An electrostatic field is a special kind of matter through which charged bodies interact.

Coulomb's Law:force of interaction F between two fixed point charges q 1 and q 2 is directly proportional to the magnitudes of these charges and inversely proportional to the square of the distance r between them:

Where ( e 0 – electrical constant);

e- the dielectric constant of the medium, showing how many times the force of interaction of charges in this medium is less than in vacuum.

Electric fields that are created by stationary electric charges are called electrostatic.

Electrostatic field strength at a given point there is a physical quantity determined by the force acting on a test point positive charge q 0 placed at that field point, i.e.:

The electrostatic field can be represented graphically using lines of force .Power line - this is such a line, the tangent at each point to which coincides in direction with the vector of the electrostatic field strength at a given point (Fig. 1, 2).

If the field is created by a point charge, then the lines of force are radial straight lines coming out of a positive charge (Fig. 2, a), and included in the negative charge (Fig. 2, b).

Rice. 1 Fig. 2

With the help of lines of force, it is possible to characterize not only the direction, but also the magnitude of the strength of the electrostatic field, associating it with the density of the lines of force. A greater density of field lines corresponds to a greater magnitude of tension (Fig. 1, 2). Quantitatively, the number of lines of force penetrating a single area located perpendicular to the lines of force is put in correspondence with the magnitude of the intensity of the electrostatic field. In this case, a certain charge q, which creates a field, corresponds to a certain number N lines of force coming out (for ) from the charge or entering (for ) into the charge, namely: .

Electrostatic field strength vector flux through an arbitrary platform S characterized by the number of lines of force penetrating a given area S.

If the site S is perpendicular to the lines of force (Fig. 3), then the flow F E tension vector through this area S: .

Rice. 3 Fig. 4

Rice. 3

If the site S is located non-perpendicular to the lines of force of the electrostatic field (Fig. 4), then the vector flux through this area S:

,

where α is the angle between the stress vectors and the normal to the site S.

To find a flow F E tension vector through an arbitrary surface S, it is necessary to divide this surface into elementary areas dS(Figure 5), define the elementary stream dФ Е through each platform dS according to the formula:

,

and then all those elementary streams dФ Е add up, which leads to integration:

,

where α is the angle between the stress vectors and the normal to the given elementary area dS.

If we introduce a vector (Fig. 5) as a vector equal in size to the area of ​​the site dS and directed along the normal vector to the site dS, then the value , where a - the angle between the vectors and can be written as dot product vectors and , that is, as , and the resulting ratio for the flow of the vector will take the form:

.

The Ostrogradsky-Gauss theorem for an electrostatic field.

The Ostrogradsky - Gauss theorem for an electrostatic field relates the magnitude of the flux F E electrostatic field strength vector in a vacuum through an arbitrary closed surface S with charge value q enclosed inside the given closed surface S(Fig. 6).

Rice. 6

Since all lines of force leaving the charge (for ) or entering the charge (for ) penetrate an arbitrary closed surface S enclosing this charge (Fig. 6), then the magnitude of the flux F E vector through this surface S will be determined by the number N lines of force leaving the charge (for ) or entering the charge (for ):

.

This ratio is Ostrogradsky-Gauss theorem for an electrostatic field.

Since the flow is considered positive if the lines of force come out of the surface S, and negative for lines included in the surface S, then if inside an arbitrary closed surface S there is not one, but several n) of unlike charges, then Ostrogradsky-Gauss theorem for an electrostatic field is formulated as follows:

electrostatic field strength vector flux in a vacuum through an arbitrary closed surface is equal to the algebraic sum of the charges enclosed inside this surface, divided by e 0:

.

Topic 2. The work of the forces of the electrostatic field. Potential

If in the electrostatic field created by a point charge q, another test charge moves q 0 from point 1 exactly 2 along an arbitrary trajectory (Fig. 7) , then the work of the forces of the electrostatic field is performed.

elementary work dA force on elementary displacement is equal to: .

