Question number 1
Electric field. To explain the nature of the electrical interactions of charged bodies, it is necessary to admit the presence in the space surrounding the charges of a physical agent that carries out this interaction. In accordance with short-range theory, which asserts that force interactions between bodies are carried out through a special material environment surrounding the interacting bodies and transmitting any changes in such interactions in space with a finite speed, such an agent is electric field.
The electric field is created by both stationary and moving charges. The presence of an electric field can be judged, first of all, by its ability to exert a forceful effect on electric charges, moving and stationary, as well as by the ability to induce electric charges on the surface of conducting neutral bodies.
The field created by stationary electric charges is called stationary electric, or electrostatic field. It is a special case electromagnetic field, through which force interactions are carried out between electrically charged bodies, moving in the general case in an arbitrary manner relative to the frame of reference.
Electric field strength. A quantitative characteristic of the force action of an electric field on charged bodies is the vector quantity E called electric field strength.
E= F / q NS.
It is determined by the ratio of strength F acting from the field on a point test charge q pr, placed at the considered point of the field, to the value of this charge.
The concept of "test charge" assumes that this charge does not participate in the creation of an electric field and is so small that it does not distort it, that is, it does not cause a redistribution in space of the charges that create the field under consideration. In the SI system, the unit of tension is 1 V / m, which is equivalent to 1 N / C.
The field strength of a point charge. Using Coulomb's law, we find an expression for the strength of the electric field created by a point charge q in a homogeneous isotropic medium at a distance r from charge:
In this formula r- radius vector connecting charges q and q pr. From (1.2) it follows that the intensity E point charge fields q at all points of the field is directed radially from the charge at q> 0 and to the charge at q< 0.
Superposition principle. The intensity of the field created by a system of stationary point charges q 1 , q 2 , q 3, ¼, q n, is equal to the vector sum of the electric field strengths created by each of these charges separately:
, where r i- the distance between the charge q i and the considered point of the field.
Superposition principle, allows you to calculate not only the field strength of the system of point charges, but also the field strength in systems where there is a continuous charge distribution. The charge of a body can be represented as the sum of elementary point charges d q.
Moreover, if the charge is distributed with linear density t, then d q= td l; if the charge is distributed with surface density s, then d q= d l and d q= rd l if the charge is distributed with bulk density r.
Question number 2
Electric induction vector flux. The flux of the electric induction vector is determined similarly to the flux of the electric field strength vector
dF D = D d S
In the definitions of flows, some ambiguity is noticeable due to the fact that for each surface, two normals of the opposite direction can be specified. For a closed surface, the outer normal is considered positive.
Gauss's theorem. Consider a point positive electric charge q located inside an arbitrary closed surface S (Fig. 1.3). The flux of the induction vector through an element of the surface dS is equal to 
Component dS D = dS cosa of surface element d S in the direction of the induction vector D considered as an element of a spherical surface of radius r, in the center of which there is a charge q.
Taking into account that dS D / r 2 is equal to the elementary solid angle dw, under which the surface element dS is visible from the point where the charge q is located, we transform expression (1.4) to the form dF D = q dw / 4p, whence, after integration over the entire space surrounding the charge, i.e., within the solid angle from 0 to 4p, we get
The flux of the electric induction vector through a closed surface of arbitrary shape is equal to the charge contained inside this surface.
If an arbitrary closed surface S does not cover a point charge q, then, having constructed a conical surface with a vertex at the point where the charge is located, we divide the surface S into two parts: S 1 and S 2. Vector stream D through the surface S we find as the algebraic sum of flows through the surfaces S 1 and S 2:
.
Both surfaces from the point where the charge q is located are visible at the same solid angle w. Therefore the flows are equal
Since the external normal to the surface is used when calculating the flow through a closed surface, it is easy to see that the flow Ф 1D< 0, тогда как поток Ф 2D >0. Total flux Ф D = 0. This means that the flux of the electric induction vector through a closed surface of arbitrary shape does not depend on the charges located outside this surface.
If the electric field is created by a system of point charges q 1, q 2, ¼, q n, which is covered by a closed surface S, then, in accordance with the principle of superposition, the flux of the induction vector through this surface is defined as the sum of the fluxes created by each of the charges. The flux of the electric induction vector through a closed surface of arbitrary shape is equal to the algebraic sum of the charges covered by this surface:
It should be noted that the charges q i do not have to be pointwise; a necessary condition is that the charged area must be completely covered by the surface. If in the space bounded by the closed surface S, the electric charge is distributed continuously, then it should be assumed that each elementary volume dV has a charge. In this case, on the right-hand side of the expression, the algebraic summation of charges is replaced by integration over the volume enclosed inside the closed surface S:
This expression is the most general formulation of the Gauss theorem: the flux of the electric induction vector through a closed surface of arbitrary shape is equal to the total charge in the volume covered by this surface, and does not depend on the charges located outside the surface under consideration 
.
Question number 3
Potential energy of a charge in an electric field. The work done by the forces of an electric field when moving a positive point charge q from position 1 to position 2, we represent as a change in the potential energy of this charge: 
, where W n1 and W n2 - potential energies of the charge q in positions 1 and 2. With a small movement of the charge q in the field created by a positive point charge Q, the change in potential energy is 
... With the final movement of the charge q from position 1 to position 2, located at distances r 1 and r 2 off charge Q,. If the field is created by a system of point charges Q 1 ,Q 2, ¼, Q n, then the change in the potential energy of the charge q in this field: 
... The above formulas allow you to find only the change potential energy of a point charge q rather than potential energy itself. To determine the potential energy, it is necessary to agree at what point in the field to consider it equal to zero. For the potential energy of a point charge q located in an electric field created by another point charge Q, we get
, where C Is an arbitrary constant. Let the potential energy be zero at an infinitely large distance from the charge Q(at r® ¥), then the constant C= 0 and the previous expression takes the form. In this case, the potential energy is defined as the work of moving a charge by field forces from a given point to an infinitely remote.In the case of an electric field created by a system of point charges, the potential energy of the charge q:
.
Potential energy of a system of point charges. In the case of an electrostatic field, potential energy serves as a measure of the interaction of charges. Let there be a system of point charges in space Q i(i = 1, 2, ... , n). Energy of interaction of all n charges will be determined by the ratio, where r ij - the distance between the corresponding charges, and the summation is performed in such a way that the interaction between each pair of charges is taken into account once.
Potential of the electrostatic field. A field of conservative force can be described not only by a vector function, but an equivalent description of this field can be obtained by defining a suitable scalar value at each of its points. For an electrostatic field, this value is electrostatic potential, defined as the ratio of the potential energy of the test charge q to the value of this charge, j = W NS / q, whence it follows that the potential is numerically equal to the potential energy possessed by a unit positive charge at a given point of the field. The unit of measurement for potential is Volt (1 V).
Point charge field potentialQ in a homogeneous isotropic medium with a dielectric constant e:.
Superposition principle. The potential is a scalar function; the principle of superposition is valid for it. So for the field potential of a system of point charges Q 1, Q 2 ¼, Q n we have, where r i is the distance from the point of the field with potential j to the charge Q i... If the charge is arbitrarily distributed in space, then, where r- distance from the elementary volume d x, d y, d z to point ( x, y, z), where the potential is determined; V- the volume of space in which the charge is distributed.
Potential and work of electric field forces. Based on the definition of the potential, it can be shown that the work of the forces of the electric field when moving a point charge q from one point of the field to another is equal to the product of the magnitude of this charge by the potential difference at the starting and ending points of the path, A = q(j 1 - j 2).
