In the case of equilibrium distribution, the charges of the conductor are distributed in a thin surface layer. So, for example, if a conductor is given a negative charge, then due to the presence of repulsive forces between the elements of this charge, they will be dispersed over the entire surface of the conductor.

Examination using a test plate

In order to experimentally investigate how charges are distributed on outer surface conductors use a so-called test plate. This plate is so small that when it comes into contact with the conductor, it can be considered as part of the surface of the conductor. If this plate is applied to a charged conductor, then part of the charge ($\triangle q$) will transfer to it and the magnitude of this charge will be equal to the charge that was on the surface of the conductor in area equal area plates ($\triangle S$).

Then the value is equal to:

\[\sigma=\frac(\triangle q)(\triangle S)(1)\]

is called the surface charge distribution density at a given point.

By discharging a test plate through an electrometer, one can judge the value of the surface charge density. So, for example, if you charge a conducting ball, you can see, using the above method, that in a state of equilibrium surface density The charge on the ball is the same at all its points. That is, the charge is distributed evenly over the surface of the ball. For conductors more complex shape charge distribution is more complex.

Surface density of conductor

The surface of any conductor is equipotential, but general case The charge distribution density can vary greatly in different points. The surface charge distribution density depends on the curvature of the surface. In the section that was devoted to describing the state of conductors in an electrostatic field, we established that the field strength near the surface of the conductor is perpendicular to the surface of the conductor at any point and is equal in magnitude:

where $(\varepsilon )_0$ is the electric constant, $\varepsilon $ is the dielectric constant of the medium. Hence,

\[\sigma=E\varepsilon (\varepsilon )_0\ \left(3\right).\]

The greater the curvature of the surface, the greater the field strength. Consequently, the charge density on the protrusions is especially high. Near the depressions in the conductor, equipotential surfaces are located less frequently. Consequently, the field strength and charge density in these places are lower. The charge density at a given conductor potential is determined by the curvature of the surface. It increases with increasing convexity and decreases with increasing concavity. Especially high density charge on the edges of the conductors. Thus, the field strength at the tip can be so high that ionization of the gas molecules that surrounds the conductor can occur. Gas ions opposite sign charge (relative to the charge of the conductor) are attracted to the conductor, neutralizing its charge. Ions of the same sign are repelled from the conductor, “pulling” neutral gas molecules with them. This phenomenon is called electric wind. The charge of the conductor decreases as a result of the neutralization process; it seems to flow off the tip. This phenomenon is called the outflow of charge from the tip.

We have already said that when we introduce a conductor into an electric field, a separation of positive charges (nuclei) and negative charges (electrons) occurs. This phenomenon is called electrostatic induction. The charges that appear as a result are called induced. Induced charges create an additional electric field.

The field of induced charges is directed towards opposite direction external field. Therefore, the charges that accumulate on the conductor weaken the external field.

The redistribution of charges continues until the charge equilibrium conditions for conductors are met. Such as: zero field strength everywhere inside the conductor and perpendicularity of the intensity vector of the charged surface of the conductor. If there is a cavity in the conductor, then with an equilibrium distribution of the induced charge, the field inside the cavity is zero. Electrostatic protection is based on this phenomenon. If they want to protect a device from external fields, it is surrounded by a conductive screen. In this case, the external field is compensated inside the screen by induced charges arising on its surface. This may not necessarily be continuous, but also in the form of a dense mesh.

Assignment: An infinitely long thread, charged with linear density $\tau$, is located perpendicular to an infinitely large conducting plane. Distance from the thread to the plane $l$. If we continue the thread until it intersects with the plane, then at the intersection we will obtain a certain point A. Write a formula for the dependence of the surface density $\sigma \left(r\right)\ $of induced charges on the plane on the distance to point A.

