An electrostatic field has two characteristics: power (strength) and energy (potential). Tension and potential are different characteristics of the same field point, therefore, there must be a connection between them.
The work of moving a single point positive charge from one point to another along the x axis, provided that the points are infinitely close to each other and x 1 - x 2 \u003d dx, is equal to qE x dx. The same work is equal to q(φ 1 - φ 2)= -dφq. Equating both expressions, we can write
Repeating similar reasoning for the y and z axes, we can find the vector:
where are the unit vectors of the coordinate axes x, y, z.
From the definition of a gradient it follows that
Or (12.31)
those. the field strength E is equal to the potential gradient with a minus sign. The minus sign is determined by the fact that tension vector E field is directed in the direction of decreasing potential.
The established relationship between the strength and potential allows, by the known field strength, to find the potential difference between two arbitrary points of this field.
Ø Field of a uniformly charged sphere radius R
The field strength outside the sphere is determined by the formula
The potential difference between the points r 1 and r 2 (r 1 >R; r 2 >R) is determined using the relation
We get the potential of the sphere if r 1 = R, r 2 → ∞:
Ø The field of a uniformly charged infinitely long cylinder
The field strength outside the cylinder (r > R) is determined by the formula
(τ is the linear density).
The potential difference between two points lying at a distance r 1 and r 2 (r 1 >R; r 2 >R) from the axis of the cylinder is equal to
(12.32)
Ø Field of a uniformly charged infinite plane
The field strength of this plane is determined by the formula
(σ - surface density).
The potential difference between points lying at a distance x 1 and x 2 from the plane is equal to
(12.33)
Ø Field of two oppositely charged infinite parallel planes
The field strength of these planes is determined by the formula
The potential difference between the planes is
(12.34)
(d is the distance between the planes).
Examples of problem solving
Example 12.1. Three point charges Q 1 \u003d 2nC, Q 2 \u003d 3nC and Q 3 \u003d -4nC are located at the vertices equilateral triangle with side length a=10cm. Determine the potential energy of this system.
Given: Q 1 \u003d 2nCl \u003d 2∙10 -9 C; Q 2 \u003d 3nCl \u003d 3 ∙ 10 -9 C; and Q 3 \u003d -4nCl \u003d 4∙10 -9 C; a=10cm=0.1m.
Find: U.
Solution:
The potential energy of a system of charges is equal to the algebraic sum of the interaction energies of each of the interacting pairs of charges, i.e.
U=U 12 +U 13 +U 23
where, respectively, the potential energies of one of the charges located in the field of another charge at a distance a from him, equal
; 
; 
(2)
We substitute formulas (2) into expression (1), find the desired potential energy of the system of charges
Answer: U \u003d -0.126 μJ.
Example 12.2. Determine the potential in the center of the ring with an inner radius R 1 =30cm and an outer R 2 =60cm, if the charge q=5nC is evenly distributed on it.
Given: R 1 \u003d 30 cm \u003d 0.3 m; R 2 \u003d 60 cm \u003d 0.6 m; q=5nCl=5∙10 -9 C
Find: φ.
Solution: We divide the ring into concentric infinitely thin rings with an inner radius r and an outer radius (r+dr).
The area of the considered thin ring (see figure) dS=2πrdr.
The potential at the center of the ring, created by an infinitely thin ring,
where is the surface charge density.
To determine the potential at the center of the ring, one should arithmetically add dφ from all infinitely thin rings. Then
Considering that the charge of the ring Q=σS, where S= π(R 2 2 -R 1 2) is the area of the ring, we obtain the desired potential at the center of the ring
Answer: φ=25V
Example 12.3.Two point charges of the same name (q 1 \u003d 2nC and q 2 \u003d 5nC) are in vacuum at a distance r 1 \u003d 20 cm. Determine the work A that must be done to bring them closer to a distance r 2 \u003d 5 cm.
Given: q 1 \u003d 2nCl \u003d 2∙10 -9 C; q 2 \u003d 5nCl \u003d 5∙10 -9 C ; r 1 \u003d 20cm \u003d 0.2m; r 2 \u003d 5 cm \u003d 0.05 m.
Find: A.
Solution: The work done by the forces of the electrostatic field when moving the charge Q from a point in the field with a potential φ 1 to a point with a potential φ 2.
A 12 \u003d q (φ 1 - φ 2)
When similar charges approach each other, the work is done by external forces, so the work of these forces is equal in absolute value, but opposite in sign to the work of the Coulomb forces:
A \u003d -q (φ 1 - φ 2) \u003d q (φ 2 - φ 1). (one)
Potentials of points 1 and 2 of the electrostatic field
Substituting formulas (2) into expression (1), we find the desired work that must be done in order to bring the charges closer,
Answer: A=1.35 μJ.
Example 12.4.An electrostatic field is created by a positively charged endless filament. The proton, moving under the action of an electrostatic field along the line of tension from the filament from a distance r 1 =2 cm to r 2 =10 cm, changed its speed from υ 1 =1 Mm/s to υ 2 =5 Mm/s. Determine the linear density τ of the charge of the thread.
Given: q=1.6∙10 -19 C; m=1.67∙10 -27 kg; r 1 \u003d 2 cm \u003d 2 ∙ 10 -2 m; r 2 \u003d 10cm \u003d 0.1m; r 2 \u003d 5 cm \u003d 0.05 m; υ 1 \u003d 1 Mm / s \u003d 1 ∙ 10 6 m / s; up to υ 2 \u003d 5 Mm / s \u003d 5 ∙ 10 6 m / s.
Find:τ .
Solution: The work done by the forces of the electrostatic field when moving a proton from a point of the field with a potential φ 1 to a point with a potential φ 2 goes to increase the kinetic energy of the proton
q(φ 1 - φ 2) \u003d ΔT (1)
In the case of a filament, the electrostatic field is axially symmetric, so
Or dφ=-Edr,
then the potential difference between two points located at a distance r 1 and r 2 from the thread,
(taken into account that the strength of the field created by a uniformly charged infinite thread, ).