It is convenient to write the definition as follows:
Question number 4
To establish a connection between the strength characteristic of the electric field - tension and its energy characteristic - potential consider the elementary work of the electric field forces on an infinitely small displacement of a point charge q: d A = qE d l, the same work is equal to the decrease in the potential energy of the charge q: d A = - d W NS = - q d, where d is the change in the potential of the electric field along the displacement length d l... Equating the right-hand sides of the expressions, we get: E d l= -d or in Cartesian coordinate system
E x d x + E y d y + E z d z =-d, (1.8)
where E x,E y,E z- the projection of the tension vector on the axis of the coordinate system. Since expression (1.8) is a total differential, then for the projections of the intensity vector we have
where .
The expression in parentheses is gradient potential j, i.e.
E= - grad = -Ñ.
The intensity at any point of the electric field is equal to the gradient of the potential at this point, taken with the opposite sign... A minus sign indicates that the tension E directed in the direction of decreasing potential.
Consider the electric field created by a positive point charge q(fig. 1.6). Potential of the field at a point M, the position of which is determined by the radius vector r, equals = q/ 4pe 0 e r... Direction of the radius vector r coincides with the direction of the tension vector E, and the potential gradient is directed in the opposite direction. Projection of the gradient onto the direction of the radius vector
... The projection of the potential gradient onto the direction of the vector t perpendicular to the vector r, is equal to 
,
i.e., in this direction, the potential of the electric field is constant(= const).
In the considered case, the direction of the vector r coincides with the direction
power lines. Summarizing the obtained result, it can be argued that at all points of the curve orthogonal to the lines of force, the potential of the electric field is the same... The locus of points with the same potential is the equipotential surface, orthogonal to the lines of force.
Equipotential surfaces are often used when graphing electric fields. Usually equipotentials are drawn in such a way that the potential difference between any two equipotential surfaces is the same. Here is a 2D picture of the electric field. Lines of force are shown by solid lines, equipotentials - by dashed lines.
Such an image allows you to say in which direction the vector of the electric field strength is directed; where the tension is more, where it is less; where the electric charge, placed at one point or another of the field, will begin to move. Since all points of the equipotential surface are at the same potential, moving the charge along it does not require work. This means that the force acting on the charge is always perpendicular to the displacement.
Question number 5
If the conductor is informed of an excess charge, then this charge distributed over the surface of the conductor... Indeed, if inside the conductor select an arbitrary closed surface S, then the flux of the electric field strength vector through this surface should be equal to zero. Otherwise, an electric field will exist inside the conductor, which will lead to the movement of charges. Therefore, in order for the condition
The total electric charge inside this arbitrary surface must be zero.
The electric field strength near the surface of a charged conductor can be determined using the Gauss theorem. To do this, select on the surface of the conductor a small arbitrary area d S and, considering it as a base, construct a cylinder on it with generator d l(fig. 3.1). On the surface of the conductor, the vector E directed normal to this surface. Therefore, the flow of the vector E through the lateral surface of the cylinder due to the smallness of d l is zero. The flux of this vector through the lower base of the cylinder, which is inside the conductor, is also zero, since there is no electric field inside the conductor. Therefore, the flow of the vector E through the entire surface of the cylinder is equal to the flow through its upper base d S ":, where E n is the projection of the electric field strength vector onto the external normal n to site d S. 
According to Gauss's theorem, this flux is equal to the algebraic sum of electric charges covered by the surface of the cylinder, referred to the product of the electrical constant and the relative permittivity of the medium surrounding the conductor. There is a charge inside the cylinder, where is the surface charge density. Hence 
and, that is, the strength of the electric field near the surface of a charged conductor is directly proportional to the surface density of electric charges located on this surface.
Experimental studies of the distribution of excess charges on conductors of various shapes have shown that the distribution of charges on the outer surface of the conductor depends only on the shape of the surface: the greater the curvature of the surface (the smaller the radius of curvature), the higher the surface charge density.
In the vicinity of areas with small radii of curvature, especially near the tip, due to high strength values, gas ionization occurs, for example, air. As a result, ions of the same name with the charge of the conductor move in the direction from the surface of the conductor, and ions of the opposite sign to the surface of the conductor, which leads to a decrease in the charge of the conductor. This phenomenon is called drainage of charge.
Excess charges on the inner surfaces of closed hollow conductors absent.
If a charged conductor is brought into contact with the outer surface of an uncharged conductor, then the charge will be redistributed between the conductors until their potentials become equal.
If the same charged conductor touches the inner surface of the hollow conductor, then the charge is transferred to the hollow conductor completely.
In conclusion, let us note one more phenomenon inherent only in conductors. If an uncharged conductor is placed in an external electric field, then its opposite parts in the direction of the field will have charges of opposite signs. If, without removing the external field, the conductor is separated, then the separated parts will have opposite charges. This phenomenon is called electrostatic induction.
Question number 8
All substances, in accordance with their ability to conduct electric current, are subdivided into conductors, dielectrics and semiconductors... Conductors are substances in which electrically charged particles - charge carriers- are able to move freely throughout the volume of the substance. Conductors include metals, solutions of salts, acids and alkalis, molten salts, and ionized gases.
We will restrict consideration solid metal conductors having crystal structure... Experiments show that with a very small potential difference applied to a conductor, the conduction electrons contained in it come into motion and move through the volume of metals almost freely.
In the absence of an external electrostatic field, the electric fields of positive ions and conduction electrons are mutually compensated, so that the strength of the resulting internal field is zero.
When a metal conductor is introduced into an external electrostatic field with a strength E 0 Coulomb forces, directed in opposite directions, begin to act on ions and free electrons. These forces cause a displacement of charged particles inside the metal, and mainly free electrons are displaced, and positive ions located in the nodes of the crystal lattice practically do not change their position. As a result, an electric field with an intensity of E ".
The displacement of charged particles inside the conductor stops when the total field strength E in a conductor, equal to the sum of the strengths of the external and internal fields, will become equal to zero: 
We represent the expression relating the strength and potential of the electrostatic field in the following form:
where E- the intensity of the resulting field inside the conductor; n- the internal normal to the surface of the conductor. From the equality to zero of the resulting tension E it follows that in within the volume of a conductor, the potential has the same value: .
The results obtained allow us to draw three important conclusions:
1. At all points inside the conductor, the field strength, that is, the entire volume of the conductor equipotential.
2. With a static distribution of charges along the conductor, the intensity vector E on its surface should be directed along the normal to the surface, otherwise, under the action of the tangent to the surface of the conductor, the components of the intensity of the charges should move along the conductor.
3. The surface of the conductor is also equipotential, since for any point on the surface
Question number 10
If two conductors have such a shape that the electric field they create is concentrated in a limited area of space, then the system formed by them is called capacitor, and the conductors themselves call covers capacitor.
Spherical capacitor. Two conductors in the form of concentric spheres with radii R 1 and R 2 (R 2 > R 1), form a spherical capacitor. Using Gauss's theorem, it is easy to show that the electric field exists only in the space between the spheres. The strength of this field 
,
where q- electric charge of the inner sphere; is the relative dielectric constant of the medium filling the space between the plates; r is the distance from the center of the spheres, and R 1 r R 2. Potential difference between plates 
and the capacitance of the spherical capacitor.
Cylindrical condenser represents two conducting coaxial cylinders with radii R 1 and R 2 (R 2 > R 1). Neglecting the edge effects at the ends of the cylinders and assuming that the space between the plates is filled with a dielectric medium with a relative permittivity, the field strength inside the capacitor can be found by the formula: 
,
where q- charge of the inner cylinder; h- the height of the cylinders (covers); r- distance from the axis of the cylinders. Accordingly, the potential difference between the plates of a cylindrical capacitor and its capacitance is 
. .
Flat capacitor. Two flat parallel plates of the same area S located at a distance d from each other, form flat capacitor... If the space between the plates is filled with a medium with a relative dielectric constant, then when a charge is imparted to them q the electric field strength between the plates is equal, the potential difference is equal to. Thus, the capacitance of a flat capacitor.