Let's consider some point B on the plane. An infinitely long charged thread at point B creates an electrostatic field; a conducting plane is in the field; induced charges are formed on the plane, which in turn create a field that weakens the external field of the thread. The normal component of the plane field (induced charges) at point B will be equal to the normal component of the thread field at the same point if the system is in equilibrium. Select on thread elementary charge($dq=\tau dx,\ where\ dx-elementary\ piece\ thread\ $), we find at point B the tension created by this charge ($dE$):

Let's find the normal component of the filament field strength element at point B:

where $cos\alpha $ can be expressed as:

Let us express the distance $a$ using the Pythagorean theorem as:

Substituting (1.3) and (1.4) into (1.2), we get:

Let us find the integral from (1.5) where the limits of integration are from $l\ (distance\ to\ the nearest\ end\ of\ the thread\ from\ the\ plane)\ to\ \infty $:

On the other hand, we know that the field of a uniformly charged plane is equal to:

Let us equate (1.6) and (1.7) and express the surface charge density:

\[\frac(1)(2)\cdot \frac(\sigma)(\varepsilon (\varepsilon )_0)=\frac(\tau )(4\pi (\varepsilon )_0\varepsilon )\cdot \frac (1)((\left(r^2+x^2\right))^((1)/(2)))\to \sigma=\frac(\tau )(2\cdot \pi (\left (r^2+x^2\right))^((1)/(2))).\]

Answer: $\sigma=\frac(\tau )(2\cdot \pi (\left(r^2+x^2\right))^((1)/(2))).$

Example 2

Assignment: Calculate the surface charge density that is created near the Earth's surface if the Earth's field strength is 200$\ \frac(V)(m)$.

We will assume that the dielectric conductivity of air is $\varepsilon =1$ like that of a vacuum. As a basis for solving the problem, we will take the formula for calculating the voltage of a charged conductor:

Let us express the surface charge density and obtain:

\[\sigma=E(\varepsilon )_0\varepsilon \ \left(2.2\right),\]

where the electric constant is known to us and is equal in SI $(\varepsilon )_0=8.85\cdot (10)^(-12)\frac(F)(m).$

Let's carry out the calculations:

\[\sigma=200\cdot 8.85\cdot (10)^(-12)=1.77\cdot (10)^(-9)\frac(Cl)(m^2).\]

Answer: The surface charge distribution density of the Earth's surface is equal to $1.77\cdot (10)^(-9)\frac(C)(m^2)$.

Electrostatics. Application of the Ostrogradsky–Gauss theorem to calculate fields in vacuum

Coulomb's law allows you to calculate the field of any system of charges, i.e., find its intensity at any point by summing vectorially the intensities created by individual charges (since intensity vectors obey the principle of superposition). Tension is called vector physical quantity, characterizing the force of action electrostatic field to a positive charge. The direction of the tension vector coincides with this force. For problems that have symmetry, the calculations can be greatly simplified; in these cases, it is convenient to use the Ostrogradsky–Gauss theorem for the flow of the intensity vector through some closed surface (Fig. 1.1). Let all charges Q i be concentrated inside a closed surface with area S.

On a surface element with area dS, the charges create a corresponding intensity, and the total

tension is equal to .

Flow Ф of the intensity vector through the closed surface under consideration

The flows of tension vectors (scalars) are summed algebraically. Taking into account the values ​​of Ф i, we can rewrite:

Where (- unit vector external normal to the surface element with area dS); – projection of the vector; Q i – charges located inside the surface.

The Ostrogradsky–Gauss theorem is formulated as follows. The vector flux through any closed surface is proportional to the total charge located inside this surface.

There are three possible cases when the flux of the tension vector through a closed surface vanishes:

A) algebraic sum charges inside the surface is zero, ;

b) there are no charges inside the surface, but there is a field associated with external charges; c) there is no field or internal charges.

Charges can be distributed in different ways, and they can be brought into the space under consideration, move in it and be removed from it, which is why they are called free charges.