Substituting expression (2) into formula (1) and taking into account that 
, we get
Where is the desired linear charge density of the thread
Answer: τ = 4.33 µC/m.
Example 12.5.An electrostatic field is created in vacuum by a ball with a radius R=8cm, uniformly charged with a bulk density ρ=10nC/m 3 . Determine the potential difference between two points of this field, lying from the center of the ball at distances: 1) r 1 =10cm and r 2 =15cm; 2) r 3 \u003d 2cm and r 4 \u003d 5cm ..
Given: R=8cm=8∙10 -2 m; ρ=10nC/m 3 =10∙10 -9 nC/m 3; r 1 \u003d 10 cm \u003d 10 ∙ 10 -2 m;
r 2 \u003d 15 cm \u003d 15 ∙ 10 -2 m; r 3 \u003d 2cm \u003d 2 ∙ 10 -2 m; r 4 \u003d 5 cm \u003d 5 ∙ 10 -2 m.
Find:1) φ 1 - φ 2; 2) φ 3 - φ 4.
Solution: 1) The potential difference between two points lying at a distance r 1 and r 2 from the center of the ball.
(1)
where is the field strength generated by a uniformly charged ball with a volume density ρ at any point outside the ball at a distance r from its center.
Substituting this expression into formula (1) and integrating, we obtain the desired potential difference
2) The potential difference between two points lying at a distance r 3 and r 4 from the center of the ball,
(2)
where is the field strength generated by a uniformly charged ball with a volume density ρ at any point lying inside the ball at a distance r from its center.
Substituting this expression into formula (2) and integrating, we obtain the desired potential difference
Answer: 1) φ 1 - φ 2 \u003d 0.643 V; 2) φ 3 - φ 4 \u003d 0.395 V
7.5. The principle of superposition of electrostatic fields
7.5.2 Superposition principle for potential
Superposition principle for potential allows you to calculate the potential of the field formed by several charged objects.
The potential φ of the resulting electrostatic field formed by several charges at a given point in space is calculated as the sum of the potentials of the fields formed by each of the charges separately:
φ \u003d φ 1 + φ 2 + ... + φ n,
where φ 1 - the potential of the field formed by the first charge; φ 2 - the potential of the field formed by the second charge; …; φ n is the potential of the field formed by the nth charge.
In order to calculate the potential of the field created by several charges Q 1, Q 2, ..., Q n at a given point in space, use the following algorithm:
1) write down the potentials of the fields formed by each of the charges Q 1, Q 2, ..., Q n (separately) taking into account the sign of the charges:
φ 1 , φ 2 , …, φ n ,
where φ 1 - the potential of the field formed by the first charge; φ 2 - the potential of the field formed by the second charge; …; φ n is the potential of the field formed by the nth charge;
2) calculate the potential of the resulting field as the algebraic sum of the potentials written above:
φ \u003d φ 1 + φ 2 + ... + φ n.
Example 12. Two point charges q 1 \u003d 5 μC and q 2 \u003d -2 μC are at points (5; 0) and (0; 2) of the rectangular coordinate system xOy, where the coordinates x, y are expressed in meters. Calculate the potential of the resulting field at the origin of the coordinate system if the permittivity of the medium is equal to one.
Solution . The figure shows the coordinate system and charges located at points with given coordinates. The potential of the resulting electrostatic field at the origin of the coordinate system is the algebraic sum
φ \u003d φ 1 + φ 2,
where φ 1 - the potential of the field formed by the first charge; φ 2 - the potential of the field formed by the second charge.
Let us calculate the potential of the resulting field at the origin of the coordinate system using the algorithm:
1) the potentials of the fields created by each of the charges separately are determined by the following formulas:
φ 1 = k q 1 r 1 ,
where k is the coefficient of proportionality, k = 9.0 ⋅ 10 9 N ⋅ m 2 /Cl 2; q 1 - charge located at the point with coordinates (5; 0); r 1 - distance from the charge q 1 to the origin of the coordinate system, r 1 = 5 m;
φ 2 \u003d k q 2 r 2,
where q 2 is the charge (taking into account the sign) located at the point with coordinates (0; 2); r 2 - distance from the charge q 2 to the origin of the coordinate system, r 2 = 2 m;
φ = φ 1 + φ 2 = φ 1 − | φ 2 | = k q 1 r 1 − k | q 2 | r2.
The calculation gives the desired value of the potential:
φ = 9 ⋅ 10 9 ⋅ 5 ⋅ 10 − 6 5 − 9 ⋅ 10 9 ⋅ 2 ⋅ 10 − 6 2 = 0 V.
At the origin, the potential of the resulting field is zero.
Example 13. Three vertices of a square with a side of 60 cm contain positive charges of 0.30 μC each. Find the potential of the resulting field at the fourth vertex of the square. The permittivity of the medium in which the system of charges is located is equal to unity.
Solution . The figure shows a square with three vertices containing identical positive charges. The potential of the resulting field must be determined at the vertex A .
Let's calculate the potential of the resulting field at the fourth vertex of the square using the algorithm:
1) the potentials of the fields formed at point A by charges q 1, q 2 and q 3 separately are determined by the following formulas:
- field formed by the charge q 1, -
φ 1 = k q 1 r 1 = k q a ,
where k is the coefficient of proportionality, k = 9.0 ⋅ 10 9 N ⋅ m 2 /Cl 2; q 1 = q; r 1 - distance from q 1 to point A , r 1 = a ;
- field formed by the charge q 2, -
φ 2 \u003d k q 2 r 2 \u003d k q a 2,
where q 2 \u003d q; r 2 - distance from q 2 to point A, r 2 = a 2;
- field formed by the charge q 3, -
φ 3 \u003d k q 3 r 3 \u003d k q a,
where q 3 \u003d q; r 3 - distance from q 3 to point A, r 3 = a;
2) the potential of the resulting field is the algebraic sum of the potentials written above
φ = φ 1 + φ 2 + φ 3 = k q a + k q a 2 + k q a = k q a (2 + 1 2) = k q a ⋅ 4 + 2 2 .