Series and parallel connection of capacitors.
At serial connection n capacitors, the total capacity of the system is
Parallel connection n capacitors form a system, the electrical capacity of which can be calculated as follows:
Question number 11
Energy of a charged conductor. The surface of the conductor is equipotential. Therefore, the potentials of those points at which point charges d q, are the same and equal to the potential of the conductor. Charge q located on a conductor can be considered as a system of point charges d q... Then the energy of the charged conductor
Taking into account the definition of capacity, you can write
Any of these expressions defines the energy of a charged conductor.
Energy of a charged capacitor. Let the potential of the capacitor plate, on which the charge is located, + q, is equal, and the potential of the plate on which the charge is located is q, is equal. The energy of such a system
The energy of a charged capacitor can be represented as
Electric field energy. The energy of a charged capacitor can be expressed in terms of quantities characterizing the electric field in the gap between the plates. Let's do this using the example of a flat capacitor. Substitution of the expression for the capacitance into the formula for the capacitor energy gives
Private U / d equal to the field strength in the gap; work S· d represents the volume V occupied by the field. Hence, 
If the field is uniform (which takes place in a flat capacitor at a distance d much smaller than the linear dimensions of the plates), then the energy contained in it is distributed in space with constant density w... Then bulk energy density electric field is equal to
Taking into account the ratio, we can write
In an isotropic dielectric, the directions of the vectors D and E coincide and
Substitute the expression, we get
The first term in this expression coincides with the energy density of the field in vacuum. The second term represents the energy spent on the polarization of the dielectric. Let us show this by the example of a non-polar dielectric. The polarization of a non-polar dielectric is that the charges that make up the molecules are displaced from their positions under the action of an electric field E... Per unit volume of the dielectric, the work expended on the displacement of charges q i by d r i is
The expression in brackets is the dipole moment per unit volume or the polarization of the dielectric R... Hence, .
Vector P related to vector E ratio. Substituting this expression into the formula for work, we get
After integrating, we determine the work expended on the polarization of a unit volume of the dielectric.
Knowing the energy density of the field at each point, one can find the energy of a field enclosed in any volume V... To do this, you need to calculate the integral: 
Energy density of the electrostatic field
Using (66), (50), (53), we transform the formula for the capacitor energy as follows:, where is the volume of the capacitor. Let's split the last expression by: 
... The quantity has the meaning of the energy density of the electrostatic field.
Question number 12
A dielectric placed in an external electric field polarizes under the influence of this field. The polarization of a dielectric is the process of acquiring a nonzero macroscopic dipole moment.
The degree of polarization of a dielectric is characterized by a vector quantity called polarization or polarization vector (P). Polarization is defined as the electrical moment of a unit volume of a dielectric,
where N- the number of molecules in the volume. Polarization P often called polarization, meaning by this a quantitative measure of this process.
In dielectrics, the following types of polarization are distinguished: electronic, orientational and lattice (for ionic crystals).
Electronic type of polarization characteristic of dielectrics with non-polar molecules. In an external electric field, positive charges inside the molecule are displaced in the direction of the field, and negative in the opposite direction, as a result of which the molecules acquire a dipole moment directed along the external field
The induced dipole moment of the molecule is proportional to the strength of the external electric field, where is the polarizability of the molecule. The polarization value in this case is equal to, where n- concentration of molecules; - the induced dipole moment of the molecule, which is the same for all molecules and the direction of which coincides with the direction of the external field.
Orientation type of polarization characteristic of polar dielectrics. In the absence of an external electric field, molecular dipoles are randomly oriented, so that the macroscopic electric moment of the dielectric is zero.
If such a dielectric is placed in an external electric field, then a moment of forces will act on the dipole molecule (Fig. 2.2), tending to orient its dipole moment in the direction of the field strength. However, full orientation does not occur, since the thermal motion tends to destroy the action of the external electric field.
This polarization is called orientation polarization. The polarization in this case is equal to, where<p> is the average value of the component of the dipole moment of the molecule in the direction of the external field.
Lattice polarization characteristic of ionic crystals. In ionic crystals (NaCl, etc.) in the absence of an external field, the dipole moment of each unit cell is zero (Fig. 2.3.a), under the influence of an external electric field, positive and negative ions are displaced in opposite directions (Fig. 2.3.b) ... Each cell of the crystal becomes a dipole, the crystal is polarized. This polarization is called lattice... In this case, the polarization can also be defined as, where is the value of the dipole moment of the unit cell, n- the number of cells per unit volume.
The polarization of isotropic dielectrics of any type is related to the field strength by the relation, where - dielectric susceptibility dielectric.
Question number 13
The polarization of the medium has a remarkable property: the flux of the polarization vector of the medium through an arbitrary closed surface is numerically equal to the value of uncompensated "bound" charges inside this surface, taken with the opposite sign:
(1). In the local formulation, the described property is described by the relation
(2), where is the bulk density of "bound" charges. These relations are called the Gauss theorem for the polarization of the medium (polarization vector) in integral and differential forms, respectively. If the Gauss theorem for the electric field strength is a consequence of the Coulomb law in the "field" form, then the Gauss theorem for polarization is a consequence of the definition of this quantity.
Let us prove relation (1), then relation (2) turns out to be valid by virtue of the mathematical theorem of Ostrogradskiy-Gauss.
Consider a dielectric made of non-polar molecules with a volume concentration of the latter equal to. We believe that under the action of an electric field, positive charges have shifted from the equilibrium position by an amount, and negative charges by an amount. Each molecule has acquired an electrical moment 
, and a unit volume has acquired an electrical moment. Consider an arbitrary sufficiently smooth closed surface in the described dielectric. Let us assume that the surface is drawn in such a way that, in the absence of an electric field, it does not "cross" individual dipoles, that is, the positive and negative charges associated with the molecular structure of the substance "compensate" each other.
Note, by the way, that relations (1) and (2) for and are satisfied identically.
Under the influence of an electric field, the surface area element will be crossed by positive charges from the volume in quantity. For negative charges, we have respectively the values and. The total charge transferred to the "outer" side of the surface area element (recall that is the outer normal to with respect to the volume enclosed by the surface) is equal to
Properties of the polarization vector of the medium
Having integrated the obtained expression over a closed surface, we obtain the value of the total electric charge that has left the considered volume. The latter allows us to conclude that an uncompensated charge remains in the considered volume - equal in magnitude to the charge gone. As a result, we have: 
Thus, the Gauss theorem for a vector field in an integral formulation is proved.
To consider the case of a substance consisting of polar molecules, it is sufficient in the above reasoning to replace the quantity by its average value.
The proof of the validity of relation (1) can be considered complete.
Question number 14
Two types of electric charges can be present in a dielectric medium: "free" and "bound". The first of them are not related to the molecular structure of the substance and, as a rule, can move relatively freely in space. The latter are associated with the molecular structure of the substance and, under the action of an electric field, can shift from an equilibrium position, as a rule, over very short distances.
Direct use of the Gauss theorem for a vector field when describing a dielectric medium is inconvenient because the right-hand side of the formula
(1) contains both the amount of "free" and the amount of "bound" (uncompensated) charges inside the closed surface.
If relation (1) is added term by term with the relation , we get 
, (2)
where is the total "free" charge of the volume covered by the closed surface. Relation (2) determines the advisability of introducing a special vector
As a convenient calculated quantity characterizing the electric field in a dielectric medium. The vector used to be called the electric induction vector or the electric displacement vector. The term "vector" is now being used. The integral form of the Gauss theorem is valid for a vector field: 
and, accordingly, the differential form of the Gauss theorem:
where is the volume density of free charges.
