revolve around the shell. If this is the case, then another value of electrostatic energy should be taken for the additional proton. It must be assumed that the excess energy of C 11 over B 11 is just equal to

i.e., it is equal to the energy required for another proton to appear outside the shell. This number is 5/6 of the value predicted by equation (8.23), so the new value for the radius will be 5/6 of (8.24). It agrees much better with direct measurements.

The agreement in numbers leads to two conclusions. First: the laws of electricity seem to operate at such small distances as 10 -1 3 see second: we were convinced of a remarkable coincidence - the non-electric part of the forces of interaction of a proton with a proton, a neutron with a neutron, and a proton with a neutron is the same.

§ 5. Energy in an electrostatic field

Let us now consider other ways of calculating electrostatic energy. All of them can be obtained from the main relation (8.3) by summing (over all pairs) the mutual energies of each pair of charges. First of all, we want to write an expression for the charge distribution energy. As usual, we assume that each volume element dV contains an element of charge pdv. Then equation (8.3) will be written as follows:

Note the appearance of the 1/2 multiplier. It arose due to the fact that in the double integral over dV 1 and by dV 2 each pair of charge elements was counted twice. (There is no convenient notation for the integral in which each pair is calculated only once.) Then note that the integral over dV 2 in (8.27) is simply the potential at point (1), i.e.

so that (8.27) can be written as

And since the point (2) fell out, we can simply write

This equation can be interpreted as follows. Potential charge energy rdV is equal to the product of this charge and the potential at the same point. The whole energy is therefore equal to the integral of jrdV. But besides that, there is a 1/2 multiplier. It is still needed because the energies are counted twice. The mutual energy of two charges is equal to the charge of one of them on the potential of the other at this point. Or the charge of the other on the potential from the first at the second point. So for two point charges one can write

Note that the same can be written like this:

The integral in (8.28) corresponds to the addition of both terms in brackets of expression (8.29). That's why the multiplier 1/2 is needed.

An interesting question is: where is the electrostatic energy located? True, one can ask in response: does it really matter?

Does this question make sense? If there is a pair of interacting charges, then their combination has some energy. Is it really necessary to clarify that the energy is concentrated on this charge, or on that one, or on both at once, or between them? All these questions are meaningless, because we know that in fact only the total, total energy is conserved. The idea that energy is concentrated somewhere, not really necessary.

Well, let's still assume that the fact that energy is always concentrated in a certain place (like thermal energy) is really there's a meaning. Then we could our principle of conservation of energy expand, combining it with the idea that if energy changes in some volume, then this change can be taken into account by observing the inflow or outflow of energy from the volume. You understand that our original statement about the conservation of energy will still hold perfectly if some energy disappears in one place and appears somewhere far away in another, and nothing happens in between these places (nothing - that means no special events will occur). Therefore, we can now move on to expanding our ideas about the conservation of energy. Let's call this extension the principle local(local) energy conservation. Such a principle would proclaim that the energy within any given volume changes only by an amount equal to the influx (or loss) of energy into (or out of) the volume. Indeed, such a local conservation of energy is quite possible. If this is so, then we will have a much more detailed law at our disposal than a simple statement about the conservation of total energy. And, as it turns out, in nature energy is indeed stored locally, in each place separately, and formulas can be written to show where energy is concentrated and how it flows from place to place.

There is also physical there is reason to demand that we be able to point out exactly where the energy is. According to the theory of gravity, any mass is a source of gravitational attraction. And by law E=ts 2 we also know that mass and energy are quite equivalent to each other. Therefore, any energy is a source of gravitational force. And if we couldn't know where the energy is, we wouldn't be able to know where the mass is. We could not say where the sources of the gravitational field are located. And the theory of gravity would become incomplete.

Of course, if we limit ourselves to electrostatics, then we have no way to find out where the energy is concentrated. But the complete system of Maxwellian equations of electrodynamics will provide us with incomparably more complete information (although even then, strictly speaking, the answer will not be completely definite). We will consider this issue in more detail later. And now we present only the result concerning the special case of electrostatics

Fig. 8.8. Each volume element dV=dxdydz in an electric field contains energy(e 0 /2) E 2 dV.

