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Let us consider several examples of the calculation of electrostatic fields using the Gauss theorem.
1.7.1. The field of an infinite uniformly charged rectilinear yarn
Consider a uniformly charged infinitely long thread. The linear charge density is .
The charge, evenly distributed over the filament, has a symmetry - it is symmetrical about the axis.
The filament has an infinite length, so any elementary charge dq 1 we can compare another elementary charge dq 2, located symmetrically about a certain point in the electrostatic field.
Since the distance from the elementary charges to this point is the same, the stress modules E 1 and E 2 are the same. Therefore, the resulting tension
E = E 1 +E 2 is directed perpendicular to the filament (see the figure).
Obviously, at other points located at the same distance from the filament, the tension will have the same magnitude and direction.
Elementary charges and a point in the field were chosen randomly, so the derivation is valid for all other elementary charges, as well as for all points of the field.
This means that the electric field created by the charged filament is symmetrical about the axis of the filament. In other words, the symmetry of the field is identical with the symmetry of the charge that creates the field.
Thus, the stress vectors at all points of the surrounding space are perpendicular to the filament and the moduli of tension at identical distances from the filament are the same.
The calculation of the field strength using Gauss's theorem should begin with obtaining an expression for the flux of the vector E .
In turn, the flow expression should begin with the choice of the shape of the closed surface and its position relative to the source of the field.
Calculation of the flow will be as simple as possible if we choose a surface whose symmetry is identical to the symmetry of the charge field that creates it.
In this case it is convenient to use a closed surface with axial symmetry.
Such a surface is a cylinder whose axis coincides with the filament. Let the height of the cylinder be l, and the radius of the base - r.
The flux of the strength vector of the field created by the filament is added from the flow through the end surfaces of the cylinder and the flow through the side surface.
The flow through the end surfaces is zero, since the stress vectors are perpendicular to the filament and, correspondingly, the angle between the vectors E and n is equal to 90 0,
.
Flow through the side surface
.
Since all the points of the lateral surface are located at equal distances from the filament, the stress modules at all points on the lateral surface of the cylinder are the same,
.
This is the form of the expression for the flux of the vector of the calculated strength.
The next step in calculating the strength of the electrostatic field is the calculation of the total charge enveloped by a closed surface.
Surface-charged charge s, you can find it like this:
.
Then, by the Gauss theorem,
.
.
Thus, the intensity of the electric field created by a uniformly charged thread is directly proportional to the linear density of the filament charge and inversely proportional to the distance from the filament to the point of interest.
Pay attention - the strength is inversely proportional to the first degree of distance from the filament (the intensity of the field of the point charge is inversely proportional to the square of the distance from the charge).
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21. A long straight wire located in a vacuum carries a charge uniformly distributed over the entire length of the wire with a linear density of 2 nC / m. Determine the strength E of the electrostatic field at a distance r = 1 m from the wire.
22. The inner cylindrical conductor of a long rectilinear coaxial wire of radius R 1 = 1.5 mm is charged with a linear density τ 1 = 0.2 nC / m. The outer cylindrical conductor of this wire of radius R 2 = 3 mm is charged with a linear density of τ 2 = -0.15 nC / m. The space between the conductors is filled with rubber (ε = 3). Determine the intensity of the electrostatic field at points lying from the wire axis at distances: 1) r 1 = 1 mm; 2) r 2 = 2 mm; 3) r 3 = 5 mm.
23. The electrostatic field is created positively charged with a constant surface density σ = 10 nC / m 2 of infinite density. What kind of work should be done in order to transfer the electron along the tension line from a distance r 1 = 2 cm to r 2 = 1 cm?
24. The electrostatic field is created by a positively charged infinite filament with a constant linear density τ = 1 nC / cm. What speed will the electron acquire, approaching under the action of the field to the filament along the tension line from a distance r 1 = 2 cm to r 2 = 1 cm?
25. Equal charges Q = 100 nC are located at the vertices of the square with side a = 10 cm. Determine the potential energy of this system.
26. In the Bohr model of the hydrogen atom, the electron moves along a circular orbit with a radius r = 52.8 pm, in the center of which is a proton. Determine: 1) the speed of an electron in orbit; 2) the potential energy of the electron in the field of the nucleus, expressing it in electron volts.
27. A ring of radius r = 5 cm from a thin wire carries a uniformly distributed charge Q = 10 nC. Determine the potential φ of the electrostatic field: 1) in the center of the ring; 2) on the axis passing through the center of the ring, at a point distant by a distance a = 10 cm from the center of the ring.
28. On the ring with an internal radius of 80 cm and an outer one - 1 m, a charge of 10 nC is equally distributed. Determine the potential in the center of the ring.
29. A metal sphere with a radius of 5 cm carries a charge Q = 10 nC. The opp electrostatic field: 1) on the surface of the ball; 2) at a distance a= 2 cm from its surface. Construct a plot of φ (r) .
