Let's consider a method for determining the value and direction of the tension vector E at each point of the electrostatic field created by a system of stationary charges q 1 , q 2 , ..., Q n .
Experience shows that the principle of independence of the action of forces discussed in mechanics (see §6) is applicable to Coulomb forces, i.e. resultant force F, acting from the field on the test charge Q 0, equal to the vector sum of forces F i applied to it from the side of each of the charges Q i:
According to (79.1), F=Q 0 E And F i ,=Q 0 E i, where E is the strength of the resulting field, and E i is the field strength created by the charge Q i. Substituting the last expressions into (80.1), we get
Formula (80.2) expresses the principle of superposition (imposition) of electrostatic fields, according to which tension E the resulting field created by the system of charges is equal to geometric sum field strengths created at a given point by each of the charges separately.
The principle of superposition is applicable to calculate the electrostatic field of an electric dipole. Electric dipole- a system of two equal in modulus opposite point charges (+ Q, - Q), distance l between which there is significantly less distance to the considered points of the field. A vector directed along the dipole axis (a straight line passing through both charges) from a negative charge to a positive charge and equal to the distance between them is called dipole arml . Vector
coinciding in direction with the dipole arm and equal to the product of the charge
| Q| on the shoulder l , called electric dipole moment p or dipole moment(Fig. 122).
According to the superposition principle (80.2), the tension E dipole fields at an arbitrary point
E=E + + E - ,
Where E+ and E- - field strengths created by positive and negative charges, respectively. Using this formula, we calculate the field strength along the extension of the dipole axis and at the perpendicular to the middle of its axis.
1. Field strength along the extension of the dipole axis at the point A(Fig. 123). As can be seen from the figure, the dipole field strength at the point A is directed along the dipole axis and is equal in magnitude
E A =E + -E - .
Marking the distance from the point A to the middle of the dipole axis through l, based on formula (79.2) for vacuum we can write
According to the definition of a dipole, l/2< 2. Field strength at a perpendicular raised to the axis from its middle, at the point IN(Fig. 123). Dot IN equidistant from the charges, therefore Where r"
- distance from point IN to the middle of the dipole arm. From the similarity of isosceles- of the given triangles based on the dipole arm and vector еv, we obtain E B =E +
l/
r".
(80.5) Substituting the value (80.4) into expression (80.5), we obtain Vector E B has the direction opposite to the electric moment of the dipole (vector R directed from negative to positive charge). One of the tasks that electrostatics sets for itself is the assessment of field parameters for a given stationary distribution of charges in space. And the principle of superposition is one of the options for solving such a problem. Let us assume the presence of three point charges interacting with each other. With the help of experiment it is possible to measure the forces acting on each of the charges. To find the total force with which two other charges act on one charge, you need to add the forces of each of these two according to the parallelogram rule. In this case, the logical question is: are the measured force that acts on each of the charges and the totality of forces from two other charges equal to each other, if the forces are calculated according to Coulomb’s law. The research results demonstrate a positive answer to this question: indeed, the measured force is equal to the sum of the calculated forces according to Coulomb’s law on the part of other charges. This conclusion is written in the form of a set of statements and is called the principle of superposition. Definition 1 Superposition principle: The principle of superposition of charge fields is one of the foundations for the study of such a phenomenon as electricity: its significance is comparable to the importance of Coulomb’s law. In the case when we are talking about a set of charges N (i.e., several field sources), the total force experienced by the test charge q, can be determined by the formula: F → = ∑ i = 1 N F i a → , where F i a → is the force with which it affects the charge q charge