If the relation is true (for rigid electrets it is not true), then for the vector from definition (3) it follows that
where is the dielectric constant of the medium, one of the most important electrical characteristics of a substance. In electrostatics and quasi-stationary electrodynamics, the quantity is real. When considering high-frequency oscillatory processes, the phase of the oscillation of the vector, and hence of the vector, may not coincide with the phase of oscillations of the vector, in such cases the value becomes a complex-valued value.
Let us consider the question under what conditions in a dielectric medium the appearance of an uncompensated bulk density of bound charges is possible. For this purpose, we write the expression for the polarization vector in terms of the dielectric constant of the medium and the vector:
The validity of which is easy to be convinced of. The quantity of interest can now be calculated:
(3)
In the absence of a bulk density of free charges in a dielectric medium, the quantity can vanish if
a) there is no field; or b) the medium is homogeneous or c) the vectors and are orthogonal. In the general case, it is necessary to calculate the value according to relations (3).
Question number 17
Consider the behavior of vectors E and D at the interface between two homogeneous isotropic dielectrics with permittivities and in the absence of free charges at the interface.
Boundary conditions for the normal components of vectors D and E follow from Gauss's theorem. Let us select a closed surface in the form of a cylinder near the interface, the generatrix of which is perpendicular to the interface, and the bases are at an equal distance from the interface.
Since there are no free charges at the interface between dielectrics, then, in accordance with the Gauss theorem, the flux of the electric induction vector through this surface
Allocating flows through the bases and lateral surface of the cylinder
, where is the value of the tangent component averaged over the lateral surface. Passing to the limit at (in this case, it also tends to zero), we obtain 
, or finally for the normal components of the electric induction vector. For the normal components of the field strength vector, we obtain 
... Thus, when passing through the interface between dielectric media, the normal component of the vector suffers break, and the normal component of the vector continuous.
Boundary conditions for the tangent components of vectors D and E follow from the relation describing the circulation of the electric field strength vector. We construct a closed rectangular contour of length near the interface l and heights h... Taking into account that for the electrostatic field, and going around the contour clockwise, we represent the circulation of the vector E in the following form: 
,
where is the average value E n on the sides of the rectangle. Passing to the limit at, we obtain for the tangent components E .
For the tangential components of the electric induction vector, the boundary condition has the form 
Thus, when passing through the interface between dielectric media, the tangent component of the vector continuous, and the tangent component of the vector suffers break.
Refraction of electric field lines. From the boundary conditions for the corresponding component vectors E and D it follows that when passing through the interface between two dielectric media, the lines of these vectors are refracted (Fig. 2.8). Let us expand the vectors E 1 and E 2 at the interface into normal and tangential components and determine the relationship between the angles and under the condition. It is easy to see that the same law of refraction of intensity lines and displacement lines is valid for both the field strength and the induction
.
When passing to a medium with a lower value, the angle formed by the lines of tension (displacement) with the normal decreases, therefore, the lines are located less frequently. When switching to an environment with a larger line of vectors E and D, on the contrary, thicken and move away from the normal.
Question number 6
The theorem on the uniqueness of the solution of electrostatics problems (the locations of the conductors and their charges are given).
If the location of the conductors in space and the total charge of each of the conductors are given, then the vector of the electrostatic field strength at each point is uniquely determined. Dock: (by contradiction)
Let the charge on the conductors be distributed as follows:
Suppose that not only such, but also a different distribution of charges is possible:
(that is, it differs as little as you like on at least one conductor)
This means that at least at one point in space another vector E will be found, i.e. near the new density values, at least at some points of E, it will be fine. That. under the same initial conditions, with the same conductors, we obtain a different solution. Now let's change the sign of the charge to the opposite.
(it is necessary to change the sign on all conductors at once)
In this case, the form of the field lines will not change (it does not contradict either the Gauss theorem or the circulation theorem), only their direction and the vector E will change.
Now let's take a superposition of charges (a combination of two variants of charges):
(i.e. put one charge on top of another, and charge in the 3rd way)
If it does not coincide at least somewhere with, then at least in one place we will get some
3) we take the lines to infinity, without shorting them on the conductor. moreover, the closed contour L is closed at infinity. But even in this case, bypassing along the field line will not give zero circulation.
Conclusion: it means that it cannot be any other than zero, so the distribution of charges is established in a unique way -> the uniqueness of the solution, i.e. E - we find in a unique way.
Question number 7
Ticket 7. Theorem on the uniqueness of the solution of electrostatic problems. (the locations of the conductors and their potentials are given). If the location of the conductors and the potential of each of them are given, then the strength of the electrostatic field at each point is found uniquely.
(Berkeley Course)
Everywhere outside the conductor, the function must satisfy the partial differential equation:, or, otherwise, (2)
Obviously, W does not satisfy the boundary conditions. At the surface of each conductor, the function W is equal to zero, since and take the same value at the surface of the conductor. Therefore, W is a solution to another electrostatic problem, with the same conductors, but provided that all conductors are at zero potential. If so, then it can be argued that the function W is equal to zero at all points in space. If this is not the case, then it must have a maximum or minimum somewhere. The path W has an extremum at the point P; then consider a ball centered at this point. We know that the average value over the sphere of a function satisfying the Laplace equation is equal to the value of the function at the center. It is unfair if the center is the maximum or minimum of this function. Thus, W cannot have a maximum or minimum; it must be equal to zero everywhere. Hence it follows that =
Question number 28
Trm. about the circulation of the I.
I is the magnetization vector. I = = N p 1 m = N n i 1 S \ c
DV = Sdl cosα; di mol = i 1 mol NSdl cosα = cIdl cosα, N is the number of mol-l per 1 cm 3. Near the contour, we consider the substance to be homogeneous, that is, all dipoles, all molecules have the same magnetic moment. For counting, let's take a molecule whose core is located directly on the contour dl. It is necessary to calculate how many atoms will cross the cylinder 1 time => These are those whose centers lie inside this very imaginary cylinder. Thus, we are only interested in i pier - i.e. current crossing the surface supported by the contour.
Question number 9
Lecture 8. Energy of the electric field
The concept of electric field energy is inextricably linked with the concepts of its accumulation and consumption. Hence, it follows that the accumulators of this energy - electric capacitors - should also be considered. At the same time, it is essential that schoolchildren understand how much energy can be concentrated in a relatively small volume of a modern capacitor. Of particular importance are experiments showing in what processes this energy can be used for practical needs.
The study of electrical capacity and capacitors makes it possible to compare primitive, but fundamentally important methods of electrostatics with the capabilities of modern electrical measuring instruments. These include, in particular, digital multimeters widely used in everyday life, which allow measuring capacities from picofarad units. Therefore, you can first estimate the capacitance and dielectric constant using electrostatic methods, and then more accurately measure these quantities using a multimeter.
An interesting methodological problem is the substantiation of the advisability of introducing the concept of electrical capacity of a solitary conductor and the development of an optimal method for the formation of this concept.
It is unlikely that it will be possible to form the concept of the energy of an electric field in full in physics lessons. Therefore, in the classes of specialized education, extracurricular student research is necessary.
8.1. Electrical capacity of a solitary conductor
During their research, students, of course, noticed that conductors can store and store electrical charges. This property of conductors is characterized by electrical capacity. Let us find out how the potential of a solitary conductor depends on its charge. The potential can be measured relative to an infinitely distant point. In practice, it is more convenient to measure the potentials of charged bodies relative to the ground.
We put a hollow conducting ball on the rod of the electrometer, and connect the body of the electrometer to ground. We will use the electrometer as an electrostatic voltmeter that measures the potential of the ball relative to the ground, or, which is the same, the potential difference between the ball and the ground. 