30. The hollow sphere carries a uniformly distributed charge. Determine the radius of the ball if the potential at the center of the ball is φ 1 = 200 V, and at a point lying from its center at a distance r = 50 cm, φ 2 = 40 V.
31. The electrostatic field is created by a positive point charge. Determine the numerical value and direction of the gradient of the potential of this field if at a distance r = 10 cm from the charge the potential is φ = 100 V.
32. The electrostatic field is created by an infinite plane, charged uniformly with a surface density σ = 5 nC / m 2 Determine the numerical value and direction of the gradient of the potential of this field.
33. The electrostatic field is created by an infinite straight filament charged uniformly with a linear density of τ = 50 pC / cm. Determine the numerical value and direction of the potential gradient at a point at a distance r = 0.5 m from the filament.
34. Determine the linear density of an infinitely long charged filament, if the work of field forces on the displacement of the charge Q = 1 nC from a distance r 1 = 5 cm and r 2 = 2 cm in a direction perpendicular to the filament is 50 μJ.
35. The electrostatic field is created by a positively charged infinite filament Proton, moving from the filament under the action of the field along the line of tension from a distance r 1 = 1 cm to r 2 = 5 cm, changed its speed from 1 to 10 Mm / s. Determine the linear density of the filament charge.
36. The electrostatic field is created by an infinite plane, uniformly charged with a surface density of sigma = 1 nC / m 2. Determine the potential difference between two points of this field, lying at a distance x 1 = 20 cm and x 2 = 50 cm from the plane.
37. Determine the surface density of charges on the plates of a flat mica (ε = 7) capacitor, charged to a potential difference U = 200 V, if the distance between its plates is d = 0.5 mm.
38. The electrostatic field is created by a uniformly charged sphere with a surface of radius R = 10 cm with a total charge Q = 15 nC. Determine the potential difference between two points of this field, lying at distances r 1 = 5 cm and r 2 = 15 cm from the surface of the sphere.
39. The electrostatic field is created by a sphere of radius R = 5 cm, uniformly charged with a surface density of sigma = 1 nC / m 2. Determine the potential difference between two points of the field, lying at distances r 1 = 10 cm and r 2 = 15 cm from the center of the sphere.
40. The electrostatic field is created by a uniformly charged ball of radius R = 1 m with a total charge Q = 50 nC. Determine the potential difference for points lying from the center of the ball at distances 1) r 1 = 1.5 m and r 2 = 2 m; 2) r 1 "= 0.3 m and r 2" = 0.8 m.
Linear density of charge - the charge per unit length:
Consequently,
.
Potential difference between points 1 and 2 fields that are at a distance r 1 and r 2 from the cylinder axis:
3. The field of a charged spherical surface
It can be seen that the expression for 
turned out to be the same as for a point charge.
Potential difference
A ball, which is a dielectric, can be internally uniformly charged with a bulk density 
. Vector Flow through a surface of radius r
R
(R Is the radius of the sphere) is 
Charge within a sphere of radius ris equal to:
.
By the Gauss theorem
and 
Outside the uniformly charged sphere, the expression for E A will be the same as that received for the hollow sphere 
, only the quantity qwill equal
V:
The potential difference for points lying at a distance r R from the center of the ball:
and for points lying at a distance r R from the center of the ball:
2. Conductors in an electric field.
Conductors They call bodies that conduct electricity well, in which there are free electric charges capable of moving through the entire volume of the conductor.
Conditions for the equilibrium of charges on a conductor:
|
|
Let us find the quantity E near the surface of the conductor: draw a cylindrical surface section dS with a generatrix perpendicular to the surface of the conductor and parallel to the vector From here: |
, because the flow is only through one cylinder base.
Because inside the conductor E=0,
and in the immediate vicinity of the surface 
, this means that when passing from the conductor to the space behind the conductor (into the air), the value 
varies from 0 to .
The average value of the field strength at the surface of the conductor is equal to:
The strength with which the conductor field acts on a charge located on its surface dS, is equal to:
The pressure experienced by the conductor surface and caused by the excess charges on its surface is:
When an uncharged conductor is placed in an electric field, the charges on it come into motion - excessive electric charges of opposite signs appear on opposite surfaces.
The charges that arise on the surface create their own field, which is exactly equal to the external field, but opposite in direction - there is no field inside the conductor (in the cavity).
Redistribution of charges in a conductor under the action of an external field occurs until the lines of force are perpendicular to the surface of the conductor.
The zeroing of the field strength in the cavity of the conductor is used to realize electrical protection, and it turned out that the electrical protection is obtained sufficiently good not only in the case of a solid metal shell, but also in the case of using a fine metal mesh.
Connecting the conductor of any body to the ground is called grounding. When grounding charged conductors, including human bodies, they lose their charge and their potential will equal the potential of the earth. The earthing of the housings of instruments and apparatus facilitates their safe operation, It excludes the possibility for the personnel to be under voltage of the apparatus and ground.