q i if there is no other N - 1 charge. Using the principle of superposition using the law of interaction between point charges, it is possible to determine the force of interaction between charges present on a body of finite dimensions. For this purpose, each charge is divided into small charges d q (we will consider them point charges), which are then taken in pairs; the interaction force is calculated and finally the vector addition of the resulting forces is carried out. Field interpretation: The field strength of two point charges is the sum of the intensities created by each of the charges in the absence of the other. For general cases, the principle of superposition with respect to tensions has the following notation: E → = ∑ E i → , where E i → = 1 4 π ε 0 q i ε r i 3 r i → is the intensity of the i-th point charge, r i → is the radius of the vector drawn from the i-th charge to a certain point in space. This formula tells us that the field strength of any number of point charges is the sum of the field strengths of each of the point charges, if there are no others. Engineering practice confirms compliance with the superposition principle even for very high field strengths. The fields in atoms and nuclei have a significant strength (of the order of 10 11 - 10 17 V m), but even in this case the principle of superposition was used to calculate energy levels. In this case, the results of calculations coincided with experimental data with great accuracy. However, it should also be noted that in the case of very small distances (on the order of ~ 10 - 15 m) and extremely strong fields, the superposition principle is probably not satisfied. Example 1 For example, on the surface of heavy nuclei at a strength of the order of ~ 10 22 V m, the superposition principle is satisfied, and at a strength of 10 20 V m, quantum mechanical nonlinearities of interaction arise. When the charge distribution is continuous (i.e. there is no need to take into account discreteness), the total field strength is given by the formula: E → = ∫ d E → . In this entry, integration is carried out over the charge distribution region: The principle of superposition makes it possible to find E → for any point in space for a known type of spatial charge distribution. Identical point charges q are given, located at the vertices of a square with side a. It is necessary to determine what force is exerted on each charge by the other three charges. Solution
In Figure 1 we illustrate the forces affecting any of the given charges at the vertices of the square. Since the condition states that the charges are identical, it is possible to choose any of them for illustration. Let's write down the summing force affecting the charge q 1: F → = F 12 → + F 14 → + F 13 → . The forces F 12 → and F 14 → are equal in magnitude, we define them as follows: F 13 → = k q 2 2 a 2 . Drawing 1 Now let’s set the direction of the O X axis (Figure 1), design the equation F → = F 12 → + F 14 → + F 13 →, substitute the force modules obtained above into it and then: F = 2 k q 2 a 2 · 2 2 + k q 2 2 a 2 = k q 2 a 2 2 2 + 1 2 . Answer: the force exerted on each of the given charges located at the vertices of the square is equal to F = k q 2 a 2 2 2 + 1 2. Example 3 An electric charge is given, distributed uniformly along a thin thread (with linear density τ). It is necessary to write down an expression that determines the field strength at a distance a from the end of the thread along its continuation. Thread length – l .
Drawing 2 Solution
Our first step will be to highlight a point charge on the thread d q. Let us compose for it, in accordance with Coulomb’s law, a record expressing the strength of the electrostatic field: d E → = k d q r 3 r → . At a given point, all tension vectors have the same direction along the OX axis, then: d E x = k d q r 2 = d E . The condition of the problem is that the charge has a uniform distribution along the thread with a given density, and we write the following: Let's substitute this entry into the previously written expression for the electrostatic field strength, integrate and get: E = k ∫ a l + a τ d r r 2 = k τ - 1 r a l + a = k τ l a (l + a) . Answer: The field strength at the indicated point will be determined by the formula E = k τ l a (l + a) . If you notice an error in the text, please highlight it and press Ctrl+Enter Electrostatics Electrostatics- a section of the study of electricity that studies the interaction of stationary electric charges and the properties of a constant electric field. 1.Electric charge.
Electric charge is intrinsic property bodies or particles, characterizing their ability to electromagnetic interactions. The unit of electric charge is the coulomb (C)- an electric charge passing through the cross-section of a conductor at a current strength of 1 ampere in 1 second. Exists elementary (minimum) electric charge
The carrier of an elementary negative charge is electron
.
Its mass Fundamental properties of electric charge established experimentally: There are two types: positive
And negative
.