Let's calculate:
φ = 9.0 ⋅ 10 9 ⋅ 0.30 ⋅ 10 − 6 60 ⋅ 10 − 2 ⋅ 4 + 2 2 = 12 ⋅ 10 3 V = 12 kV.
The potential of the electrostatic field at the fourth vertex of the square is 12 kV.
Example 14. Two concentric spheres with radii of 0.25 and 0.50 m are uniformly charged with charges of -0.80 and 0.50 μC, respectively. Find the potential of a field point located at a distance of 1.0 m from the center of the spheres. The charge system is in a vacuum.
Solution . Let's make an illustration to the condition of the problem. Concentric spheres have a common center, a sphere of smaller radius 1 is charged with a negative charge, and a sphere of larger radius 2 is positive.
The potential of the electrostatic field at point M is the algebraic sum of the potentials of the fields formed by the first φ 1 and second φ 2 spheres:
φ \u003d φ 1 + φ 2.
Let's calculate the potential of the resulting field using the algorithm:
1) the potentials of the fields formed at point M by charges q 1 and q 2 distributed over the surface of the inner and outer spheres, respectively, are separately determined by the following formulas:
- field formed by the charge q 1, -
φ 1 = k q 1 r 1 = k q 1 l ,
where k is the coefficient of proportionality, k ≈ 9 ⋅ 10 9 N ⋅ m 2 /Cl 2; q 1 - charge distributed over the surface of the inner sphere, q 1 = −|q 1 |; r 1 - distance from the center of the spheres to the point M, r 1 = l;
- field formed by the charge q 2, -
φ 2 \u003d k q 2 r 2 \u003d k q 2 l,
where q 2 is the charge distributed over the surface of the outer sphere; r 2 - distance from the center of the sphere to the point M, r 2 = r 1 = l;
2) the potential of the resulting field is the algebraic sum of the potentials written above
φ = φ 1 + φ 2 = k q 1 l + k q 2 l = k l (q 1 + q 2) = k l (− | q 1 | + q 2) .
Let's calculate:
φ = 9 ⋅ 10 9 1.0 (− 0.80 + 0.50) ⋅ 10 − 6 = − 2.7 ⋅ 10 3 V = − 2.7 kV.
The potential of the resulting electrostatic field at point M is −2.7 kV. The result does not depend on the radii of the spheres.
The potential φ at any point of the electrostatic field is a physical quantity determined by the potential energy of a single positive charge placed at this point. The potential of the field created by a point charge Q is
Potential - a physical quantity, which is determined by the work of moving a single positive electric charge when it is removed from a given point of the field to infinity. This work is numerically equal to the work done by external forces (against the forces of the electrostatic field) in moving a unit positive charge from infinity to a given point in the field.
The unit of potential is the volt (V): 1 V is equal to the potential of such a point in the field at which a charge of 1 C has a potential energy of 1 J (1 V = 1 J/C). Considering the dimension of the volt, it can be shown that the unit of electrostatic field strength introduced earlier is indeed 1 V/m: 1 N/Cl=1 N m/(Cl m)=1 J/(Cl m)=1 V/m.
From formulas (3) and (4) it follows that if the field is created by several charges, then the potential of the given field of the system of charges is equal to the algebraic sum of the potentials of the fields of all these charges:
The strength at any point of the electric field is equal to the potential gradient at this point, taken with the opposite sign. The minus sign indicates that the intensity E is directed in the direction of decreasing potential.
E = - grad phi = - N phi.
To establish a connection between the power characteristic of the electric field - the strength and its energy characteristic - the potential, consider the elementary work of the electric field forces on an infinitely small displacement of a point charge q: dA = q E dl, the same work is equal to the loss potential energy charge q: dA = - dWp = - q dphi, where d phi is the change in the potential of the electric field over the travel length dl. Equating the right parts of the expressions, we get: E dl = -d phi or in the Cartesian coordinate system
Ex dx + Ey dy + Ez dz = -d fi
where Ex, Ey, Ez are the projections of the intensity vector on the axes of the coordinate system. Since the expression is a total differential, then for the projections of the intensity vector we have
The expression in brackets is the gradient of the phi potential.
Superposition principle as a fundamental property of fields. General expressions for the strength and potential of the field created at a point with a radius vector by a system of point charges located at points with coordinates. (See item 4)
If we consider the principle of superposition in the most general sense, then according to it, the sum of the impact of external forces acting on a particle will be the sum of the individual values of each of them. This principle applies to various linear systems, i.e. systems whose behavior can be described by linear relations. An example is a simple situation when a linear wave propagates in some particular medium, in which case its properties will be preserved even under the influence of disturbances arising from the wave itself. These properties are defined as a specific sum of the effects of each of the harmonic components.
The superposition principle can also take other formulations that are completely equivalent to the one given above:
· The interaction between two particles does not change when a third particle is introduced, which also interacts with the first two.
· The interaction energy of all particles in a many-particle system is simply the sum of the energies of pair interactions between all possible pairs of particles. There are no multiparticle interactions in the system.
· The equations describing the behavior of a multiparticle system are linear in the number of particles.
6 The circulation of the tension vector is the work done by electrical forces when moving a unit positive charge along a closed path L
Since the work of the electrostatic field forces in a closed loop is zero (the work of the potential field forces), therefore, the circulation of the electrostatic field strength in a closed loop is zero.
Field potential. The work of any electrostatic field when moving a charged body in it from one point to another also does not depend on the shape of the trajectory, as well as the work of a uniform field. On a closed trajectory, the work of the electrostatic field is always zero. Fields with this property are called potential fields. In particular, the electrostatic field of a point charge has a potential character.
The work of a potential field can be expressed in terms of a change in potential energy. The formula is valid for any electrostatic field.