Figure 7 shows that .

Then ().

Work A when moving charge q 0 along the path from the point 1 to the point 2 :

That is, the work done when moving a charge from a point 1 v

point 2 in an electrostatic field does not depend on the trajectory of movement, but is determined only by the positions of the start and end points. So electrostatic field point charge is potential.

The work done by the forces of the electrostatic field when moving the charge q 0 from point 1 exactly 2 , is expressed as follows:

,

where φ 1 and φ 2 – electrostatic field potentials at points 1 and 2 .

The potential of the electrostatic field is determined up to an arbitrary additive constant WITH, that is, for the field of a point charge q:

.

Then , .

Potential difference two points 1 and 2 in an electrostatic field is determined by the work done by the forces of the electrostatic field when moving a test point charge q 0 from point 1 exactly 2 :

.

Relationship between strength and potential of an electrostatic field

Tension and Potential φ electrostatic fields are interconnected as follows:

= – grad φ

or , where

are the unit vectors of the coordinate axes Oh,Oy, Oz, respectively.

The minus sign in the above formula means that the vector of the electrostatic field strength is directed to direction of maximum decrease capacity j.

For a graphical representation of the potential distribution of the electrostatic field are used equipotential surfaces, that is, surfaces at all points of which the potential j has the same meaning.

For example, for a field created by a point charge q, potential j is determined by the expression: , and the equipotential surfaces are concentric spheres (Fig. 8).

This figure shows that in the case of a point charge, the field lines of force (dashed lines in the figure) normal(perpendicular) to the equipotential surfaces (solid lines in the figure).

This property normal relative position lines of force and equipotential surfaces of an electrostatic field is common to any case of an electrostatic field.

Thus, knowing the location of the electrostatic field lines, it is possible to construct the equipotential surfaces of this electrostatic field and, conversely, from the known location of the electrostatic field equipotential surfaces, it is possible to construct the electrostatic field lines.

A magnetic field

Topic 3. Magnetic field. Biot-Savart-Laplace law

An electric current creates a field that acts on a magnetic needle. The arrow is oriented tangentially to a circle lying in a plane perpendicular to the conductor with current (Fig. 9).

Main characteristic magnetic field is the induction vector. It is assumed that the magnetic field induction vector is directed towards the north pole of a magnetic needle placed at a given point of the field (Fig. 9).

Similar to the electric field, the magnetic field can also be represented graphically using lines of force (magnetic field lines).

force line- this is such a line, the tangent to which at each point coincides in direction with the magnetic field induction vector. The lines of force of the magnetic field, in contrast to the lines of force of the electrostatic field, are closed and cover current-carrying conductors. The direction of the lines of force is given by the rule of the right screw (rule of the gimlet): the head of the screw, screwed in in the direction of the current, rotates in the direction of the lines Fig. 9

magnetic induction (Fig. 9).

For several sources of a magnetic field, according to the principle of superposition of magnetic fields, the induction of the resulting magnetic field is equal to the vector sum of the inductions of all individual magnetic fields:

The induction vector of the magnetic field created by a current-carrying conductor can be determined using Biot-Savart-Laplace law. At the same time, it is necessary to take into account that Biot-Savart-Laplace law allows you to find the module and direction of only the magnetic field induction vector created by an element of a conductor with current. Therefore, to determine the induction vector of the magnetic field created by a current-carrying conductor, it is necessary to first divide this conductor into elements of the conductor, for each element using Biot-Savart-Laplace law find the induction vector , and then, using the principle of superposition of magnetic fields, vectorially add all the found induction vectors .

This theorem is only a consequence of Coulomb's law and the principle of superposition of electric fields. Here is her wording:

Tension vector flow electric field through a closed surface in vacuum is equal to the algebraic sum of the electric charges enclosed inside this surface, divided by the electric constant  0 .

Let's start the proof of the theorem with the simplest case: we calculate the flux of the field strength vector of a point charge Q.

