In accordance with the theory of short-range interaction, interactions between charged bodies that are distant from each other are carried out through fields (electromagnetic) created by these bodies in the space surrounding them. If fields are created by stationary particles (bodies), then the field is electrostatic. If the field does not change over time, then it is called stationary. The electrostatic field is stationary. This field is a special case electromagnetic field. Power characteristics electric field serves as a tension vector, which can be defined as:
where $\overrightarrow(F)$ is the force acting from the field on a stationary charge q, which is sometimes called “test”. In this case, it is necessary that the “test” charge be small so that it does not distort the field, the strength of which is measured with its help. From equation (1) it is clear that the intensity coincides in direction with the force with which the field acts on a unit positive “test charge”.
The electrostatic field strength does not depend on time. If the intensity at all points of the field is the same, then the field is called uniform. Otherwise the field is not uniform.
Power lines
For graphic image electrostatic fields use the concept of lines of force.
Definition
Lines of force or field strength lines are lines whose tangents at each point of the field coincide with the directions of the strength vectors at these points.
Power lines electrostatic fields are open. They start on positive charges and end on negative ones. Sometimes they can go to infinity or come from infinity. The field lines do not intersect.
The electric field strength vector obeys the principle of superposition, namely:
\[\overrightarrow(E)=\sum\limits^n_(i=1)((\overrightarrow(E))_i(2)).\]
The resulting field strength vector can be found as the vector sum of the strengths of its constituent “individual” fields. 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 ($\tau =\frac(dq\ )(dl)$ is the linear charge distribution density), 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.
Field strength
The field strength in a dielectric is equal to the vector sum of the field strengths that create free charges ($\overrightarrow(E_0)$) and bound charges ($\overrightarrow(E_p)$):
\[\overrightarrow(E)=\overrightarrow(E_0)+\overrightarrow(E_p)\left(4\right).\]
Very often in examples we come across the fact that the dielectric is isotropic. In this case, the field strength can be written as:
\[\overrightarrow(E)=\frac(\overrightarrow(E_0))(\varepsilon )\ \left(5\right),\]
where $\varepsilon$ is the relative dielectric constant of the medium at the field point under consideration. Thus, from (5) it is obvious that the electric field strength in a homogeneous isotropic dielectric is $\varepsilon $ times less than in vacuum.
The electrostatic field strength of a system of point charges is equal to:
\[\overrightarrow(E)=\frac(1)(4\pi (\varepsilon )_0)\sum\limits^n_(i=1)(\frac(q_i)(\varepsilon r^3_i))\overrightarrow (r_i)\ \left(6\right).\]
In the SGS system, the field strength of a point charge in vacuum is equal to:
\[\overrightarrow(E)=\frac(q\overrightarrow(r))(r^3)\left(7\right).\]
Assignment: The charge is uniformly distributed over a quarter circle of radius R with linear density $\tau $. Find the field strength at point (A), which would be the center of the circle.
Let us select an elementary section ($dl$) on the charged part of the circle, which will create a field element at point A, for it we will write an expression for the intensity (we will use the CGS system), in this case the expression for $d\overrightarrow(E)$ has the form :
The projection of the vector $d\overrightarrow(E)$ onto the OX axis has the form:
\[(dE)_x=dEcos\varphi =\frac(dqcos\varphi )(R^2)\left(1.2\right).\]
Let us express dq in terms of the linear charge density $\tau $:
Using (1.3) we transform (1.2), we get:
\[(dE)_x=\frac(2\pi R\tau dRcos\varphi )(R^2)=\frac(2\pi \tau dRcos\varphi )(R)=\frac(\tau cos\varphi d\varphi )(R)\ \left(1.4\right),\]
where $2\pi dR=d\varphi $.
Let's find the complete projection $E_x$ by integrating expression (1.4) over $d\varphi $, where the angle changes $0\le \varphi \le 2\pi $.