7-11 If the lines of force of a uniform electric field of strength penetrate some area S, then the flux of the intensity vector (we used to call the number of lines of force through the area) will be determined by the formula:
where En is the product of the vector and the normal to the given area (Fig. 2.5).
Rice. 2.5
The total number of lines of force passing through the surface S is called the flow of the intensity vector FU through this surface.
In vector form, you can write - the scalar product of two vectors, where the vector .
Thus, the vector flow is a scalar, which, depending on the angle α, can be either positive or negative.
Consider the examples shown in Figures 2.6 and 2.7.
 | |||
| Rice. 2.6 | Rice. 2.7 | ||
For Figure 2.6, surface A1 is surrounded by a positive charge and the flow here is directed outward, i.e. The A2– surface is surrounded by a negative charge, and here it is directed inward. The total flow through surface A is zero.
For Figure 2.7, the flux will be non-zero if the total charge inside the surface is non-zero. For this configuration, the flux through surface A is negative (count the number of field lines).
Thus, the intensity vector flux depends on the charge. This is the meaning of the Ostrogradsky-Gauss theorem.
Gauss theorem
The experimentally established Coulomb's law and the principle of superposition make it possible to completely describe the electrostatic field of a given system of charges in vacuum. However, the properties of the electrostatic field can be expressed in a different, more general form, without resorting to the concept of the Coulomb field of a point charge.
Let's introduce a new physical quantity, which characterizes the electric field - the flux Φ of the vector of the electric field strength. Let some sufficiently small area ΔS be located in the space where the electric field is created. The product of the vector module and the area ΔS and the cosine of the angle α between the vector and the normal to the site is called the elementary flux of the intensity vector through the site ΔS (Fig. 1.3.1):
Let us now consider some arbitrary closed surface S. If we divide this surface into small areas ΔSi, determine the elementary fluxes ΔΦi of the field through these small areas, and then sum them up, then as a result we get the vector flow Φ through the closed surface S (Fig. 1.3.2 ):
Gauss' theorem states:
The flow of the electrostatic field strength vector through an arbitrary closed surface is equal to the algebraic sum of the charges located inside this surface, divided by the electric constant ε0.
where R is the radius of the sphere. The flux Φ through the spherical surface will be equal to the product of E and the area of the sphere 4πR2. Consequently,
Let us now surround the point charge with an arbitrary closed surface S and consider an auxiliary sphere of radius R0 (Fig. 1.3.3).
Consider a cone with a small solid angle ΔΩ at the vertex. This cone selects a small area ΔS0 on the sphere, and an area ΔS on the surface S. The elementary flows ΔΦ0 and ΔΦ through these areas are the same. Really,
In a similar way, one can show that if the closed surface S does not enclose a point charge q, then the flow Φ = 0. Such a case is shown in fig. 1.3.2. All lines of force of the electric field of a point charge penetrate the closed surface S through and through. There are no charges inside the surface S, therefore, in this region, the lines of force do not break and do not originate.
The generalization of the Gauss theorem to the case of an arbitrary distribution of charges follows from the principle of superposition. The field of any charge distribution can be represented as a vector sum of electric fields of point charges. The flow Φ of a system of charges through an arbitrary closed surface S will be the sum of the flows Φi of the electric fields of individual charges. If the charge qi turned out to be inside the surface S, then it makes a contribution to the flow equal to if this charge turned out to be outside the surface, then the contribution of its electric field to the flow will be equal to zero.
Thus, the Gauss theorem is proved.
Gauss's theorem is a consequence of Coulomb's law and the superposition principle. But if we accept the statement contained in this theorem as an initial axiom, then Coulomb's law will turn out to be its consequence. Therefore, Gauss's theorem is sometimes called an alternative formulation of Coulomb's law.
Using the Gauss theorem, in a number of cases it is easy to calculate the electric field strength around a charged body if the given charge distribution has some kind of symmetry and the general structure of the field can be guessed in advance.
An example is the problem of calculating the field of a thin-walled, hollow, uniformly charged long cylinder of radius R. This problem has axial symmetry. For reasons of symmetry, the electric field must be directed along the radius. Therefore, to apply the Gauss theorem, it is advisable to choose a closed surface S in the form of a coaxial cylinder of some radius r and length l, closed at both ends (Fig. 1.3.4).
For r ≥ R, the entire flow of the intensity vector will pass through side surface a cylinder whose area is 2πrl, since the flow through both bases is zero. Applying the Gauss theorem gives:
This result does not depend on the radius R of the charged cylinder, so it is also applicable to the field of a long uniformly charged filament.
To determine the field strength inside a charged cylinder, it is necessary to construct a closed surface for the case r< R. В силу симметрии задачи поток вектора напряженности через боковую поверхность гауссова цилиндра должен быть и в этом случае равен Φ = E 2πrl. Согласно теореме Гаусса, этот поток пропорционален заряду, оказавшемуся внутри замкнутой поверхности. Этот заряд равен нулю. Отсюда следует, что электрическое поле внутри однородно заряженного длинного полого цилиндра равно нулю.
Similarly, Gauss's theorem can be applied to determine the electric field in a number of other cases where the charge distribution has some kind of symmetry, for example, symmetry about the center, plane or axis. In each of these cases, it is necessary to choose a closed Gaussian surface of an expedient form. For example, in the case of central symmetry, it is convenient to choose a Gaussian surface in the form of a sphere centered at a point of symmetry. With axial symmetry, a closed surface must be chosen in the form of a coaxial cylinder closed at both ends (as in the example discussed above). If the distribution of charges does not have any symmetry and the general structure of the electric field cannot be guessed, the application of the Gauss theorem cannot simplify the problem of determining the field strength.
Consider another example of a symmetrical distribution of charges - the definition of the field of a uniformly charged plane (Fig. 1.3.5).
In this case, it is advisable to choose the Gaussian surface S in the form of a cylinder of some length, closed at both ends. The axis of the cylinder is directed perpendicular to the charged plane, and its ends are located at the same distance from it. Due to symmetry, the field of a uniformly charged plane must be directed along the normal everywhere. Applying the Gauss theorem gives:
|
where σ is the surface charge density, i.e., the charge per unit area.