With a test ball, touching the conductor of the electricity source, we transfer a certain charge inside the ball q... The electrostatic voltmeter needle will deflect to indicate a certain potential. Let's repeat the experiment, imparting charges to the hollow ball 2 q, 3q... We find that the voltmeter needle deviates, showing values 2, 3 ...
Thus, the charge ratio Q of the conducting body to its potential remains constant and characterizes electrical capacity conductor:
Let's replace the hollow sphere of the electrometer with another one, for example, of a smaller size, and repeat the experiment. We observe that when the same charges are imparted to it q, 2q, 3q, ... the voltmeter shows values that increase in proportion to the charge, but higher than in the previous series of experiments. Hence, the capacity C = Q/ this ball is less.
In the SI system, the capacitance is expressed in farads: 1 F = 1 C / 1 V.
8.2. Spherical conductor capacity
Let in a medium with dielectric constant there is a spherical conductor of radius R... If the potential at infinity is considered equal to zero, then the potential of the charged sphere
Then the electric capacity of a sphere with a radius R there is 
Thus, the capacity of a solitary conducting sphere is proportional to its radius.
Simple experiments show that bodies carrying an electric charge can be considered solitary if the surrounding bodies do not cause a significant redistribution of the charge on them.
8.3. Capacitor
Let's make a capacitor from two identical conducting plates arranged in parallel, and connect it to an electrometer that acts as a voltmeter. We put a hollow conducting sphere on the rod of the electrometer. We charge one of the plates with a test ball, transferring the charge to it q from an electrified ebony stick or other source of electricity. In this case, the voltmeter will show some voltage U between the plates.
We will transfer equal charges to the inside of the hollow sphere, and hence to the capacitor plate. In this case, we will see that the voltmeter readings increase by equal values. This means that the system of two conducting plates has a capacity
and can act as a capacitor - an electric charge storage. We emphasize that here q- the charge of one of the capacitor plates.
8.4. Capacity of a flat capacitor
Let us calculate theoretically the electric capacity of a flat capacitor. The intensity of the field created by one of its plates 
where is the surface charge density on the plate. According to the principle of superposition, the electric field strength between the plates of the capacitor is twice as large (see study 5.7): 
Since the field is uniform, the potential difference between the plates located at a distance d from each other is equal to 
Hence, the capacity of a flat capacitor is:
Let us confirm the theory by experiment. To do this, we assemble a flat capacitor, charge it and connect the plates with an electrostatic voltmeter. Leaving the charge of the capacitor unchanged, we will change its other parameters, observing the voltmeter, the readings of which are inversely proportional to the capacitance of the capacitor:
Increasing distance d between the plates of the capacitor leads to a proportional increase in the voltage between them, which means that the capacitance of the capacitor WITH ~ 1/d... By shifting the plates relative to each other so that they remain parallel, we will increase the overlapping area of the plates. S... In this case, the voltage between them decreases to the same extent, i.e. the capacitance of the capacitor grows: WITH ~ S... Fill the gap between the plates with a dielectric with a dielectric constant and see that the voltmeter readings decrease by a factor, i.e. WITH ~ .
Since the charge of the system remained unchanged, it can be concluded that the capacitance of the capacitor is directly proportional to the overlapping area of the plates, inversely proportional to the distance between them and depends on the properties of the medium, i.e. WITH ~ S/d, which confirms formula (8.2). The value of the electrical constant 0 is obtained by measuring in experiments U, q, d, S, and calculating the capacity once by formula (8.1), and another - by formula (8.2).
8.5. Parallel connection of capacitors
With parallel connection of two capacitors with capacities WITH 1 and WITH 2 voltages across them are the same and equal U, and charges q 1 and q The two are different. It is clear that the total battery charge is equal to the sum of the capacitor charges q = q 1 + q 2, and its capacity:
(8.3)
8.6. Series connection of capacitors
We connect an electrostatic voltmeter with a hollow sphere to a battery of two series-connected capacitors. Let us inform the plate of the first capacitor connected to the voltmeter with a charge + q... By induction, the second plate of this capacitor will acquire a charge - q, and the plate of the second capacitor connected to it by a conductor is charge + q... As a result, both capacitors will carry the same charge. q... In this case, the voltages across the capacitors are different. It is clear that the sum of the voltages across each of the capacitors is equal to the total battery voltage:
But U = q/WITH, U 1 = q/WITH 1 , U 2 = q/WITH 2, so the battery capacity is determined by the formula
8.7. Energy of a flat capacitor
Let us inform one of the plates of a flat capacitor to charge q such a value that the potential difference between the plates becomes equal to U... If the distance between the plates d, then the electric field strength in the capacitor E = U/d.
One of the plates of the capacitor with a charge q is located in the uniform electric field created by the second plate with strength E/ 2, therefore it is acted upon by the force of attraction to the second plate f = qE/ 2. Potential charge energy q in this field is equal to the work that the electric field does when the capacitor plates come close to each other:
Substituting into this equality the value Ed = U and using formula (8.1), we find that the energy of the electric field between the plates of the capacitor: 
(8.5)
8.8. Energy of an arbitrary capacitor
The resulting formula is valid not only for a flat, but in general for any capacitor. Indeed, the voltage across a capacitor of a given capacity is directly proportional to its charge U = q / C. If the charge has changed by a small amount q, then the electric field has done the work A = Uq... The total work of the field is obviously equal to the area under the graph:
The situation will not change if a solitary conductor is used instead of a capacitor. Its potential (relative to infinity) is = q / C, therefore, the energy of the electric field
8.9. Experimental determination of the energy stored in a capacitor
The energy of the capacitor will be measured by the thermal effect. Place a thin metal spiral in the test tube. We close the test tube with a stopper with a capillary tube, inside which there is a drop of water. We got gas thermometer- a device in which the displacement of a drop in the tube is proportional to the amount of heat released in the test tube. We connect a capacitor to the spiral through a discharge gap of two metal balls, parallel to which we connect an electrometer with a hollow ball. To charge the capacitor, we will use any source of electricity and a metal ball on an insulating handle.
We charge the capacitor to a certain voltage and, bringing the balls closer together, we discharge it through the spiral. In this case, the drop in the tube will move a certain distance. Since the discharge occurs quickly, the process of heating the air in the test tube can be considered adiabatic, i.e. occurring without heat exchange with the environment.
Let's wait until the air in the test tube cools down and the drop returns to its original position. Let's increase the voltage by two and then three times. After the discharges, the droplet will move a distance that is four and nine times greater than the initial one, respectively. We replace the capacitor with another, the capacity of which is twice as large, and charge it to its original voltage. Then, during the discharge, the drop will move two times further.
Thus, experience confirms the validity of formula (8.5) W = CU 2/2, according to which the energy stored in a capacitor is proportional to its capacity and the square of the voltage.
8.10. Electric field energy density
Let us express the energy of the electric field between the capacitor plates in such a formula that there are no quantities that characterize the capacitor itself, and only quantities that characterize the field remain. It is clear that this can only be achieved in one way: to calculate the field energy per unit volume. Since the voltage across the capacitor U = Ed, and its capacity then substitution of these expressions into the formula (8.5) gives:
The magnitude Sd represents volume V electric field in the capacitor. Therefore, the energy density of the electric field is 
proportional to the square of its intensity.
Research 8.1. Measuring the capacitance of a flat capacitor using a multimeter
Information. In recent years, many different types of digital multimeters have become available. These devices, in principle, allow you to measure voltage, current, resistance, temperature, capacity, inductance, and determine the parameters of transistors. The list of values measured by the multimeter is determined by the type of multimeter. We are now interested in multimeters that can measure capacitance; these include, for example, devices of the M890G and DT9208A types. For definiteness, in what follows, we will mean the last device.
Problem. How to experimentally confirm the validity of the theoretically obtained formula for the capacitance of a capacitor?