If the charge dQ is continuously distributed in some small volume dV. In this case, the concept of volumetric charge density is introduced

ρ = dQ/dV (expressed in coulombs per cubic meter). If the charges are continuously distributed over the surface of the conductor, then the concept of surface density σ = dQ/dS is introduced, where dS is the area of ​​the conductor surface element on which the elementary charge dQ is located. The unit of surface density is 1 C/m2. If the charges are uniformly distributed along the line, in this case the concept of linear charge density λ = dQ/dl is introduced, where dl is the length of the line segment on which the charge dQ is distributed. The unit of linear density is 1 C/m.

The voltage vector on the surface of a charged conductor is always perpendicular to the surface (for example, for a charged ball, Fig. 1.2), since otherwise the charges would move along the surface under the influence of the tangential component of the voltage. Thus, at the surface of the conductor

and inside a solid conductor

Rice. 1.2. Field of a charged metal ball

If the charges are distributed over the volume of the dielectric with volume density ρ, then the Ostrogradsky–Gauss theorem is written as:

where dV is a volume element; V is the volume limited by the surface S.

When the charges are distributed over the surface of the conductor, and the integration surface coincides with the latter, then

.

Then the voltage on the surface of the conductor is proportional to the surface charge density:

Positive field point charge has spherical symmetry relative to the point at which it is located, and is characterized by tension directed along radii drawn from this point and equal to

i.e., it obeys Coulomb’s law (for a negative charge the vector is directed towards this point). The field of a charged metal ball is subject to the same laws. The charge on the ball is distributed evenly over the surface. Then for a metal ball with radius R 0 the field strength is determined in accordance with formula (1.2).

If inside a charged ball or other metal conductor there is a cavity into which no charges are introduced, then the field inside this cavity cannot be created by charges located on the surface of the conductor. Since the field inside the cavity is not associated with any charges, it is absent, i.e. E field = 0.

Of practical interest is the field created by a long uniformly charged wire (cylinder) with radius R 0 (Fig. 1.3). By choosing the integration surface in the form of a coaxial cylinder of radius R and height h and introducing the linear charge density

We are convinced that, due to cylindrical symmetry, the tension on the side surface of the cylinder is everywhere the same in magnitude and directed along the radii, and there is no tension flow through the bases.

In this case, the field strength varies in inverse proportion to the first power of the distance. On the surface of the wire we get

Let us now find the field strength of a boundless flat metal plate (Fig. 1.4). Let the plate be uniformly charged. As the surface of integration we choose the surface

rectangular parallelepiped, two faces of area S are parallel to the charged plate. The surface charge density is

σ = Q /2S, since the plate has two sides and the charge is distributed on both sides. Due to symmetry, the flux of the tension vector for the faces is nonzero. Hence,

For two parallel plates (Fig. 1.5), having the same charge density in absolute value, using the superposition principle we obtain: a) for the field between the plates

b) for the field outside the plates

.

We can conclude that the charges are collected on the sides of the plates facing each other with a surface density σ1 = σ. The tension determined by expression (1.3) does not depend on the distance and is the same at all points. Such fields are called homogeneous. There are no real infinite wires and plates, but the resulting formulas retain their value for regions sufficiently close to charged bodies (the distance to the field point under study should be much less than the linear size of the charged body). The distribution of tension lines can be obtained experimentally by placing electrodes of one shape or another in a liquid dielectric (vaseline oil) and pouring fine dielectric powder (quinine) onto the surface of the oil. In this case, the powder particles are located approximately along the tension lines.

The Ostrogradsky–Gauss theorem can be used not only in integral form, connecting the values ​​of intensity E at some points of the field with charges located at other points, but also in differential form. Let us connect the quantities related to the same point in the field.

Let there be tension at some point A with coordinates (x,y,z) where i , j , k are direction vectors in Cartesian system coordinates

Select near point A (Fig. 1.6) cuboid infinitesimal volume dV = dx`dy`dz .