The resulting expression for the electric field of a uniformly charged plane is also applicable in the case of flat charged areas of a finite size. In this case, the distance from the point at which the field strength is determined to the charged area must be significantly less than the size of the area.
And schedules for 7 - 11
1. The intensity of the electrostatic field created by a uniformly charged spherical surface.
Let a spherical surface of radius R (Fig. 13.7) bear a uniformly distributed charge q, i.e. the surface charge density at any point on the sphere will be the same.
a. We enclose our spherical surface in a symmetric surface S with radius r>R. The intensity vector flux through the surface S will be equal to
According to the Gauss theorem
Consequently
c. Let us draw through the point B, located inside the charged spherical surface, the sphere S with radius r 2. Electrostatic field of the ball. Let we have a ball of radius R, uniformly charged with bulk density. At any point A, lying outside the ball at a distance r from its center (r> R), its field is similar to the field of a point chargelocated at the center of the ball. Then outside the ball and on its surface (r=R) At point B, lying inside the ball at distances r from its center (r>R), the field is determined only by the charge enclosed inside the sphere of radius r. The intensity vector flow through this sphere is equal to on the other hand, according to the Gauss theorem From a comparison of the last expressions it follows where is the permittivity inside the sphere. The dependence of the field strength created by a charged sphere on the distance to the center of the ball is shown in (Fig. 13.10) Let the plane have an infinite extent and the charge per unit area is equal to σ. From the laws of symmetry it follows 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 must be the same. Let us limit a part of the charged plane to an imaginary cylindrical box, so that the box is cut in half and its generators are perpendicular, and two bases, each having an area S, are parallel to the charged plane (Figure 1.10). total vector flow; tension is equal to the vector times the area S of the first base, plus the vector flow through the opposite base. The flux of tension through the side surface of the cylinder is equal to zero, since the lines of tension do not cross them. In this way, In this way, 12. Field of a uniformly charged sphere. Let the electric field be created by the charge Q, uniformly distributed over the surface of a sphere of radius R(Fig. 190). To calculate the field potential at an arbitrary point located at a distance r from the center of the sphere, it is necessary to calculate the work done by the field when moving a unit positive charge from a given point to infinity. Earlier we proved that the field strength of a uniformly charged sphere outside it is equivalent to the field of a point charge located at the center of the sphere. Therefore, outside the sphere, the potential of the field of the sphere will coincide with the potential of the field of a point charge φ
(r)=Q 4πε
0r . (1) In particular, on the surface of a sphere, the potential is equal to φ
0=Q 4πε
0R. There is no electrostatic field inside the sphere, so the work to move a charge from an arbitrary point inside the sphere to its surface is zero A= 0, therefore, the potential difference between these points is also equal to zero Δ φ
= -A= 0. Therefore, all points inside the sphere have the same potential, which coincides with the potential of its surface φ
0=Q 4πε
0R . So, the distribution of the field potential of a uniformly charged sphere has the form (Fig. 191) φ
(r)=⎧⎩⎨Q 4πε
0R, npu r<RQ 4πε
0r, npu r>R . (2) Please note that there is no field inside the sphere, and the potential is different from zero! This example is a vivid illustration of the fact that the potential is determined by the value of the field from a given point to infinity. Gauss theorem for vector can be successfully used as an effective tool for calculating the strength and potential of the electric field of a certain charge distribution, when the integral on the left can be turned into the product of the surface area over which the integration is performed by the value of the vector component normal to the surface, that is, when It is clear that in order to calculate vector this will suffice, firstly, when vector perpendicular to the surface. Therefore, the integration surface must be equipotential surface calculated field. Her form need to know in advance. Finally, secondly, at all points of this - equipotential - surface, the component normal to it must have the same value, otherwise, it cannot be taken out from under the integral sign and it will be possible to find only the average value on the equipotential surface. We emphasize that from the fact that the surface is equipotential, namely, from the fact that it does not at all follow that at points on this surface. Looking ahead, we point out that, for example, the surface of a charged conductor, under the condition of an equilibrium charge distribution on it, is always equipotential, but if it is not a ball, but a body of complex shape, then in the vicinity of protrusions (points) the field strength can be orders of magnitude greater than in the vicinity of depressions on the surface. The persistence requirement is a separate requirement. From what has been said above, it follows that the Gauss theorem can quickly and easily lead to the result (vector ) only in the case when the charge distribution that creates the field has a high degree symmetry, respectively, the shape of the equipotential surfaces of the field is known in advance and there is confidence that on these surfaces. If all this is the case, then the solution looks like this: spherical symmetry With a spherically symmetric charge distribution, the field generated by it is also spherically symmetrical. Vector (and scalar) fields with this symmetry are also called central fields. The centrally symmetric field can generally be written as Here -
radius vector starting at the center of symmetry of the field r- its module, - the radial component of the field strength, depending only from the distance to its center of symmetry. The potential of such a field depends only on and And, moreover, as follows from, under arbitrary normalization, the field potential has the form Thus, the applicability conditions are satisfied and we can use this relation. Let us take as an equipotential spherical surface of some current radius r, its area. In view of the assumed continuity of the charge distribution, we use the expression: where is the bulk charge density. Again, taking into account the spherical symmetry of the charge distribution - depends only on , it is natural to take an infinitely thin spherical layer with an inner radius and an outer radius as an element of volume. The volume of such a layer , as a result we get Finally, for any spherically symmetric charge distribution, when , we get Continuation of calculations requires specifying the type of dependence of the charge density on the modulus of the radius vector . The field is uniform over the volume of the charged sphere The charge distribution uniform over the volume of a ball of radius (Fig. 1.41) means that its charge density has the form Rice. 1.41. Lines of force of the electric field of a uniformly charged ball It should not be forgotten that by the condition there are no charges outside the ball. Since at the point the charge density changes