Exercise. Develop a demonstration experiment that allows you to confirm the validity of formula (8.2) for the capacity of a flat capacitor with an air dielectric in the lesson.
Execution option. 
Assemble a flat capacitor from the round plates included in the ESD kit and connect a multimeter to it. Use a ruler to measure the diameter of the plates and the distance between them. Using the formula (8.2), calculate the capacitance of the capacitor and compare the resulting value with the measured one. In a demo experiment, for example, the following results can be obtained: diameter of capacitor plates D= 0.23 m, distance between plates d= 0.01 m, capacity calculated by the formula: 
the multimeter shows the same value.
Change the distance between the plates, the overlapping area of the capacitor plates and introduce different dielectrics between them. In this case, the capacitance values measured by the multimeter change accordingly. Together with the students, analyze the results of the experiment and draw a conclusion regarding the validity of formula (8.2).
Research 8.2. Determination of dielectric constant by measuring capacitance
Exercise. Using a digital multimeter, determine the dielectric constants of various substances.
Execution option. Assemble a flat air dielectric capacitor, measure the distance d between plates and container WITH 0 capacitor. Measure the thickness l plane-parallel dielectric plate, carefully insert the dielectric between the plates and the multimeter, measure the capacitance WITH... According to the formula 
calculate the dielectric constant of the substance. Show students how this formula works. Measure the dielectric constants of glass, plexiglass, vinyl plastic, textolite, polyethylene, etc. Compare the resulting values with the table values.
Research 8.3. Parallel and series connection of capacitors
Exercise. Using a digital multimeter, confirm the validity of formulas (8.3) and (8.4) for the capacity of parallel and series-connected capacitors.
Execution option. 
Select radio-technical capacitors with a capacity from tens of picofarads to tens of nanofarads and use a multimeter to determine their capacities. Please note that the measured values generally do not match those indicated on the capacitor housings. This is due to the fact that the permissible error in the capacitance of radio-technical capacitors reaches 20%. Connect the capacitors in parallel, measure the resulting capacitance and make sure that it is equal to the sum of the capacitances of each of the capacitors. Then connect the capacitors in series and make sure that the reciprocal of the resulting capacitance is equal to the sum of the reciprocal of the capacitances of the connected capacitors.
Students can be offered quantitative tasks for calculating the capacity of various capacitor banks, followed by verification of the solution in a real experiment.
Research 8.4. Electric field work
Exercise... When a charged body is brought to light balls lying on the surface, they begin to bounce. Using this phenomenon, experimentally show that the work of an electric field to move a charge is proportional to the potential difference that this charge has passed: A = qU.
Execution option. 
Attach a fixed flat electrode horizontally at the bottom of a plastic bottle, and a movable electrode in parallel above it. Glue a scale with millimeter divisions to the wall of the bottle. Place a foam ball wrapped in thin aluminum foil between the electrodes. Connect the electrodes to a high-voltage source. When voltage is applied to the electrodes, the ball will start bouncing. By increasing the voltage, make the ball bounce to a height h equal to the distance d between the electrodes. In this case, the work of the electric field to move the charged ball A = qU = mgh... Double the voltage and make sure the height h will also double. Learn from experience.
Note that the potential difference is expressed in terms of the electric field strength by the formula U = Ed... Since, according to the conditions of experience, h = d, then a force constant in absolute value acts on the ball detached from the lower electrode from the side of the electric field F = Eq = mg.
Research 8.5. Electrostatic motor
Exercise. Use the electric wind phenomenon (see Study 7.7) to build a working model of an electrostatic motor.
Execution option. The first to manufacture an electrostatic motor was one of the founders of the theory of electricity, the outstanding American scientist B. Franklin. So called Franklin wheel available in any physics classroom (photo above).
At home, schoolchildren can make the simplest model of such an engine if a Segner wheel-shaped figure cut from aluminum foil is put on one of the electrodes of a piezoelectric source (photo below). By periodically pressing the source lever, they will be able to bring the resulting Franklin wheel into continuous rotation.
The photo shows a much more powerful electrostatic motor, which is able to rotate even the fan impeller. The device is assembled on a plastic bottle.
Research 8.6. Energy of a charged capacitor
Exercise. Students will remember for a long time the property of a capacitor to accumulate electrical energy if a capacitor is assembled right before their eyes and demonstrated at work. Suggest an easy way to make a capacitor that captures the imagination of schoolchildren.
Execution option. Prepare two duralumin plates, for example, 15-15 cm in size. Cut a rectangle about 20-20 cm in size from a thick plastic foil and put it between the plates to assemble the capacitor. Switch on the high-voltage source, set the voltage to 10 kV and, bringing the electrodes of the source closer together, show a spark jumping between them. Then, from the same source at the same voltage, charge the capacitor assembled on the demonstration table. Discharge the capacitor and show that a much more powerful spark is produced than between the source electrodes. Pay attention to the need to comply with safety regulations when working with capacitors.
Research 8.7. Battery of galvanic cells
Problem. Students are familiar with individual cells and batteries of galvanic cells, which are widely used in everyday life. Students know that these devices are voltage-based and capable of producing electrical current. However, the voltage of these sources does not exceed several volts, while voltages of thousands and tens of thousands of volts are used in electrostatics. Therefore, the charges on the electrodes of galvanic sources practically do not manifest themselves in any way. How can one experimentally prove that the terminals of batteries of galvanic cells actually have electric charges, the physical nature of which is the same as those found in electrostatic experiments?
Exercise. Set up an experiment to detect charges on the terminals of a battery of galvanic cells and determine their sign.
Execution option. 
The set for electrometers includes a disk capacitor, which is two metal discs 100 mm in diameter, the working surfaces of which are covered with a thin layer of varnish. One of the disks has a mount for attaching to the electrometer rod, the other is equipped with an insulating handle.
Using the specified equipment and referring to the photograph, complete the task.
Research 8.8. Estimating the energy of a charged capacitor
Information. In research 2.7, you became convinced that the energy of the electric field can be estimated from the flash of an incandescent lamp, which occurs during the discharge of the charged bodies creating the field. Indeed, during a discharge, the potential energy of stationary charges is converted into kinetic energy of moving charges, the charges are neutralized, and the field disappears. The movement of free charges along the conductor causes it to heat up.
Exercise. Prepare two batteries of 4.5 V, two electrolytic capacitors with a capacity of 1000 μF, designed for an operating voltage of at least 12 V, and four bulbs for a pocket torch for a voltage of 1 V. Prove that the energy of a charged capacitor is proportional to its capacity and the square of the voltage.
Questions for self-control
1. What is the method of introducing and forming the concept of electrical capacity of a conductor and a system of conductors?
2. How can the validity of the formula for the capacity of a flat capacitor be substantiated in a demonstration experiment?
3. How appropriate is the demonstration directly in the classroom of the essence of the method for determining the dielectric constant of a substance?
4. Propose a technique for introducing and forming the concept of the energy density of an electric field.
5. Develop a student research series on the experimental rationale for building electrostatic motors.
6. List the most striking experiments demonstrating the accumulation of electrical energy by capacitors.
7. How to prove that batteries of galvanic cells used in everyday life are not fundamentally different from electrostatic sources of electricity?
8. What experiments can confirm that the energy stored in a capacitor is proportional to its capacity and the square of the voltage?
Literature
E. I. Butikov, Kondratyev A.S. Physics: Textbook. manual: In 3 kn. Book. 2. Electrodynamics. Optics. - M .: Fizmatlit, 2004.
Demonstration experiment in physics in high school high school... T. 2. Electricity. Optics. Physics of the Atom: Ed. A.A. Pokrovsky. - M .: Education, 1972.
Mayer V.V., Mayer R.V. Electricity. Academic research: Teacher and student library. - M .: FML, 2007.