Like charges repel, unlike charges attract. Electric charge invariant- its value does not depend on the reference system, i.e. depending on whether it is moving or at rest. Electric charge discrete- the charge of any body is an integer multiple of the elementary electric charge e. Electric charge additive- the charge of any system of bodies (particles) is equal to the sum of the charges of the bodies (particles) included in the system. Electric charge obeys charge conservation law
: In this case, a closed system is understood as a system that does not exchange charges with external bodies. Electrostatics uses a physical model - point electric charge- a charged body, the shape and dimensions of which are unimportant in this problem. 2.Coulomb's law
Law of interaction of point charges - Coulomb's law: interaction force F between two stationary point charges, located in a vacuum, is proportional to the charges and
inversely proportional to the square of the distance r between them: Force
is directed along a straight line connecting interacting charges, i.e. is central, and corresponds to attraction (F<0) в случае разноименных зарядов и отталкиванию (F> 0) in the case of charges of the same name. In vector form, the force acting on the charge from: Per charge q 2 charge side
force acts - electrical constant,
one of the fundamental physical constants: Where farad (F)- unit of electrical capacity (clause 21). If the interacting charges are in an isotropic medium, then the Coulomb force Where - dielectric constant of the medium- dimensionless quantity showing how many times the interaction force F between charges in a given medium is less than their interaction force
in a vacuum: Dielectric constant of vacuum. Dielectrics and their properties will be discussed in more detail below (section 15). Any charged body can be considered How totality point charges, similar to how in mechanics any body can be considered a collection of material points. That's why electrostatic force, with which one charged body acts on another, is equal to geometric sum of forces, applied to all point charges of the second body from the side of each point charge of the first body. It is often much more convenient to assume that the charges distributed continuously in a charged body
- along some lines(for example, in the case of a charged thin rod), surfaces(for example, in the case of a charged plate) or volume. They use the concepts accordingly linear, surface and volume charge densities. Volume density of electric charges Where dq- charge of a small element of a charged body with volume dV. Surface density of electric charges
Where dq- charge of a small section of a charged surface with an area dS. Linear density of electric charges Where dq- charge of a small section of a charged line length dl. 3.
An electrostatic field is a field created by stationary electric charges. The electrostatic field is described by two quantities: potential(energy scalar field characteristic) and tension(power vector field characteristic). Electrostatic field strength- vector physical quantity determined by the force acting per unit positive charge placed at a given point in the field: The unit of electrostatic field strength is newton per coulomb(N/Cl): 1 N/Kp=1 V/m, where V (volt) is the unit of electrostatic field potential. Point charge field strength in vacuum (and in dielectric) where is the radius vector connecting a given field point with charge q. In scalar form: Vector directioncoincides with the direction of the sipa, acting on a positive charge. If the field is created positive
charge, then the vector
directed along the radius vector from the charge into outer space(repulsion of test positive charge). If the field is created negative charge, then the vector
directed towards the charge(attraction). Graphically, the electrostatic field is represented using tension lines- lines whose tangents at each point coincide with the direction of the vector E(Fig.(a)). Lines of tension are assigned direction coinciding with the direction of the tension vector. Since at a given point in space the tension vector has only one direction, then the tension lines never intersect. For uniform field(when the tension vector at any point is constant in magnitude and direction) the tension lines are parallel to the tension vector. If the field is created by a point charge, then the intensity lines are radial straight lines, going out out of charge, if it is positive, And inbox into it, if the charge is negative(Fig. (b)). 4. Flow vector .
So that with the help of tension lines it is possible to characterize not only the direction, but also tension value electrostatic field, they are carried out with a certain thickness: the number of tension lines penetrating a unit surface area perpendicular to the tension lines must be equal to the vector modulus .
Then the number of tension lines penetrating an elementary area dS, equals called tension vector flow
through the platform dS. Here dS = dS- a vector whose modulus is equal to dS, and the direction of the vector coincides with the direction
to the site. Flow vector
through an arbitrary closed surface S: The principle of superposition of electrostatic fields. Considered in mechanics, we apply to Coulomb forces principle of independent action of forces- resulting the force acting from the field on the test charge is equal to vector sum sip applied to it from the side of each of the charges creating an electrostatic field. Tension resulting field created by the system of charges is also equal to geometric
the sum of the intense fields created at a given point by each of the charges separately. This formula expresses principle of superposition (imposition) of electrostatic fields
.
It allows you to calculate the electrostatic fields of any system of stationary charges, presenting it as a collection of point charges. Let us recall the rule for determining the magnitude of the vector of the sum of two vectors
And
: 6. Gauss's theorem.
Calculation of the field strength of a system of electric charges using the principle of superposition of electrostatic fields can be significantly simplified using the Gauss theorem, which determines the flow of the electric field strength vector through any closed surface. Consider the flow of the tension vector through a spherical surface of radius G, covering a point charge q, located at its center This result is valid for any closed surface of arbitrary shape enclosing a charge. If the closed surface does not cover the charge, then the flow through it is zero, since the number of tension lines entering the surface is equal to the number of tension lines leaving it. Let's consider general case arbitrary surface surrounding n charges. According to the superposition principle, the field strength ,
created by all charges is equal to the sum of the intensities created by each charge separately. That's why Gauss's theorem for an electrostatic field in a vacuum: the flux of the electrostatic field strength vector in a vacuum through an arbitrary closed surface is equal to the algebraic sum of the charges contained inside this surface divided by. If the charge is distributed in space with a volume density ,
then Gauss's theorem: 7. Circulation of the tension vector.