abruptly: the limit "on the left" is non-zero Accordingly, the field grows linearly with increasing distance to the center of the ball, which is explained simply: the surface area, and the charge inside it In the second case, the integral is "cut off from above" at: The last expression takes into account that , where is the total charge of the ball. Thus, outside the ball, its field is the field of a point charge equal to the total charge of the ball and placed in the center of this ball: Both expressions can be combined into one formula. If we use the full charge of the ball, we get: If instead of the total charge of the ball, charge density is used as a parameter, these formulas will take the following form (Fig. 1.42): Rice. 1.42. The distribution of the electric field strength of a uniformly charged ball Formulas and express the same dependence, their convenience is determined by what parameters are set: or . From these formulas it is clearly seen that the field strength on the surface of the ball is continuous, that is, it has no discontinuity. This is due to the fact that in this case, the discontinuity of the charge density on the surface of the ball of the first kind is of finite magnitude: from to zero. Therefore, both in and in the upper and lower formulas, signs of non-strict inequalities are placed. In what cases the field strength can tolerate discontinuity will be clear from the following example. The field potential is easy to find by substituting, for example, from to and performing integration. We get: where and are constants of integration, which are found from the following considerations. The constant is determined from the normalization condition, for example, to zero at infinity Where . The constant is determined from the condition of continuity of the potential on the surface of the ball, that is, for: We note that the requirement of potential continuity is often referred to as “joining” of two solutions at the interface. In this case, this is the boundary between two areas: the area where there is a charge (inside the ball), and the area where it is not (outside the ball). Already now it can be noted that the potential is continuous in all cases, except for one: the so-called "double layer". Imagine a surface on one side of which a positive charge is distributed with a density, and on the other side of which a negative charge is distributed with a density. Such a surface is called a double layer; on this surface, the potential suffers a discontinuity. Such a (flat) surface can be obtained by infinitely bringing together two plates of a flat capacitor. The same can be done for a capacitor of any shape, for example, spherical or cylindrical. In all other cases the potential is continuous. Substituting the obtained values of the integration constants into, we write the final result in the form With this normalization, the potential at the center of the ball is different from zero and is equal to The results obtained are illustrated in Figure 1.43 below. Rice. 1.43. Strength (1) and potential (2) of the electric field of a uniformly charged ball of radius R in units of strength and potential on its surface (r = R) Field of a uniformly charged spherical surface In this case of a uniform charge distribution over a spherical surface, as in the previous one, spherical symmetry takes place, therefore general formulas obtained above are applicable here as well. However, they must be treated with some caution for the following reason. The volume charge density included in the right-hand side behaves in this case in the following interesting way: Indeed, there is a charge only on surfaces, that is, at , everywhere inside, that is, at and everywhere outside, that is, when there are no charges. What volumetric the charge density at the points of the surface goes to infinity (+∞ in the case of a positive charge and –∞ in the case of a negative one) can be shown as follows. The figure next to it shows a section of some surface along which superficial charge density. To determine the value voluminous charge density at some point on the surface, consider a cylinder (Fig. 1.45), the upper base of which is above the surface, and the lower one is below the surface. The area of the bases of the cylinder is , the height is , and the volume is . The charge inside the cylinder, the volumetric charge density, by definition, is equal to the limit of the ratio of the charge inside a certain volume to the value of this volume when the latter tends to zero (with all the reservations regarding the volume of "physically infinitely small"). We get Rice. 1.45. Surface charge density It is important that the density on the surface is equal to infinity. Functions of this kind (everywhere, except for one point - zero, and at this single point - infinity) belong to the class of so-called generalized functions, are called Dirac functions in honor of the physicist Dirac, who first introduced such a function into the everyday life of physics to meet the needs quantum mechanics. We will not study in detail here and use functions of this kind in calculations. Our goal is to show that the consideration of formally infinitely thin charged surfaces leads to the appearance of discontinuities (infinite) in the volume charge density, which, in turn, generates infinite discontinuities in the electric field strength on such a charged surface. We emphasize that the field potential remains continuous in this case. The way out is simple. For all, we use the first of the formulas with , we obtain that the field is absent everywhere inside a uniformly charged spherical shell: . For all, the second formula from As in the case of a charged sphere uniform in volume, outside a uniformly charged spherical shell, its field is the field of a point charge placed at the center of this shell and equal to its total charge. In this case, of course. The final result is this: On the spherical surface itself, the field strength in this case suffers a discontinuity. The dependence of the radial component of the field on the distance to the center of the spherical surface is shown in Fig. . 1.46. The dependence of the potential on the distance to the center of the spherical shell can be obtained by integrating. When normalized to zero at infinity, the result looks like this: The dependence is shown in fig. 1.47. Rice. 1.47. Potential of a uniformly charged sphere A uniform (uniform) charge distribution over an infinitely long cylindrical surface (Fig. 1.48) has cylindrical, translational and mirror symmetry. This means the following. When such a charge distribution is rotated around the axis of the cylindrical surface through any angle, it coincides with itself. With a shift (transfer, translation) of such a charge distribution at any distance along the axis of symmetry, it also coincides with itself. And, finally, if we draw a plane perpendicular to the axis through any point on the axis of symmetry, and reflect in this plane the “upper” part of the charge distribution as in a mirror, then the reflection of the “upper” part will coincide with the “lower” and vice versa, the reflection of the “lower” matches the top. In other words, this charge distribution is invariant under these transformations. Consequently, the electric field created by this charge distribution must be invariant (coincide with itself) under the indicated transformations. Rice. 1.48. Infinitely long cylindrical surface We introduce a cylindrical coordinate system: we direct the axis along the axis of