The strength of this field is well known (see 1.3)

Taking into account the spherical symmetry of the field, we first choose as a Gaussian closed surface a sphere with a radius r, centered at the point where the charge is Q(Fig. 2.5., 1). The flow of the stress vector through this surface is easy to calculate

Here we take into account that:

Rice. 2.5.

Taking into account the last remark, we write the flow (2.7) in the following form:

(2.8)

Thus, for the first simplest case, the Gauss theorem turned out to be valid. What follows from this?

The result obtained allows us to conclude that the found flux does not depend on the radius of the Gaussian surface. This is easy to understand: after all, with increasing distance from the charge Q surface area growsproportionately square of the radius, and the field strength decreasinginversely radius square.

Recall, in addition, that the flux of the intensity vector is equal to the number of lines of force penetrating the Gaussian surface. The independence of the flow from the radius of the surface means that the lines of force of the field of a point charge, starting at a positive charge, extend further to infinity without interruption. From here - further conclusions.

The flow of the field strength vector of a point charge through any closed surface (Fig. 2.5, 2), covering a point chargeQ, is equal to the ratio

This conclusion is undoubted, since the flow is equal to the former unchanged number of lines of force penetrating a closed surface.

The flow of the tension vector through an arbitrary closed surface that does not cover an electric charge is equal to zero (Fig. 2.5, 3).

This conclusion is also easy to understand, since the number of field lines flowing into the Gaussian surface is equal to the number of lines leaving it. Therefore, the total flow through this surface is zero.

Now we can turn to the consideration of the general case: let an arbitrary closed surface S covers N point charges (Fig. 2.6.). Let us calculate the flow of the total field intensity vector through this surface S, taking into account that, in accordance with the superposition principle, the resulting field is equal to the vector sum of the individual fields

Rice. 2.6.

So, using the definition of the flow, we calculate it through an arbitrary closed surface S.

(2.9)

The result obtained is a proof of the Gauss theorem: the flow of the vector of the electrostatic field strength in vacuum through any closed surface is proportional to the algebraic sum of the charges enclosed inside this surface.

The rigorous derivation of the Ostrogradsky-Gauss theorem is quite complicated, we will deduce it for a particular case, which can be generalized quite convincingly. The Ostrogradsky-Gauss theorem allows you to determine the flow of the intensity vector from any number of charges. To begin with, let us determine the flux of the intensity vector through a spherical surface, in the center of which a point charge will be located.

Hence it follows that from each point charge comes the flow of the intensity vector, which is equal to the value q/εε 0 . From a generalization this provision the Ostrogradsky–Gauss theorem is derived for the general case – the total flow of the intensity vector through a closed surface of arbitrary shape is equal to the algebraic sum of electric charges enclosed inside this surface, divided by the absolute permittivity ε a = εε 0 , that is:

Where: n is the number of charges, q i is the charge sharpened inside the surface.

In the Gaussian system, this equation will look like:

For the flow of the electric displacement vector N D (induction vector), a similar formula can be obtained:

That is, the induction flux through a closed arbitrary surface is equal to the algebraic sum of electric charges that are covered by this surface.

If we take some closed surface that does not cover the charge q, then each line of tension (or induction) will cross it twice - once it will enter the surface, and the other time it will leave it. Because of this phenomenon, the algebraic sum of induction lines passing through a closed surface, the number of which determines the total induction flow N D through this surface, will be equal to zero (N D = 0).

Before considering several special cases of applying the Ostrogradsky-Gauss theorem to determine the strengths of various electrostatic fields, we introduce the concept of charge density.

is a physical quantity that characterizes the charge distribution along a line (thread) or a thin cylindrical body and is numerically equal to the ratio of the charge to the length of the thread element:

And with a uniform charge distribution along the entire length, the linear density:

In SI, the unit of measure for the linear charge density τ is 1 C/m.