Rice. 1.6. On the Ostrogradsky–Gauss theorem

The volumetric charge density in it is equal to ρ. It depends on the coordinates of the selected field point p = f (x,y,z). Flow vector through the right

. In the same way for the top and bottom edges we get ,

and for the back and front faces . Let us apply the Ostrogradsky–Gauss theorem to this volume:

,we finally get the expression . In vector analysis, the amount worth

In this form the theorem is applicable to individual points of the field.

The Ostrogradsky–Gauss theorem is not a consequence of Coulomb's law. It is one of the main theorems of vector analysis, connecting the volume integral with the surface integral. In physics, this theorem applies to central forces, depending on the distance according to the law R n, where n is any number. Thus, Coulomb's law is a special case of the Ostrogradsky–Gauss theorem.

Let's consider the work of electrostatic forces when moving a particle with charge q from one field point to another along an arbitrary path 1A 2 (Fig. 1.7):

where E i is the projection of the direction vector dl. This work will depend only on the position of the initial and end points path, and not from its form, i.e. the field is potential:

where φ1, φ2 are the potentials of the initial and final points of the trajectory. Potential is a scalar characteristic of a field point. U = φ1 – φ2 – potential difference or change potential energy single positive charge, transported in an electrostatic field.

Thus, the work of electrostatic forces is proportional to the potential difference U at the starting and ending points of the path. The unit of potential and potential difference is the Volt (V).

The work of electrostatic forces along any closed path is zero:

This integral is called the circulation of the tension vector. Equality to zero circulation means that there are no closed lines of tension in the electrostatic field: they begin and end on charges (positive or negative, respectively) or go to infinity.

In an electrostatic field, it is possible to construct (Fig. 1.7) surfaces that represent a set of points of equal potential (equipotential surfaces). Let us prove that the tension lines are normal to these surfaces. If you move a charge along equipotential surface, then the work will be zero. But the field strength at the surface can be different from zero. Therefore, from the definition of elementary work

it follows that when , therefore, and the vector dl is directed tangentially to the surface.

Consequently, at all points of a surface of equal potential, the tension is directed normal to this surface. From calculating the fields of symmetrical conductors using the Ostrogradsky–Gauss theorem, it is clear that the surface of a conductor in an electrostatic field is always equipotential.

The electrostatic field strength is related to the potential at each point of the field by the relation

1.2.The concept of charge density

To simplify mathematical calculations of electrostatic fields, the discrete structure of charges is often neglected. It is assumed that the charge is distributed continuously and introduces the concept of charge density.

Let us consider various cases of charge distribution.

1.Charge is distributed along the line. Let there be a charge in an infinitesimal area
. Let's enter the value

. (1.5)

Magnitude called linear charge density. Her physical meaning– charge per unit length.

2. The charge is distributed over the surface. Let us introduce the surface charge density:

. (1.6)

Its physical meaning is the charge per unit area.

3. The charge is distributed throughout the volume. Let's introduce bulk density charge:

. (1.7)

Its physical meaning is a charge concentrated in a unit volume.

A charge concentrated on an infinitesimal portion of a line, surface, or in an infinitesimal volume can be considered a point charge. The field strength created by it is determined by the formula:

. (1.8)

To find the field strength created by the entire charged body, you need to apply the principle of field superposition:

. (1.9)

In this case, as a rule, the problem is reduced to calculating the integral.

1.3. Application of the superposition principle to the calculation of electrostatic fields. Electrostatic field on the axis of a charged ring

Statement of the problem . Let there be a thin ring of radius R, charged with a linear charge density τ . It is necessary to calculate the electric field strength at an arbitrary point A, located on the axis of the charged ring at a distance x from the plane of the ring (Fig.).

Let us choose an infinitesimal element of the length of the ring dl; charge dq, located on this element is equal to dq= τ· dl. This charge creates at a point A electric field strength
. The modulus of the tension vector is equal to:

. (1.10)

According to the principle of field superposition, the electric field strength created by the entire charged body is equal to the vector sum of all vectors
:

. (1.11)

Let's expand the vectors
into components: perpendicular to the axis of the ring (
) and rings parallel to the axis (
).