symmetry, - the distance to the axis of symmetry, - the azimuthal angle, the angle of rotation around the axis of symmetry, - as before, the field potential. It follows from the symmetry properties that the field potential cannot depend either on the coordinate - the translational symmetry will be broken, or on the coordinate - the axial (cylindrical) symmetry will be broken. It remains only the dependence on - the distance to the axis of the cylinder. In this way: Respectively the electric field strength vector is directed along radial lines perpendicular to the symmetry axis (Fig. 1.49), and its value depends only on the distance to the axis. Potential surfaces are cylinders coaxial with a charged cylindrical surface. Rice. 1.49. The electric field strength vector is directed along radial straight lines Using these circumstances, we will integrate on the left side of the Gauss theorem over the closed surface of a cylinder with base radius and height coaxial with the considered charged cylindrical surface of radius . The flow through the bases of the cylinder is equal to zero due to the fact that on the bases , and the flow through its side surface is equal to the product of its area: . Accordingly, the total (through the entire closed surface of the considered cylinder) vector flow is equal to At , the charge inside the cylinder is equal to where is the linear charge density numerically equal to the charge per unit length of the cylindrical surface. According to the Gauss theorem whence for we get When inside the cylinder, through the surface of which the vector flow is calculated, there are no charges, and therefore the field is equal to zero. Combining these two results, we finally get (Fig. 1.50): Due to the surface nature of the charge distribution (see the previous calculation for more details) on the most charged surface, that is, at , the radial component of the field suffers a break. Integration (1.51) (see also (1.49)), the requirement of continuity of the potential at , and normalization of , lead to the following dependence of the potential on the distance to the axis of the cylindrical surface: In this case, when an infinitely large modulo charge is distributed over an infinitely long cylinder, this refers to those cases where the normalization to zero at infinity is meaningless. As can be seen from (1.52), the dependence of the potential on the distance to the axis is logarithmic, normalization to zero at infinity, in the language of formulas (1.52), means that , but, then the potential will be infinitely large in absolute value at any final distance from the axis of the charged surface, which is meaningless. The choice of the final distance from the axis of symmetry at which it is convenient to consider the potential equal to zero does not cause difficulties and is determined by the specifics of the problem. For example, nothing prevents us from putting , then the potential everywhere inside and on the most charged surface will be equal to zero. The field is infinite evenly charged plane Let the surface charge density be From the analysis of symmetry, it is quite obvious that the potential at any point outside the plane can depend only on the distance from this point to the plane. Let us direct the axis of the Cartesian coordinate system perpendicular to the plane, and let the axes and let belong to the plane itself, then Moreover, due to mirror symmetry, the field "in front" of the plane differs from the field "behind" the plane only the direction of the vector . This means that the dependence on must be odd, and the dependence of the potential on must be even. By virtue of these considerations, we take a closed surface - the one for which we will write the Gauss theorem - of the following form (Figure 1.51). Rice. 1.51. Electric field of a charged plane This is a cylinder with a side surface perpendicular to the plane and with bases parallel to the plane. The height of the cylinder, the area of the bases. Taking into account the oddness of the dependence , it is convenient to place the bases of the cylinder at the same distance from the plane, then the contribution of the bases to the flow will be the same. The field strength at the bases, firstly, is perpendicular to them, secondly, is co-directed with the outer normal, thirdly, it is the same at all their points in absolute value The contribution to the flow of the vector from the lateral surface is equal to zero, since on the lateral surface . Therefore, the total flow through the entire closed cylindrical surface is equal to Inside the considered cylindrical surface there is a charge where is the charge density on the plane. According to the Gauss theorem hence the module field strength of the charged plane equals We emphasize that the result obviously does not depend on how far from the plane the bases of the considered cylinder are located. It follows that on each side of the plane, the electric field created by it is uniform. Using the previously introduced axis perpendicular to the charged plane, the field on both sides of the plane can be described by one formula, suitable for any sign of the charge on the plane Here, is the unit vector of the axis. Integrating taking into account for the dependence on the field potential of the plane, it is easy to obtain: The potential in is normalized by the condition . Here, as in the example with an infinitely long charged cylindrical surface, the potential grows with distance to infinity, so the normalization to zero at infinity is meaningless. Field lines of the field of the charged plane are shown in fig. 1.52 and 1.53. Rice. 1.52. Field of a positively charged plane Rice. 1.53. The field of a negatively charged plane Flat capacitor field Let us determine the strength of the field created by two infinite parallel planes charged uniformly and oppositely. The charge densities on the planes are identical and equal in modulus, respectively: and (an ideal flat capacitor). With the help of fig. 1.54 it is easy to figure out that in the gap between the planes, the fields created by them are directed in one direction, therefore, inside the total field is twice the field from each of the planes. Outside of the planes, the fields they create are directed in opposite directions, respectively, the total field from both planes is zero (Fig. 1.55). Rice. 1.54. Electric field of a flat capacitor Rice. 1.55. Electric field of oppositely charged planes In Appendix 6, an example with the movement of a charged particle in a constant electric field is analyzed. The field potential of a charged disk As has been noted more than once, knowing the potential of the field of a point charge and using the principle of superposition, in principle always, it is possible to calculate the potential of the field created by any distribution of charges. Let us find, for example, the potential of the electric field created on the axis of a thin disk of radius R, uniformly charged with a surface charge density (Fig. 1.57). Due to axial symmetry, at points on the axis, two field strength components perpendicular to the axis are equal to zero: , it remains to find - the field component directed along the axis. Can be expanded in a series, limited to the first two terms of the expansion Coulomb's law and the dimension of space The space in which we live has three dimensions. In other words, three coordinates (for example, in Cartesian or spherical systems) are needed to specify the position of a point BUT(Fig. 1.58). It turns out that the number 3 is