Energy is contained in the space where there is an electric field. This, apparently, is quite reasonable, because it is known that, when accelerating, charges radiate electric fields. And when light or radio waves propagate from point to point, they carry their energy with them. But there are no charges in these waves. So I would like to place the energy where there is an electromagnetic field, and not where there are charges that create this field. Thus, we describe energy not in terms of charges, but in terms of the fields they create. Indeed, we can show that equation (8.28) numerically coincides with

This formula can be interpreted by saying that in that place in space where there is an electric field, energy is also concentrated; density ee (the amount of energy per unit volume) is equal to

This idea is illustrated in Fig. 8.8.

To show that equation (8.30) agrees with our laws of electrostatics, we begin by introducing into equation (8.28) the relationship between r and j obtained in Chap. 6:

Having written the integrand component-by-component, we

we will see that

And our energy integral is then equal to

Using the Gauss theorem, the second integral can be turned into a surface integral:

We calculate this integral for the case when the surface extends to infinity (so that the integral over the volume becomes an integral over all space), and all charges are located at a finite distance from each other. The easiest way to do this is to take the surface of a sphere of huge radius centered at the origin. We know that away from all charges j changes as 1/R, and Cj as 1/R 2 . (And even faster if the total charge is zero.) The surface area of a large sphere grows only as R 2 , so the surface integral decreases as the radius of the sphere increases as

(1/R)(1/R 2)/R 2 = (1/R). So, if our integration captures the whole space (R® Ґ), then the surface integral vanishes, and we find

We see that it is possible to represent the energy of an arbitrary charge distribution as an integral of the energy density concentrated in the field.

§ 6. Energy of a point charge

The new relation (8.35) tells us that even for a single point charge q there is some electrostatic energy. The field in this case is given by the expression

so that the energy density at a distance r from the charge is

The volume element can be taken as a spherical layer with thickness dr, equal in area to 4pr 2 . The total energy will

The upper limit r=Ґ does not lead to difficulties. But since the charge is point, then we intend to integrate up to zero (r=0), which means infinity in the integral. Equation (8.35) states that the field of one point charge contains an infinite amount of energy, although we started with the idea that there is only energy between point charges. In our original form for the energy of a set of point charges (8.3), we did not include any energy of interaction of a charge with itself. What then happened? And the fact that, passing in equation (8.27) to a continuous distribution of charges, we counted the interaction of any infinitesimal charge with all other infinitesimal charges. The same accounting was kept in equation (8.35), so that when we apply it to final point charge, we include in the integral the energy that would be needed to accumulate this charge from infinitesimal parts. Indeed, you may have noticed that we could also obtain the result following from equation (8.36) from expression (8.11) for the energy of a charged ball, by setting its radius to zero.

We are forced to conclude that the idea that energy is concentrated in a field does not agree with the assumption that point charges exist. One way to overcome this difficulty is to say that elementary charges (such as the electron) are really not points at all, but small charge distributions. But the opposite can also be said: the wrongness is rooted in our theory of electricity at very small distances, or in our idea of the conservation of energy in each place separately. But each such point of view still meets with difficulties. And they have never yet been overcome; they exist to this day. A little later, when we get acquainted with some additional concepts, such as the momentum of the electromagnetic field, we will talk in more detail about these basic difficulties in our understanding of nature.

7. Electric field energy

(Examples of problem solving)

Interaction energy of charges

Example 1

Determine the electrical energy of the interaction of point charges located at the vertices of a square with sides a(see Fig.2).

Solution.

In Fig. 3, all pairwise interactions of charges are conditionally depicted by bidirectional arrows. Taking into account the energies of all these interactions, we get:

Example 2

Determine the electrical energy of the interaction of a charged ring with a dipole located on its axis, as shown in Fig.4. Known distances a, l, charges Q, q and ring radius R.

Solution.

Where
- the charge of an infinitely small fragment of the ring, - the distance from this fragment to the charge q. Since everything the same and equal
, That

Similarly, we find the interaction energy of a point charge - q with charged ring:

Summing up W 1 and W 2 , we obtain for the energy of interaction of the ring with the dipole:

Electrical energy of charged conductors

Example 3

Determine the work of electric forces when the radius of a uniformly charged sphere is reduced by a factor of 2. sphere charge q, its initial radius R.