Shilov V.F. On the priority measures for the material and technical renovation of the physics room. - Educational physics, 2000, No. 4.
Let two charges q 1 and q 2 be at a distance r from each other. Each of the charges, being in the field of another charge, has a potential energy P. Using П = qφ, we define
P 1 = W 1 = q 1 φ 12 P 2 = W 2 = q 2 φ 21
(φ 12 and φ 21 are, respectively, the field potentials of charge q 2 at the point where charge q 1 and charge q 1 are located at the point where charge q 2 is located).
According to the definition of the potential of a point charge
Hence.
or 
Thus,
The energy of the electrostatic field of the system of point charges is
(12.59)
(φ і is the potential of the field created by n -1 charges (except for q i) at the point where the charge q i is located).
The energy of a solitary charged conductor
A solitary uncharged conductor can be charged to the potential φ, repeatedly transferring portions of the charge dq from infinity to the conductor. The elementary work that is done against the field forces, in this case, is equal to
The charge transfer dq from infinity to the conductor changes its potential by
(C is the electrical capacity of the conductor).
Hence,
those. when the charge dq is transferred from infinity to the conductor, we increase the potential energy of the field by
dП = dW = δA = Cφdφ
By integrating this expression, we find the potential energy of the electrostatic field of a charged conductor with an increase in its potential from 0 to φ:
(12.60)
Applying the ratio 
, we obtain the following expressions for the potential energy:
(12.61)
(q is the charge of the conductor).
Energy of a charged capacitor
If there is a system of two charged conductors (capacitor), then the total energy of the system is equal to the sum of the intrinsic potential energies of the conductors and the energy of their interaction:
(12.62)
(q is the charge of the capacitor, C is its electrical capacity.
WITH 
taking into account that Δφ = φ 1 –φ 2 = U is the potential difference (voltage) between the plates), we obtain the formula
(12.63)
The formulas are valid for any shape of the capacitor plates.
A physical quantity that is numerically equal to the ratio of the potential energy of the field contained in an element of volume to this volume is calledvolumetric energy density.
For a uniform field, the bulk energy density
(12.64)
For a flat capacitor, the volume of which is V = Sd, where S is the area of the plate, d is the distance between the plates,
But 
,
then
(12.65)
(12.66)
(E is the strength of the electrostatic field in a medium with a dielectric constant ε, D = ε ε 0 E is the electric displacement of the field).
Consequently, the volumetric energy density of a uniform electrostatic field is determined by the strength E or the displacement D.
It should be noted that the expression 
and 
are valid only for an isotropic dielectric, for which the relation p = ε 0 χE is satisfied.
Expression 
corresponds to field theory - the theory of short-range action, according to which the energy carrier is the field.
Consider the process of charging a solitary conductor. For its charge to reach the value Q, we will impart the charge to the conductor in portions d q, transferring them from an infinitely distant point 1
to the surface of the conductor to a point 2
(fig. 3.14). To transfer a new portion of the charge to the conductor 
external forces must perform work against the forces of the electric field:. Since the conductor is secluded (point 1
is infinitely far from the guide), then 
... Point potential 2
is equal to the potential of the conductor . That's why 
... If charge is transferred to the conductor q then its potential 
... Full work of external forces to charge the conductor to the value of the charge Q will be equal
.
According to the law of conservation of energy, the work of external forces on charging the conductor increases the energy of the generated electrostatic field, i.e. the conductor stores a certain energy:
.
(3.13)
Consider the process of charging a capacitor from an EMF source. During the charging process, the source transfers charges from one plate to another, and the external forces of the source do the work to increase the energy of the capacitor:
,
where Q- the charge of the capacitor after charging. Then the energy of the electric field created by the capacitor is defined as
.
(3.14)
Expression (3.14) allows you to write the value of the energy of the electrostatic field in two ways:
and 
.
Comparison of the two ratios allows us to ask the question: what is the carrier of electrical energy? Charges (first formula) or field (second formula)? Both recorded equalities are in excellent agreement with the experimental results, i.e. the calculation of the field energy can be carried out equally correctly by both formulas. However, this is observed only in electrostatics, i.e. when the energy of the field of stationary charges is calculated. When considering the theory of the electromagnetic field in the future (Chapter 8), we will see that the electric field can be created not only by stationary charges. An electrostatic field is a special case of an electromagnetic field that exists in space in the form of an electromagnetic wave. Its energy is distributed in space with a certain density. Let's introduce the concept volumetric field energy density in the following way.
We transform the last equality (3.14) for the case of a flat capacitor, using the relationship between the potential difference and the strength of the uniform field:
where 
- the volume of the condenser, i.e. the volume of the part of the space in which the electric field is created.
The volumetric energy density of the field is the ratio of the energy of the field, enclosed in a small volume of space to this volume:
.
(3.15)
Therefore, the energy of a uniform electric field can be calculated as follows: 
.
This conclusion can be extended to the case of an inhomogeneous field as follows:
,
(3.16)
where 
- such an elementary volume of space, within which the field can be considered homogeneous.
For example, let's calculate the energy of the electric field created by a solitary metal ball with a radius R charged charge Q, and located in a medium with a relative dielectric constant . Repeating the reasoning of the example from Section 2.5, we obtain the modulus of the field strength in the form of a function 
:
Then the expression for the bulk density of the field energy will take the form:
(fig. 3.15). The volume of this layer 
... Then the field energy is defined as follows:We would get a similar result if we calculated the energy of a charged ball by formula (3.13), using (3.6):
.
However, it should be remembered that this method is inapplicable if it is necessary to find the energy of the electric field, which is not contained in the entire volume of the field, but only in its part. Also, the calculation method according to formula (3.13) cannot be used to determine the field energy of a system for which the concept of “capacity” is inapplicable.
If a conductor is placed in an external electrostatic field, then it will act on its charges, which will begin to move. This process proceeds very quickly, after its completion, an equilibrium distribution of charges is established, in which the electrostatic field inside the conductor turns out to be equal to zero. On the other hand, the absence of a field inside the conductor indicates the same potential value at any point of the conductor, and also that the field strength vector on the outer surface of the conductor is perpendicular to it. If this were not the case, a component of the intensity vector would appear, directed tangentially to the surface of the conductor, which would cause the movement of charges, and the equilibrium distribution of charges would be violated.
If we charge a conductor in an electrostatic field, then its charges will be located only on the outer surface, since, in accordance with the Gauss theorem, due to the equality of the field strength inside the conductor to zero, the integral of the electric displacement vector D on a closed surface coinciding with the outer surface of the conductor, which, as was established earlier, should be equal to the charge inside the named surface, i.e., zero. This raises the question of whether we can communicate to such a conductor any, arbitrarily large, charge.To get an answer to this question, we will find the relationship between the surface charge density and the strength of the external electrostatic field.
Let us choose an infinitesimal cylinder crossing the "conductor - air" boundary so that its axis is oriented along the vector E ... We apply the Gauss theorem to this cylinder. It is clear that the flux of the electric displacement vector along the lateral surface of the cylinder will be equal to zero due to the equality of the field strength inside the conductor to zero. Therefore, the total flow of the vector D through the closed surface of the cylinder will be equal only to the flow through its base. This flow, equal to the product D∆S, where ∆S- base area, equal to the total charge σ∆S inside the surface. In other words, D∆S = σ∆S, whence it follows that
D = σ, (3.1.43)
then the strength of the electrostatic field at the surface of the conductor
E = σ /(ε 0 ε) , (3.1.44)
where ε - dielectric constant of the medium (air) that surrounds the conductor.
Since there is no field inside a charged conductor, the creation of a cavity inside it will not change anything, that is, it will not affect the configuration of the arrangement of charges on its surface. If now a conductor with such a cavity is grounded, then the potential at all points of the cavity will be zero. Based on this electrostatic protection measuring instruments from the influence of external electrostatic fields.