closely related to the form of Coulomb's law. We have seen that the Ostrogradsky-Gauss theorem follows from the Coulomb law. The converse is also true, Coulomb's law can be deduced from the Ostrogradsky-Gauss theorem. But this theorem is more general than Coulomb's law. In particular, it is applicable to spaces with dimension , where it does not have to be equal to three. So, in two-dimensional space, our area plays the role of volume. Indeed, a sphere is a locus of points in space that are equidistant from the center. According to this definition, a two-dimensional sphere is a circle with a radius of the dimensional world proportional to the dimensional world. When we obtain from here the inverse square law (Coulomb's law). When we find In fact, we are already familiar with this behavior of the electric field. It is this law (10.17) that we derived for the field of an infinite charged cylinder. If you think carefully and remember the location of the lines of force of the cylinder, it becomes clear that nothing depends on the coordinate along the axis of the cylinder. Thus, this system simulates an electric field in a two-dimensional world. Now it is easier to understand that a charged plane imitates a point charge in a one-dimensional world: everything depends on only one coordinate - the distance to the plane. But we found above that the electric field does not depend on this distance. And from formula (10.49) at it also follows that the strength grad ) should give an expression for the strength of the electric field. Interesting conclusions follow from this. Since the potentials grow at infinity in one- and two-dimensional worlds, an infinitely large amount of work is needed to separate two attracting charges. This means that in low-dimensional worlds only finite motion of two attracting bodies (charges, masses) is possible. Recall that motion in a bounded region of space is called finite. Therefore, in worlds with it is impossible to ionize an atom, it is impossible to launch a satellite beyond solar system etc. In such a world there would be no chemical reactions, galaxies and stars could not have evolved. In a word, life there would be stagnantly boring. One would expect a more pleasant pastime in the multidimensional worlds. Alas, this turns out to be an illusion. Study of the equation of motion leads to the conclusion that at essentially there is no finite motion: it is realized only for circular orbits, and even then it is unstable - the slightest perturbation leads to the fall of an electron (planet) on an attracting center or its (her) runaway to an infinitely long distance. It turns out that in such a world, atoms, planetary systems and everything else could not have formed at all. There is no stability in the worlds of higher dimension - this is an alternative to "stagnant" low-dimensional worlds. Only at is possible both stable finite and infinite motions. It turns out that three-dimensional space is the only convenient form of existence and movement of matter, at least of its known types, which we study in physics. Additional Information http://hea.iki.rssi.ru/~nik/astro/spher.htm - spherical coordinate system; http://edu.ioffe.ru/register/?doc=physica/lect3.ch2.tex - finite motion, Kepler's problem. Formula can be used to find the potential function of a given potential field a(x, y, z)=P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k. To do this, we fix the starting point M 0 (x 0, y 0, z 0) and connect it to the current point M (x, y, z) of the polyline M 0 AVM, the links of which are parallel to the coordinate axes, namely, M 0 A? Ox , AB? Oy, BM? Oz (Fig. 6.2). Then formula (6.6) takes the form where x, y, z are the coordinates of the current point on the links of the polyline, along which the integration is carried out. Example 6.7. Prove that the vector field a= (e + z)i + (x + z)j + (x + y)k is potential, and find its potential. Solution. 1st way. A necessary and sufficient condition for the potentiality of the field a(M) is that rot a(M) is equal to zero. In our case i.e. the field is potential. We find the potential of this field using formula (6.10). Let us take the origin of coordinates O(0, 0, 0) as the initial fixed point. Then we get where C is an arbitrary constant. 2nd way. By definition, the potential is such a scalar function for which grad φ=a. This vector equality is equivalent to three scalar equalities: Integrating (6.12) over x, we get where f(y, z) is an arbitrary differentiable function of y and z. Differentiating both sides of (6.12) with respect to y and taking (6.11) into account, we obtain a relation for finding the as yet undefined function f(y, z). We have Integrating (6.16) over y, we have where F(z) - while undefined function from z. Substituting (6.17) into (6.11) we get Differentiating the last equality with respect to z and taking into account relation (6.12), we obtain an equation for finding F(z): From here, so. 3rd way. By the definition of the total differential of a function, we have Substituting here instead of the partial derivatives , , their expressions from (6.10), (6.11), (6.12), we obtain dφ =(y + z)dx + (x + z)dy + (x + y)dz or, after simple transformations, dφ=(ydx+xdy)+(zdx+xdz)+(ydz+zdy)=d(xy)+d(xz)+d(yz)=d(xy +xz +yz). dφ=d(xy + yz + zх). Hence it follows that In the case when the region Ω is star-shaped with the center at the origin O(0, 0, 0), the potential φ(M) of the vector field a=a(M) at the point M(x, y, z) can be found by the formula where r(M)=xi + yj + zk is the radius vector of the point M(x, y, z), and the point (tx,ty,tz) for , the segment OM runs through the straight line passing through the points O and M. Example 6.8. Find the vector field potential a = yzi + xzj + xyk. Solution. It is easy to see that rot a 0, i.e., the given vector field is potential. This field is defined in the entire three-dimensional space, which is stellar with the center at the origin O(0, 0, 0), so to find its potential, we use formula (6.12). Since in this case a( )=a(tx, ty, tz)= t 2 yzi + t 2 xzj + t 2 xyk, then the scalar product of vectors a() and r(M) equals (a( ), r(M))=t 2 (xyz+xyz+xyz)=3t 2xyz. Seeking Potential
(13.10)
(13.11)
(13.12)
On the other hand, according to the Gauss theorem
(13.15)
Outside the plate, the vectors from each of them are directed in opposite directions and cancel each other out. Therefore, the field strength in the space surrounding the plates will be equal to zero E=0.
.
It remains to choose a surface according to the symmetry of the charge distribution and calculate the charge inside .
.
.
, and the limit "on the right" is equal to zero 
, the calculation will have to be carried out in two stages: first for a spherical surface of radius (it lies inside the ball), and then for a spherical surface of radius (it encloses the ball). In the first case
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.
.
Rice. 1.44. Electric field strength of a uniformly charged sphere
Rice. 1.46. Dependence of the field on the distance to the center of the spherical shell
Rice. 1.50. Electric field strength of a uniformly charged cylindrical surface
. Such a charge distribution over an infinite plane is characterized by the fact that its form does not depend on: a) rotation through any angle around any axis perpendicular to the plane, b) shift through any distance along a straight line lying in the plane and any direction. Finally, c) reflection of a given charge distribution in a mirror coinciding with the plane itself will leave it unchanged.