If the charge dq is distributed over some volume dV, then it is obvious that the volume charge density will be numerically equal to the ratio of the charge to the volume element:

And with a uniform charge distribution:

In the SI system, it is measured in 1 C / m 3.

In cases where the charge dq is distributed over the surface dS and its penetration depth is negligibly small, then surface charge density will be determined by the ratio:

And if the charge q over the area S is uniformly distributed, then:

In the SI system, surface density is measured in C/m 2 .

Let's calculate , which is created by a uniformly charged spherical surface.

Suppose that a spherical surface has a radius R and a uniformly distributed charge q, that is, the surface density σ at any point of the sphere will be the same.

Let's choose point A, which is located at a distance r from the center of the sphere (figure below):

Through the point A we mentally draw a new spherical surface S, symmetrical to the charged sphere.

In this case, the flow of the intensity vector through the surface S will be equal to:

According to the Gauss theorem N E = q/εε 0 . It follows that for r>R:

If we compare this relation with the formula for the field strength of a point charge, we can conclude that outside the charged sphere, the field strength is the same as if all the available charge of the sphere were concentrated in its center.

For points that are on the surface of a charged sphere with an existing radius R, by analogy with equation (7), we can write:

If we pass through point B, which is inside a spherical charged surface, the sphere S / with radius r /

Now let's try to define field strength created by a uniformly charged filament (cylinder) of infinite length.

Let us assume that a hollow cylindrical surface with a certain radius R is charged with a constant surface density σ. Let's draw a coaxial surface of cylindrical type with radius r>R.

Through this surface, the flux of the intensity vector will be equal to:

According to the Gauss theorem:

Equating the right parts of these equations, we get:

From formula (4a), we find that the linear charge density of the cylinder is:

Using this equality, we find:

Now let's define field strength, which is created by a uniformly charged infinite plane.

If we assume that the given plane has an infinite extent and the charge per unit of the plane is equal to σ. It follows from the laws of symmetry that the field is directed everywhere perpendicular to the plane, and if there are no other external charges, then the fields on both sides of the plane should be equal in magnitude.

If we restrict a part of the charged plane 1 to an imaginary rectangular box 2 (Gaussian surface) in such a way that the box is cut in half (figure below).

Both sides of the box, which have a certain area S, must be parallel to the charged plane. The vector E is equal to the total flow of the intensity vector, multiplied by the area of ​​the first face S, plus the flow of the vector E through the opposite face. Through the remaining faces, the tension flow will be equal to zero, since they are not crossed by tension lines.

Repeating the previous reasoning and applying the Ostrogradsky-Gauss theorem, we obtain the following expression:

But E \u003d E 1 \u003d E 2. In this case, the field strength of an infinite uniform plane will be equal to:

The coordinates of the point at which the field strength is determined are not included in formula (12). This implies the conclusion that in an infinite uniformly charged plane, the electrostatic field will be uniform, and its intensity at any point of the field is the same.

And, finally, let's determine the strength of the field, which is created by two infinite parallel planes, with the same density and differently charged.

It can be seen from the figure above that between two infinite parallel planes having surface charge densities -σ and +σ, the field strength is equal to the sum of the field strengths that are created by both plates, that is:

The vectors E outside the plates are directed oppositely to each other and cancel each other out. Therefore, the electric field strength in the space that surrounds the plates will be zero (E = 0).

The principle of superposition in combination with Coulomb's law gives the key to calculating the electric field of an arbitrary system of charges, but the direct summation of the fields according to formula (4.2) usually requires complex calculations. However, in the presence of one or another symmetry of the system of charges, the calculations are greatly simplified if we introduce the concept of an electric field flux and use the Gauss theorem.