Now consider a conductor remote from other conductors, other charges and bodies. As we have established earlier, the potential of a conductor is proportional to its charge. It was experimentally found that conductors made of different materials, being charged to the same charge, have different potentials φ ... Conversely, conductors made of different materials that have the same potential have different charges. Therefore, we can write that Q = Cφ, where
C = Q / φ (3.1.45)
called electrical capacity(or simply capacity) a solitary conductor. The unit for measuring electrical capacity is farad (F), 1 F is the capacity of such a solitary conductor, the potential of which changes by 1 V when a charge equal to 1 C is imparted to it.
Since, as was established earlier, the potential of a ball of radius R in a dielectric medium with a dielectric constant ε
φ = (1 / 4πε 0) Q / εR, (3.1.46)
then, taking into account 3.1.45 for the capacity of the ball, we obtain the expression
C = 4πε 0 εR. (3.1.47)
From 3.1.47 it follows that a ball in vacuum and having a radius of about 9 * 10 9 km, which is 1400 times the radius of the Earth, would have a capacity of 1 F. This suggests that 1 F is a very large electrical capacity. The capacity of the Earth, for example, is only about 0.7 mF. For this reason, in practice, they use millifarads (mF), microfarads (μF), nanofarads (nF) and even picofarads (pF). Further, since ε Is a dimensionless quantity, then from 3.1.47 we obtain that the dimension of the electrical constant ε 0 - F / m.
Expression 3.1.47 says that a conductor can have a large capacitance only with a very large sizes... In practice, however, devices are required that, with small dimensions, would be able to accumulate large charges at relatively low potentials, that is, would have large capacities. Such devices are called capacitors.
We have already said that if a conductor or dielectric is brought closer to a charged conductor, charges will be induced on them so that charges of the opposite sign will appear on the side of the introduced body closest to the charged conductor. Such charges will weaken the field that is created by a charged conductor, and this will lower its potential. Then, in accordance with 3.1.45, we can talk about an increase in the capacity of a charged conductor. It is on this basis that capacitors are created.
Usually capacitor comprises two metal plates separated by dielectric... Its design should be such that the field is concentrated only between the plates. This requirement is satisfied two flat plates, two coaxial(having the same axis) cylinder different diameters and two concentric spheres... Therefore, capacitors built on such plates are called flat, cylindrical and spherical... In everyday practice, the first two types of capacitors are often used.
Under capacitor capacity understand the physical quantity WITH , which is equal to the charge ratio Q accumulated in the capacitor to the potential difference ( φ 1 - φ 2), i.e.
C = Q/(φ 1 - φ 2). (3.1.48)
Let's find the capacity of a flat capacitor, which consists of two plates with an area S spaced from each other at a distance d and having charges + Q and –Q... If d is small in comparison with the linear dimensions of the plates, then the edge effects can be neglected and the field between the plates can be considered uniform. Insofar as Q = σS, and, as shown earlier, the potential difference between two oppositely charged plates with a dielectric between them is φ 1 - φ 2 = (σ/ε 0 ε) d, then after substituting this expression in 3.1.48 we obtain
C= ε 0 εS / d. (3.1.49)
For cylindrical capacitor with length l and cylinder radii r 1 and r 2
C = 2πε 0 εl / ln (r 2 / r 1). (3.1.50)
From expressions 3.1.49 and 3.1.50 it is clearly seen how the capacitance of the capacitor can be increased. First of all, materials with the highest dielectric constant should be used to fill the space between the plates. Another obvious way to increase the capacitance of a capacitor is to reduce the distance between the plates, but this method has an important limitation – dielectric breakdown, i.e., an electric discharge through the dielectric layer. The potential difference at which an electrical breakdown of the capacitor is observed is called breakdown voltage... This value is different for each type of dielectric. As for increasing the area of plates of flat and length of cylindrical capacitors to increase their capacitance, there are always purely practical limitations on the size of capacitors, most often these are the dimensions of the entire device, which includes a capacitor or capacitors.
In order to be able to increase or decrease the capacitance, in practice, parallel or series connection of capacitors is widely used. When capacitors are connected in parallel, the potential difference across the capacitor plates is the same and is equal to φ 1 - φ 2, and the charges on them will be equal Q 1 = C 1 (φ 1 - φ 2), Q 2 = C 2 (φ 1 - φ 2), … Q n = C n (φ 1 - φ 2), therefore, the full charge of the battery from the capacitors Q will be equal to the sum of the listed charges ∑Q i, which in turn is equal to the product of the potential difference (φ 1 - φ 2) at full capacity С = ∑C i... Then for the total capacity of the capacitor bank we get
C = Q / (φ 1 - φ 2). (3.1.51)
In other words, when capacitors are connected in parallel, the total capacity of the capacitor bank is equal to the sum of the capacitances of the individual capacitors.
When capacitors are connected in series, the charges on the plates are equal in magnitude, and the total potential difference ∆ φ battery is equal to the sum of the potential differences ∆ φ 1 at the terminals of individual capacitors. Since for each capacitor ∆ φ 1 = Q / C i, then ∆ φ = Q / C = Q ∑ (1 / C i), whence we get
1 / C = ∑ (1 / C i). (3.1.52)
Expression 3.1.52 means that when capacitors are connected in series into a battery, the values opposite to the capacitances of individual capacitors are summed up, while the total capacitance turns out to be less than the smallest capacitance.
We have already said that the electrostatic field is potential. This means that any charge in such a field has potential energy. Let there be a conductor in a field for which the charge is known Q, capacity C and potential φ , and let us need to increase its charge by dQ... To do this, you need to do the work dA = φdQ = Сφdφ on the transfer of this charge from infinity to the conductor. If we need to charge the body from zero potential to φ , then you have to do the work, which is equal to the integral of Сφdφ within the specified limits. It is clear that integration will give the following equation
A = Сφ 2/2. (3.1.53)
This work is aimed at increasing the energy of the conductor. Therefore, for the energy of a conductor in an electrostatic field, we can write
W = Сφ 2/2 = Q φ / 2 = Q 2 / (2C). (3.1.54)
A capacitor, like a conductor, also has energy, which can be calculated using a formula similar to 3.1.55
W = С (∆φ) 2/2 = Q∆φ / 2 = Q 2 / (2C), (3.1.55)
where ∆φ – potential difference between capacitor plates, Q Is its charge, and WITH- capacity.
Substitute in 3.1.55 the expression for the capacity 3.1.49 ( C= ε 0 εS / d) and take into account that the potential difference ∆φ = Ed, we get
W = (ε 0 εS / d) (Ed 2) / 2 = ε 0 εE 2 V / 2, (3.1.56)
where V = Sd... Equation 3.1.56 shows that the energy of a capacitor is determined by the strength of the electrostatic field. From Equation 3.1.56, an expression for the bulk density of the electrostatic field can be obtained
w = W / V = ε 0 εE 2/2. (3.1.57)
Control questions
1. Where are the electric charges of a charged conductor located?
2. What is the strength of the electrostatic field inside a charged conductor?
3. What determines the strength of the electrostatic field at the surface of a charged conductor?
4. How is the protection of devices against external electrostatic interference ensured?
5. What is the electrical capacity of a conductor and what is the unit of its measurement?
6. What devices are called capacitors? What types of capacitors are there?
7. What is meant by the capacity of a capacitor?
8. What are the ways to increase the capacitance of a capacitor?
9. What is capacitor breakdown and breakdown voltage?
10. How is the capacity of a capacitor bank calculated when capacitors are connected in parallel?
11. What is the capacity of a capacitor bank when capacitors are connected in series?
12. How is the energy of a capacitor calculated?