If in the electrostatic field of a point charge q Another point charge moves from point 1 to point 2 along an arbitrary trajectory, then the force applied to the charge does work. Work of force on elementary movement dl is equal to: Work when moving a charge
from point 1 to point 2: Job
does not depend on the trajectory of movement, but determined only by the positions of the start and end points. Therefore, the electrostatic field of a point charge is potential,
and electrostatic forces - conservative.
Thus, the work of moving a charge in an electrostatic along any closed circuit L equal to zero: If the transferred charge unit
,
then the elementary work of field forces on the path
equal to
, where is the projection of the vector
to the direction of elementary movement .
Integral
called circulation of the tension vector along a given closed contour L. Vector circulation theorem
:
The circulation of the electrostatic field strength vector along any closed loop is zero A force field that has this property. called potential. This formula is correct only for electric field stationary charges (electrostatic). 8. Potential charge energy.
In a potential field, bodies have potential energy and the work of conservative forces is done due to the loss of potential energy. Therefore, work can be represented as the difference in potential charge energies q 0 at the initial and final points of the charge field q:
Potential energy of a charge located in a charge field q on distance r equal to Assuming that when the charge is removed to infinity, the potential energy goes to zero, we get: const = 0. For namesake charges potential energy of their interaction (push-off)positive, For different names charges potential energy from interaction (attraction)negative. If the field is created by the system P point charges, then the potential energy of the charge d 0, located in this field, is equal to the sum of its potential energies created by each of the charges separately: 9. Electrostatic field potential. The ratio does not depend on the test charge and is, energy characteristic of the field, called potential
:
Potential
at any point in the electrostatic field there is scalar a physical quantity determined by the potential energy of a unit positive charge placed at that point. For example, the field potential created by a point charge q, is equal 10.Potential difference
Work done by electrostatic field forces when moving a charge
from point 1 to point 2, can be represented as that is, equal to the product of the moved charge and the potential difference at the starting and ending points. Potential difference two points 1 and 2 in an electrostatic field is determined by the work done by the field forces when moving a unit positive charge from point 1 to point 2 Using the definition of the electrostatic field strength, we can write down the work
as where integration can be performed along any line connecting the start and end points, since the work of the electrostatic field forces does not depend on the trajectory of movement. If you move the charge from arbitrary point outside the field
(to infinity), where the potential energy, and therefore the potential, are equal to zero, then the work of the electrostatic field, whence Thus, another definition of potential: potential
- physical a quantity determined by the work done to move a unit positive charge when moving it from a given point to infinity. Unit of potential -
volt (V): 1V is the potential of a point in the field at which a charge of 1 C has a potential energy of 1 J (1 V = 1 JL C). The principle of superposition of potentials of electrostatic fields
:
If the field is created by several charges, then the field potential of the system of charges is equal to algebraic sum field potentials of all these charges. 11. The relationship between tension and potential.
For a potential field, there is a relationship between potential (conservative) force and potential energy: where ("nabla") - Hamilton operator
: Since and , then The minus sign indicates that the vector
directed to the side descending potential. 12. Equipotential surfaces.
To graphically display the potential distribution, equipotential surfaces are used - surfaces at all points of which the potential has the same value. Equipotential surfaces are usually drawn so that the potential differences between two adjacent equipotential surfaces are the same. Then the density of equipotential surfaces clearly characterizes the field strength at different points. Where these surfaces are denser, the field strength is greater. In the figure, the dotted line shows the lines of force, the solid lines show sections of equipotential surfaces for: positive point charge (A), dipole (b), two like charges (V), charged metal conductor of complex configuration (G). For a point charge, the potential is , so the equipotential surfaces are concentric spheres. On the other hand, tension lines are radial straight lines. Consequently, the tension lines are perpendicular to the equipotential surfaces. It can be shown that in all cases 1) vector perpendicular equipotential surfaces and 2) always directed towards decreasing potential. 13.Examples of calculations of the most important symmetrical electrostatic fields in vacuum.