Solution.

The electrical energy of a solitary conductor is determined by the formula
, Where q- the charge of the conductor,  - its potential. Considering that the potential of a uniformly charged sphere of radius R equals
, find its electrical energy:

After halving the radius of the sphere, its energy becomes equal to

The electrical forces do work

Example 4

Solution.

After connecting the balls with a thin wire, their potentials become the same

and the steady charges of the balls Q 1 and Q 2 are obtained as a result of the flow of charge from one ball to another. In this case, the total charge of the balls remains constant:

From these equations we find

,
.

The energy of the balls before connecting them with a wire is equal to

and after connecting

Substituting in the last expression the values Q 1 and Q 2 , we obtain after simple transformations

Example 5

Solution.

When mercury balls merge, their total charge and volume are preserved:

Where Q- the charge of the ball, R is its radius, r is the radius of each small mercury ball. Total electrical energy N of solitary balls is equal to

The electrical energy of the ball obtained as a result of the merger

After algebraic transformations, we get

= 4.

Example 6

metal ball radius R= 1 mm and charge q= 0.1 nC from a great distance, they slowly approach an uncharged conductor and stop when the potential of the ball becomes equal to  \u003d 450 V. What work should be done for this?

Solution.

Where q 1 and q 2 - charges of conductors,  1 and  2 - their potentials. Since the conductor is not charged according to the condition of the problem, then

Where q 1 and  1 charge and potential of the ball. When the ball and the uncharged conductor are at a great distance from each other,

and electrical energy of the system

In the final state of the system, when the potential of the ball became equal to , the electrical energy of the system:

The work of external forces is equal to the increment of electrical energy:

= -0.0225 μJ.

Example 7 .

The system consists of two concentric thin metal shells with radii R 1 and R 2 (
and corresponding charges q 1 and q 2. Find electrical energy W systems. Consider also the special case where
.

Solution.

The electrical energy of a system of two charged conductors is determined by the formula

To solve the problem, it is necessary to find the potentials of the inner ( 1) and outer ( 2) spheres. This is not difficult to do (see the corresponding section of the tutorial):

,
.

Substituting these expressions into the formula for energy, we obtain

At
the energy is

Own electric energy and interaction energy

Example 8

Solution.

The electric energy of two charged spheres distant from each other is equal to

The electrical energy of the resulting spherical capacitor:

The potential of the inner sphere,
- the potential of the external sphere. Hence,

The work of electric forces with this composition of the capacitor:

Note that the electrical energy of a spherical capacitor W 2 is equal to the work of external forces in charging the capacitor. In this case, the electric forces do work
. This work is done not only when the charged plates approach each other, but also when a charge is applied to each of the plates. That's why A EL differs from the work found above A, perfected by electric forces only when the plates approach each other.

Example 9

Solution.

Interaction energy of a point charge q with charges located on a spherical shell of radius R is equal to

The self-electric energy of the shell (the energy of interaction of the charges of the shell with each other) is equal to:

The work of electric forces during the expansion of the shell:

After transformations, we get

1.8 J

Another way to solve

We represent a point charge as a uniformly charged sphere of small radius r and charge q. The total electrical energy of the system is

Radius Sphere Potential r,

Radius Sphere Potential R. As the outer sphere expands, the electrical forces do work

After substitutions and transformations, we get the answer.

Volumetric energy density of the electric field

Example 10 .

Solution.

Volumetric energy density of the electric field

defines electrical energy
, localized in an infinitesimal volume
(E- module of the electric field strength vector in this volume,  - permittivity). To calculate the total electrical energy of a charged conducting ball, let us mentally divide the entire space into infinitely thin spherical layers concentric with the charged ball. Consider one of these layers of radius r and thickness dr(see fig.5). Its volume is

and the electrical energy concentrated in the layer

tension E field of a charged conducting ball depends, as is well known, on the distance r to the center of the ball. inside the ball
, therefore, when calculating the energy, it suffices to consider only those spherical layers whose radius r which exceeds the radius of the ball R.

At
field strength

the dielectric constant
and therefore

Where q is the charge of the ball.