The ideas about the flow of an electric field were introduced into electrodynamics from hydrodynamics. In hydrodynamics, the fluid flow through a pipe, that is, the volume of fluid N passing through the pipe section per unit time, is equal to v ⋅ S , where v is the fluid velocity, and S is the pipe cross-sectional area. If the velocity of the liquid changes over the cross section, then the integral formula N = ∫ S v → ⋅ d S → must be used. Indeed, let us single out a small area d S in the velocity field perpendicular to the velocity vector (Fig. ).

Rice. 1.4: Liquid stream

The volume of liquid flowing through this area during the time d t is equal to v d S d t . If the site is inclined to the flow, then the corresponding volume will be v d S cos θ d t , where θ is the angle between the velocity vector v → and the normal n → to the site d S . The volume of fluid flowing through the area d S per unit time is obtained by dividing this value by d t . It is equal to v d S cos θ d t , i.e. scalar product v → ⋅ d S → of the velocity vector v → by the area element vector d S → = n → d S . The unit vector n → normal to the area d S can be drawn in two directly opposite directions. one of them is conditionally taken as positive. The normal n → is drawn in this direction. The side of the area from which the normal n → comes out is called external, and the one into which the normal n → enters is called internal. The area element vector d S → is directed along the outer normal n → to the surface, and is equal in magnitude to the area of ​​the element d S = ∣ d S → ∣ . When calculating the volume of a fluid flowing through an area S of finite size, it must be developed into infinitely small areas d S , and then the integral ∫ S v → ⋅ d S → is calculated over the entire surface S .

Expressions like ∫ S v → ⋅ d S → are found in many branches of physics and mathematics. They are called the flow of the vector v → through the surface S, regardless of the nature of the vector v → . In electrodynamics, the integral

N = ∫ S E → ⋅ d S →

(5.1)

is called the flow of the electric field strength E → through an arbitrary surface S, although no real flow is associated with this concept.

Let us assume that the vector E → is represented by the geometric sum

E → = ∑ j E → j .

Multiplying this equality scalarly by d S → and integrating, we obtain

N = ∑ j N j .

where N j is the flow of the vector E → j through the same surface. Thus, it follows from the principle of superposition of the electric field strength that flows through the same surface are added algebraically.

The Gauss theorem states that the flow of the vector E → through an arbitrary closed surface is equal to the total charge Q of all particles inside this surface multiplied by 4 π:

We will prove the theorem in three stages.

1. Let's start by calculating the flow of the electric field of a single point charge q (Fig. ). In the simplest case, when the integration surface S is a sphere and the charge is at its center, the validity of the Gauss theorem is practically obvious. On the surface of the sphere, the electric field strength

E → = q r → ∕ r 3

is constant in magnitude and is directed along the normal to the surface everywhere, so that the electric field flux is simply equal to the product of E = q ∕ r 2 and the area of ​​the sphere S = 4 π r 2 . Therefore, N = 4 π q . This result does not depend on the shape of the surface surrounding the charge. To prove this, we select an arbitrary area of ​​the surface of a sufficiently small size with the direction of the outward normal n → set on it. On fig. shows one such segment of an exaggerated (for clarity) size.

The flow of the vector E → through this area is d N = E → ⋅ d S → = E cos θ d S ,

where θ is the angle between the direction E → and the outward normal n → to the site d S . Since E \u003d q ∕ r 2, and d S cos θ ∕ ​​r 2 in absolute value is an element of the solid angle d Ω \u003d d S ∣ cos θ ∣ ∕ r 2 , under which the platform d S is visible from the charge location point,

D N = ± q d Ω .

where the plus and minus signs correspond to the cos θ sign, namely: the plus sign should be taken if the vector E → makes an acute angle with the direction of the outward normal n → , and the minus sign otherwise.

2. Now consider a finite surface S enclosing some distinguished volume V . With respect to this volume, it is always possible to determine which of the two opposite directions of the normal to any element of the surface S should be considered external. The outer normal is directed outward from the volume V. Summing over the segments, up to a sign, we have N = q Ω , where Ω is the solid angle under which the surface S is visible from the point where the charge q is located. If the surface S is closed, then Ω = 4 π provided that the charge q is inside S . Otherwise, Ω = 0 . To clarify the last statement, we can again refer to Fig. .