1. Electrostatic field of an electric dipole in a vacuum.
Electric dipole(or double electric pole) is a system of two equal in magnitude opposite point charges (+q,-q), distance l between which there is significantly less distance to the considered points of the field ( l< Dipole arm
- a vector directed along the dipole axis from a negative charge to a positive charge and equal to the distance between them. Electric dipole moment p e- a vector coinciding in direction with the dipole arm and equal to the product of the charge modulus and the arm: Let r- distance to point A from the middle of the dipole axis. Then, given that r>>l. 2) Field strength at point B on the perpendicular, restored to the dipole axis from its center at r'>>l. That's why Superposition principle Let's say we have three point charges. These charges interact. You can conduct an experiment and measure the forces that act on each charge. In order to find the total force with which the second and third act on one charge, it is necessary to add the forces with which each of them acts according to the parallelogram rule. The question arises whether the measured force that acts on each of the charges is equal to the sum of the forces exerted by the other two, if the forces are calculated according to Coulomb's law. Research has shown that the measured force is equal to the sum of the calculated forces in accordance with Coulomb's law on the part of two charges. This empirical result is expressed in the form of statements: This statement is called the principle of superposition. This principle is one of the foundations of the doctrine of electricity. It is as important as Coulomb's law. Its generalization to the case of many charges is obvious. If there are several field sources (number of charges N), then the resulting force acting on the test charge q can be found as: \[\overrightarrow(F)=\sum\limits^N_(i=1)(\overrightarrow(F_(ia)))\left(1\right),\] where $\overrightarrow(F_(ia))$ is the force with which charge $q_i$ acts on charge q if there are no other N-1 charges. The principle of superposition (1) allows, using the law of interaction between point charges, to calculate the force of interaction between charges located on a body of finite dimensions. To do this, it is necessary to divide each of the charges into small charges dq, which can be considered point charges, take them in pairs, calculate the interaction force and perform a vector addition of the resulting forces. The principle of superposition has a field interpretation: the field strength of two point charges is equal to the sum of the intensities that are created by each of the charges, in the absence of the other. In general, the principle of superposition with respect to tensions can be written as follows: \[\overrightarrow(E)=\sum(\overrightarrow(E_i))\left(2\right).\] where $(\overrightarrow(E))_i=\frac(1)(4\pi (\varepsilon )_0)\frac(q_i)(\varepsilon r^3_i)\overrightarrow(r_i)\ $ is the intensity of the i-th point charge, $\overrightarrow(r_i)\ $ is the radius vector drawn from the i-th charge to a point in space. Expression (1) means that the field strength of any number of point charges is equal to the sum of the field strengths of each of the point charges, if there are no others. It has been confirmed by engineering practice that the superposition principle is observed up to very high field strengths. The fields in atoms and nuclei have very significant strengths (of the order of $(10)^(11)-(10)^(17)\frac(B)(m)$), but even for them the principle of superposition was used in calculating the energy levels of atoms and the calculation data coincided with the experimental data with great accuracy. However, it should be noted that at very small distances (of the order of $\sim (10)^(-15)m$) and extremely strong fields, the superposition principle may not hold. So, for example, on the surface of heavy nuclei the strengths reach the order of $\sim (10)^(22)\frac(V)(m)$ the superposition principle is satisfied, but at a strength of $(10)^(20)\frac(V )(m)$ arise quantum - mechanical nonlinearities of interaction. If the charge is distributed continuously (there is no need to take discreteness into account), then the total field strength is found as: \[\overrightarrow(E)=\int(d\overrightarrow(E))\ \left(3\right).\] In equation (3), integration is carried out over the charge distribution region. If the charges are distributed along the line ($\tau =\frac(dq\ )(dl)-linear\ density\ distribution\ charge$), then integration in (3) is carried out along the line. If the charges are distributed over the surface and the surface distribution density is $\sigma =\frac(dq\ )(dS)$, then integrate over the surface. Integration is carried out over volume if we are dealing with volumetric charge distribution: $\rho =\frac(dq\ )(dV)$, where $\rho$ is the volumetric charge distribution density. The principle of superposition, in principle, allows one to determine $\overrightarrow(E)$ for any point in space from a known spatial charge distribution. Example 1 Assignment: Identical point charges q are located at the vertices of a square with side a. Determine the force exerted on each charge by the other three charges. Let us depict the forces acting on one of the charges at the vertex of the square (the choice is not important, since the charges are the same) (Fig. 1). We write the resulting force acting on the charge $q_1$ as: \[\overrightarrow(F)=(\overrightarrow(F))_(12)+(\overrightarrow(F))_(14)+(\overrightarrow(F))_(13)\ \left(1.1\right ).