The total electric energy of a charged ball is determined by the integral

and the energy concentrated inside an imaginary sphere of radius nR, is equal to

Hence,

Example 11.

Solution.

Induced charges are distributed on the inner and outer surfaces of the spherical layer. Their algebraic sum is zero, so the induced charges do not create an electric field when
, Where r is the distance from the center of the system. In area
the field strength of the induced charges is also zero, since they are uniformly distributed over spherical surfaces. Thus, the electric field of the system coincides with the field of a sphere uniformly charged over the surface, with the exception of the inner region of the spherical layer, where E= 0. Figure 7 shows an approximate graph of the dependence
. Omitting detailed calculations (see example 10), we write for the electrical energy of the system:

Where
,
,
. After integration, we get

Example 12.

Initial charge q distributed uniformly over the volume of a sphere of radius R. Then, due to mutual repulsion, the charges pass to the surface of the ball. What work is done by the electrical forces? Consider the dielectric constant equal to unity.

Solution.

The work of electrical forces is equal to the loss of electrical energy:

Where W 1 is the electric energy of a sphere uniformly charged over the volume, W 2 is the energy of the same ball uniformly charged over the surface. Since the total charge is the same in both cases, the electric field outside the ball does not change when the charge passes from the volume to the surface. The electric field and energy change only inside the ball.

Using the Gauss theorem, one can derive a formula for the field strength inside a uniformly charged ball at a distance r from its center:

The electrical energy concentrated inside the ball is determined by the integral:

When all the charges passed to the surface of the ball, the electric field, and hence the energy of the electric field inside the ball, became equal to zero. Thus,

Electric charge is a physical quantity that characterizes the ability of particles or bodies to enter into electromagnetic interactions. Electric charge is usually denoted by the letters q or Q. In the SI system, electric charge is measured in Coulomb (C). A free charge of 1 C is a gigantic amount of charge, practically not found in nature. As a rule, you will have to deal with microcoulombs (1 μC = 10 -6 C), nanocoulombs (1 nC = 10 -9 C) and picocoulombs (1 pC = 10 -12 C). Electric charge has the following properties:

This factor is called electric point potential. That is: in electromagnetism, the electric potential or electrostatic potential is the field equivalent to the potential energy associated with the static electric field divided by the electric charge of the particle under test. Like a good potential, only the physical potential differences have a physical meaning. Electrostatics is a part of the study of electricity, which studies electric charges without movement, that is, at rest.

Electrostatic and electrodynamics

Electrostatic shielding makes the electric field zero. This is due to the distribution of excess electric charges in the conductor. Loads of the same signal tend to go off until they reach rest. While electrostatics studies electric charges without motion, electrodynamics studies charges in motion.

1. Electric charge is a kind of matter.

2. The electric charge does not depend on the movement of the particle and on its speed.

3. Charges can be transferred (for example, by direct contact) from one body to another. Unlike body mass, electric charge is not an inherent characteristic of a given body. The same body in different conditions can have a different charge.

Thus, electrostatics and electrodynamics are fields of study in physics that deal with different aspects of electricity. In addition to these areas, there is also electromagnetism, which studies the ability of electricity to attract and repress poles.

After equilibrium, sphere A is brought into contact with another identical sphere C, which has an electric charge of 3e. What will be the electric charge density of this region? The hydrophobic nature of polyurethane is due to the force of electrostatic repulsion between material molecules and water molecules, a physical phenomenon that occurs between bodies with electric charges of the same signal. It is correct to say that the force of electrostatic repulsion.

4. There are two types of electric charges, conventionally named positive And negative.

5. All charges interact with each other. At the same time, like charges repel each other, unlike charges attract. The forces of interaction of charges are central, that is, they lie on a straight line connecting the centers of charges.

This is an occasion to return to the examples above and ask yourself why the spring stops fast enough to oscillate, like the swing, if not kept moving. This is because there are frictions and they generate heat without us realizing it. The energy is very constant, but some is dissipated as heat.

Material, reservoir of electrical and nuclear energy

However, unlike mass, charge can be either positive or negative: force is then attractive if the charges are of opposite signs, but repulsive if they are of the same sign. In an electrical cell or other generator, positive electrical charges are distributed at the positive pole, and negative electrical charges are distributed at the opposite pole.