Obviously, the flows through the segments of a closed surface, based on equal solid angles, but facing in opposite directions, mutually cancel. It is also obvious that if the charge is outside the closed surface, then any outward-facing segment has a corresponding inward-facing segment.

3. Finally, using the principle of superposition, we arrive at the final formulation of the Gauss theorem (). Indeed, the field of the system of charges is equal to the sum of the fields of each charge separately, but only the charges inside the closed surface make a non-zero contribution to the right side of the theorem (). This completes the proof.

In macroscopic bodies, the number of charge carriers is so large that it is convenient to represent a discrete ensemble of particles as a continuous distribution by introducing the concept of charge density. By definition, the charge density ρ is the ratio Δ Q ∕ Δ V in the limit when the volume Δ V tends to a physically infinitesimal value:

where the integration on the right-hand side is performed over the volume V enclosed by the surface S .

The Gauss theorem gives one scalar equation for three components of the vector E → , so this theorem alone is not enough to calculate the electric field. A known symmetry of the distribution of charge density is required so that the problem can be reduced to a single scalar equation. The Gauss theorem allows you to find the field in cases where the integration surface in () can be chosen so that the electric field strength E is constant over the entire surface. Consider the most instructive examples.

▸ Task 5.1

Find the field of a sphere uniformly charged in volume or surfaces.

Solution: Electric field of a point charge E → = q r → ∕ r 3 tends to infinity at r→0. This fact shows the inconsistency of the representation elementary particles by point charges. If the charge q uniformly distributed over the volume of a sphere of finite radius a , then the electric field has no singularities.

It is clear from the symmetry of the problem that the electric field E → is directed everywhere radially, and its tension E = E(r) depends only on the distance r to the center of the ball. Then the flow of the electric field through the sphere of radius r is simply equal to 4 π r 2 E (Fig. ).

On the other hand, the charge inside the same sphere is equal to the total charge ball Q if r ≥ a . Equating 4 π r 2 E to the charge of the ball q multiplied by 4 π, we obtain: E (r) = q ∕ r 2 .

Thus, in outer space, a charged ball creates such a field as if all the charge were concentrated at its center. This result is valid for any spherically symmetric charge distribution.

The field inside the ball is E (r) = Q ∕ r 2 , where Q is the charge inside sulfur of radius r . If the charge is uniformly distributed over the volume of the sphere, then Q = q (r ∕ a) 3 . In this case

E (r) = q r ∕ a 3 = (4 π ∕ 3) ρ r ,

where ρ = q ∕ (4 π a 3 ∕ 3) — charge density. Inside the ball, the field decreases linearly from the maximum values ​​on the surface of the ball to zero at its center (Fig. ).

E(r) function moreover, it is everywhere finite and continuous.

If the charge is distributed over the surface of the sphere, then Q = 0 , and therefore also E = 0 . This result is also valid for the case when inside a spherical there is no charge cavity, and external charges are distributed spherically symmetrical. ▸ Task 5.2

Find the field of a uniformly charged infinite thread; thread radius a , charge per unit length ϰ .

▸ Task 5.3

Find the field of an infinite straight thread and an infinitely long uniformly charged cylinder.

▸ Task 5.4

Find the field of an infinite charged plane and uniformly charged infinite flat layer.

Solution: Due to the symmetry of the problem, the field is directed along the normal to the layer and depends only on the distance x from plane of symmetry of the plate. To calculate a field using Gauss theorem, it is convenient to choose the integration surface S in in the form of a parallelepiped, as shown in Fig. .

The last result is obtained by passing to the limit a → 0 while increasing the charge densityρ so that the value σ = ρ a remained unchanged. On opposite sides of the plane electric field strength is the same in magnitude, but opposite in direction. Therefore, when passing through charged plane, the field abruptly changes by the value 4 π σ . Note that the plate can be considered infinite if distance from is negligible compared to its size. On the very large distances compared to the dimensions of the plate, it acts like a point charge, and its field decreases back proportional to the square of the distance.