\] The forces $(\overrightarrow(F))_(12)$ and $(\overrightarrow(F))_(14)$ are equal in magnitude and can be found as: \[\left|(\overrightarrow(F))_(12)\right|=\left|(\overrightarrow(F))_(14)\right|=k\frac(q^2)(a^2 )\ \left(1.2\right),\] where $k=9 (10)^9\frac(Nm^2)((C)^2).$ We will find the force modulus $(\overrightarrow(F))_(13)$, also according to Coulomb’s law, knowing that the diagonal of the square is equal to: therefore we have: \[\left|(\overrightarrow(F))_(13)\right|=k\frac(q^2)(2a^2)\ \left(1.4\right)\] Let's direct the OX axis as shown in Fig. 1, we project equation (1.1), substitute the resulting force modules, we obtain: Answer: The force acting on each of the charges at the vertices of the square is equal to: $F=\frac(kq^2)(a^2)\left(\frac(2\sqrt(2)+1)(2)\right) .$ Example 2 Assignment: An electric charge is uniformly distributed along a thin thread with a uniform linear density $\tau$. Find an expression for the field strength at a distance $a$ from the end of the thread along its continuation. The length of the thread is $l$. Let us select a point charge $dq$ on the thread and write for it from Coulomb’s law the expression for the electrostatic field strength: At a given point, all tension vectors are directed equally, along the X axis, therefore, we have: Since the charge, according to the conditions of the problem, is uniformly distributed over the thread with a linear density $\tau $, we can write the following: Let's substitute (2.4) into equation (2.1) and integrate: Answer: The field strength of the thread at the indicated point is calculated by the formula: $E=\frac(k\tau l)(a(l+a)).$ The principle of superposition (overlay) of fields is formulated as follows: If at a given point in space various charged particles create electric fields, the strengths of which, etc., then the resulting field strength at this point is equal to: . The principle of field superposition is valid for the case when fields created by several different charges do not have any influence on each other, that is, they behave as if there are no other fields. Experience shows that for fields of ordinary intensities found in nature, this actually occurs. Thanks to the principle of superposition, to find the field strength of a system of charged particles at any point, it is enough to use the expression for the field strength of a point charge. The figure below shows how at the point A the field strength created by two point charges is determined q 1
And q 2. The electric field in space is usually represented by lines of force. The concept of lines of force was introduced by M. Faraday while studying magnetism. This concept was then developed by J. Maxwell in his research on electromagnetism. A line of force, or an electric field strength line, is a line whose tangent to each of its points coincides with the direction of the force acting on a positive point charge located at this point in the field. The figures below show the voltage lines of a positively charged ball (Fig. 1); two differently charged balls (Fig. 2); two similarly charged balls (Fig. 3) and two plates charged with charges of different signs, but identical in absolute value (Fig. 4). The tension lines in the last figure are almost parallel in the space between the plates, and their density is the same. This suggests that the field in this region of space is uniform. An electric field is called homogeneous if its strength is the same at all points in space. In an electrostatic field, the lines of force are not closed; they always begin on positive charges and end on negative charges. They do not intersect anywhere; the intersection of the field lines would indicate the uncertainty of the direction of the field strength at the intersection point. The density of field lines is greater near charged bodies, where the field strength is greater. Field strength of a charged conducting ball at a distance from the center of the ball exceeding its radius r ≥
R. is determined by the same formula as the fields of a point charge The charge of the ball is distributed evenly over its surface. Inside the conducting ball, the field strength is zero.
Superposition principle
Field interpretation of the superposition principle
Definition 2 
kg. The carrier of an elementary positive charge is proton. Its mass 
kg.
Algebraic sum of electric charges of any closed
the system remains unchanged, no matter what processes occur
within this system.
or 
. Then 
Where
- vector projection on normal
to the site dS. (Vector
- unit vector perpendicular to the site dS). Magnitude
Field interpretation of the superposition principle
Electric field lines.
Field of a charged ball.
. This is evidenced by the distribution of field lines (Fig. A), similar to the distribution of intensity lines of a point charge (Fig. b).