. (1.12)

The vector sum of the perpendicular components is zero:
, Then
. Replacing the sum with an integral, we get:

. (1.13)

From the triangle (Fig. 1.2) it follows:

=
. (1.14)

Let us substitute expression (1.14) into formula (1.13) and take out the constant values ​​outside the integral sign, we obtain:

. (1.15)

Because
, That

. (1.16)

Considering that
, formula (1.16) can be represented as:

. (1.17)

1.4.Geometric description of the electric field. Tension vector flow

To describe the electric field mathematically, you need to indicate the magnitude and direction of the vector at each point , that is, set the vector function
.

There is a visual (geometric) way to describe a field using vector lines (power lines) (Fig. 13.).

Tension lines are drawn as follows:

WITH There is a rule: electric field strength vector lines, created by the system stationary charges, can begin or end only on charges or go to infinity.

Figure 1.4 shows an image of the electrostatic field of a point charge using vector lines , and in Figure 1.5 is an image of the electrostatic field of the dipole .

1.5. Electrostatic field strength vector flow

P Let us place an infinitesimal area dS in the electric field (Fig. 1.6). Here - unit vector normal to the site. Electric field strength vector forms with the normal some angle α. Vector projection to the normal direction is equal to E n =E·cos α .

Vector flow through an infinitesimal area is called dot product

, (1.18)

The electric field strength vector flux is an algebraic quantity; its sign depends on the mutual orientation of the vectors And .

Flow vector through an arbitrary surface S finite value is determined by the integral:

. (1.20)

If the surface is closed, the integral is marked with a circle:

. (1.21)

For closed surfaces, the normal is taken outward (Fig. 1.7).

The flow of the tension vector has a clear geometric meaning: it is numerically equal to the number of lines of the vector , passing through the surface S.

Question 42. Equilibrium of charges on a conductor. Surface charges. Examples of fields near a conductor. Conductor in an external electric field.

Conductor - This solid, which contains “ free electrons”, moving within the body.

Charge carriers in a conductor are capable of moving under the influence of arbitrarily small forces. Therefore, the balance of charges on a conductor can be observed only when following conditions:

2) The vector on the surface of the conductor is directed normal to each point on the surface of the conductor.

Indeed, if the condition 1 was not performed, then mobile media electric charges, present in each conductor, under the influence of field forces would begin to move (in the conductor a electric current) and the balance would be upset.

From 1 it follows that since

Question 43. Electrical capacity of a solitary conductor. Types of capacitors, their electrical capacity and other characteristics.

Electrical capacity of a solitary conductor – a characteristic of a conductor, indicating the ability of the conductor to accumulate an electrical charge.

The capacitance of a conductor depends on its size and shape, but does not depend on the material, state of aggregation, shape and size of cavities inside the conductor. This is due to the fact that excess charges are distributed on the outer surface of the conductor. Capacitance also does not depend on the charge of the conductor or its potential.

/* Electric capacity of the ball

It follows that a solitary sphere located in a vacuum and having a radius of R=C/(4pe 0)»9×10 6 km, which is approximately 1400 times greater than radius Earth (electric capacity of the Earth WITH" 0.7 mF). Therefore, farad is very large value, therefore in practice they are used submultiples- millifarad (mF), microfarad (μF), nanofarad (nF), picofarad (pF). */

Types of capacitors, their electrical capacity and other characteristics.

Capacitor - a system consisting of two conductors (plates) separated by a dielectric layer, usually the capacitor is charged symmetrically on the plates

Question 44. Energy of capacitors. Electric field energy density.

Capacitor is a system of charged bodies and has energy.
Energy of any capacitor:

where C is the capacitance of the capacitor
q - capacitor charge
U - voltage on the capacitor plates
The energy of the capacitor is equal to the work done by the electric field when the capacitor plates are brought close together,
or equal to the work of separating positive and negative charges required when charging the capacitor.

Electric field energy density.