In a number of cases, the Gauss theorem makes it possible to find the electric field strength of extended charged bodies without resorting to the calculation of cumbersome integrals. This usually refers to bodies whose geometric shape has certain symmetry elements (ball, cylinder, plane). Let's consider some examples of applying the Gauss theorem to calculate the strength of electric fields.

Example 1. The field of a uniformly charged plane.

The electric field created by an infinitely extended uniformly charged plane is homogeneous - at every point in space outside the plane, its intensity is the same everywhere. This field is directed perpendicular to the plane in both directions (Fig. 2.5). Therefore, for the flow of the field strength vector through an arbitrarily chosen cylindrical surface based on an element of the plane ΔS, we can write: , whence , where is the surface charge density. Dimension in SI: .

Thus, the desired electric field strength uniformly charged plane .

Example 2. The field of a uniformly charged thread (cylinder).

In this case, the electric field has axial symmetry - it does not depend on the azimuthal angle φ and coordinate z and is directed along the radius vector (Fig. 2.6). Therefore, for the vector flow through the selected cylindrical surface with the axis coinciding with the charged thread, we have: , where is an element of a cylindrical surface; l is the length of an arbitrary section of the thread.

On the other hand, according to the Gauss theorem, this flow is equal to: where , is the linear charge density of the thread. From here we find:

The desired electric field strength uniformly charged filament : .

Example 3. The field of a uniformly charged sphere.

a) metal ball. At equilibrium, the charges are evenly distributed over the outer surface of the charged ball (Fig. 2.7). Therefore, when< (внутри шара) электрическое поле отсутствует: .

Outside the ball ( > ), the electric field created by charges uniformly distributed over its surface has spherical symmetry (directed along radial lines), therefore, according to the Gauss theorem:

.

We see that the electric field of a uniformly charged metal ball does not depend on the radius of the ball and coincides with the field point charge .

b) dielectric ball .

Consider a ball with a conditional permittivity ε = 1, uniformly charged in volume with a charge density (Fig. 2.8).

The dimension of the volumetric charge density in SI: .

The total charge of the ball is obviously: .

We have by the Gauss theorem:

1) inside the ball(r< R) : , where Δq = is the charge of the inner region of the ball, bounded by the chosen spherical surface of radius r. From here we find: .

2) outside the ball (r > R): , whence = ,

that is outside charged dielectric ball electric field the same , as in the case metallic ball.

Figure 2.9 shows the qualitative course of dependencies E(r) for metal and dielectric balls.

metal Fig.2.9. Addiction E(r). dielectric

1.4 Gauss's theorem. Electric induction vector.

Gauss theorem.

The calculation of the field strength of a system of electric charges using the principle of superposition of electrostatic fields can be greatly simplified using the Gauss theorem, which determines the flow of the electric field strength vector through an arbitrary closed surface.

Consider the flow of the intensity vector through a spherical surface of radius r enclosing a point charge q located in its center

This result is valid for any closed surface of arbitrary shape enclosing the charge.

If a closed surface does not contain a charge, then flow through it is zero, since the number of tension lines entering the surface is equal to the number of tension lines emerging from it.

Consider general case arbitrary surface surrounding n charges.According to the principle of superposition, the strength of the field created by all charges is equal to the sum of the strengths created by each charge separately. So

Gauss' theorem for an electrostatic field in vacuum:the flux of the electrostatic field strength vector in vacuum through an arbitrary closed surface is equal to algebraic the sum of the charges contained within this surface, divided by ε 0 .

In the general case, electric charges can be distributed with a certain bulk density, which is different in different places in space. Then the total charge of the volume V enclosed by the closed surface S is equal to and the Gauss theorem should be written as .