6. There is the smallest possible (modulo) electric charge, called elementary charge. Its meaning:

e= 1.602177 10 -19 C ≈ 1.6 10 -19 C

The electric charge of any body is always a multiple of the elementary charge:

Where: N is an integer. Please note that it is impossible to have a charge equal to 0.5 e; 1,7e; 22,7e and so on. Physical quantities that can take only a discrete (not continuous) series of values are called quantized. The elementary charge e is a quantum (the smallest portion) of the electric charge.

In addition to its manifestations in electricity, this "Coulomb" interaction is responsible for the stability of matter. Nuclei of positive electric charge attract negative electrons, which causes them to form atoms, which themselves attract each other. Moreover, when a chemical reaction occurs, the result is a reorganization of the nuclei and electrons and a modification of the Coulomb energy. This is called chemical energy. A fuel such as coal, gasoline, or hydrogen is a reservoir of chemical energy, but this energy is nothing but Coulomb energy.

In an isolated system, the algebraic sum of the charges of all bodies remains constant:

The law of conservation of electric charge states that in a closed system of bodies processes of the birth or disappearance of charges of only one sign cannot be observed. It also follows from the law of conservation of charge if two bodies of the same size and shape that have charges q 1 and q 2 (it doesn’t matter what sign the charges are), bring into contact, and then back apart, then the charge of each of the bodies will become equal:

The elastic energy of the spring, which we discussed above, is also a consequence of the Coulomb interaction. In nuclear nuclei, there are also nuclear interactions that are very close to the nearest and, therefore, are important only within these nuclei. They bind nucleons, i.e. protons and neutrons. Thus, enormous energy can be released by combining light nuclei. Huge energy is also obtained by fissioning heavy nuclei such as uranium, which is produced in the A bomb or in a nuclear reactor by nuclear fission.

electric field

w = 1 2 ε 0 E2 + 1 2 E P. (11)

IN formula (11), the first term expresses the energy density of the electric field in vacuum, and the second term expresses the energy spent on the polarization of a unit volume of the dielectric.

IN in the general case of an inhomogeneous electric field, its energy in a certain volume V can be calculated using the formula

4. Ponderomotive forces. Application of the law of conservation of energy to the calculation of ponderomotive forces.

Any charged body placed in an electric field is subject to a mechanical force. Ponderomotive forces are forces acting from the electric field on macroscopic charged bodies..

Let us determine the force of mutual attraction between oppositely charged plates of a flat capacitor (ponderomotive force) in two ways.

On the one hand, this force can be defined as the force F 2 acting on the second plate from the first

F 2= Q 2E 1, (14)

where Q 2 is the charge on the second plate, E 1 is the field strength of the first plate. The amount of charge Q 2 of the second plate is determined by the formula

Q 2 = σ 2 S , (15)

where σ 2 is the surface charge density on the second plate, and the strength E 1 of the field created by the first plate is calculated by the formula

E 1 = σ 1 , (16)

where σ 1 is the surface charge density on the first plate. We substitute formulas (16) and (15) into formula (14)

Considering that σ = D = ε 0 ε E , we obtain a formula for the force acting on one plate from the other

For the force acting per unit area of the plate, the formula will have the following form

F = ε 0 ε E 2 . (18)

Now we get the formula for the ponderomotive force using the law of conservation of energy. If a body moves in an electric field, then the ponderomotive forces

field, work A will be done. According to the law of conservation of energy, this work will be done due to the energy of the field, that is

A + W = 0 or A = W . (19)

The work of changing the distance between the plates of a charged capacitor by dx is determined by the formula

where F is the force of interaction between the plates (ponderomotive force).

The energy of a charged capacitor is determined by formula (9). When one of the plates is displaced by a distance dx, the energy of the capacitor will change by W

As you can see, formulas (18) and (22) are the same. At the same time, the use of the law of conservation of energy to calculate ponderomotive forces greatly simplifies the calculations.

Questions for self-examination:

1. Derive a formula for the energy of a solitary charged conductor and a system of conductors.

2. What is the carrier of electrical energy? What is meant by volume

interaction between the plates of a charged capacitor?