Let's deal with the projection of the tension vector onto the axis OY, and by analogy, without any special explanation, we will write:
\[(dE)_y=dEsin\varphi =\frac(\tau )(R)sin\varphi d \varphi \ \left(1.6\right).\]
We integrate expression (1.6), the angle changes $\frac(\pi )(2)\le \varphi \le 0$, we get:
Let's find the magnitude of the tension vector at point A using the Pythagorean theorem:
Answer: The field strength at point (A) is equal to $E=\frac(\tau )(R)\sqrt(2).$
Assignment: Find the electrostatic field strength of a uniformly charged hemisphere whose radius is R. The surface charge density is $\sigma$.
Let us highlight on the surface of the charged sphere elementary charge$dq$, which is located on the area element $dS.$ In spherical coordinates, $dS$ is equal to:
where $0\le \varphi \le 2\pi ,\ 0\le \theta \le \frac(\pi )(2).$
Let us write the expression for the elementary field strength of a point charge in the SI system:
We project the tension vector onto the OX axis, we get:
\[(dE)_x=\frac(dqcos\theta )(4 \pi \varepsilon_0R^2)\left(2.3\right).\]
Let us express the elementary charge through the surface charge density, we obtain:
We substitute (2.4) into (2.3), use (2.1) and integrate, we get:
It is easy to obtain that $E_Y=0.$
Therefore, $E=E_x.$
Answer: The field strength of a charged hemisphere along the surface at its center is equal to $E=\frac(\sigma)(4(\varepsilon )_0).$
Charged bodies can influence each other without contact through an electric field. The field that is created by stationary electrical particles is called electrostatic.
Instructions
If another charge Q0 is placed in the electric field created by charge Q, then it will act on it with a certain force. This characteristic is called the electric field strength E. It represents the ratio of the force F with which the field acts on a positive electric charge Q0 at a certain point in space to the value of this charge: E = F/Q0.
Depending on a specific point in space, the value of field strength E can change, which is expressed by the formula E = E (x, y, z, t). Therefore, the electric field strength is a vector physical quantities.
Since the field strength depends on the force acting on a point charge, the electric field strength vector E is the same as the force vector F. According to Coulomb's law, the force with which two charged particles interact in a vacuum is directed along a straight line that connects these charges.
Michael Faraday proposed to visually represent the field strength of an electric charge using tension lines. These lines coincide with the tension vector at all tangential points. In drawings they are usually designated by arrows.
If the electric field is uniform and its intensity vector is constant in magnitude and direction, then the intensity lines are parallel to it. If the electric field is created by a positively charged body, the lines of tension are directed away from it, and in the case of a negatively charged particle, towards it.
Please note
The tension vector has only one direction at each point in space, so the tension lines never intersect.
>>Physics: Electric field strength. Principle of field superposition
It is not enough to assert that an electric field exists. Need to enter quantitative characteristics fields. After this, electric fields can be compared with each other and their properties can continue to be studied.
An electric field is detected by the forces acting on a charge. It can be argued that we know everything we need about the field if we know the force acting on any charge at any point in the field.
Therefore, it is necessary to introduce a characteristic of the field, knowledge of which will allow us to determine this force.
If you alternately place small charged bodies at the same point in the field and measure the forces, you will find that the force acting on the charge from the field is directly proportional to this charge. Indeed, let the field be created by a point charge q 1. According to Coulomb's law (14.2) on the charge q 2 there is a force proportional to the charge q 2. Therefore, the ratio of the force acting on a charge placed at a given point in the field to this charge for each point in the field does not depend on the charge and can be considered as a characteristic of the field. This characteristic is called electric field strength. Like force, field strength is vector quantity; it is denoted by the letter . If a charge placed in a field is denoted by q instead of q 2, then the tension will be equal to:
Hence the force acting on the charge q from the electric field side, is equal to:
Field strength of a point charge. Let's find the electric field strength created by a point charge q 0. According to Coulomb's law, this charge will act on a positive charge q with a force equal to
if at a given point in space various charged particles create electric fields whose strengths
etc., then the resulting field strength at this point is equal to the sum of the strengths of these fields:
Moreover, the field strength created by an individual charge is determined as if there were no other charges creating the field.
Thanks to the principle of superposition, to find the field strength of a system of charged particles at any point, it is enough to know expression (14.9) for the field strength of a point charge. Figure 14.8 shows how the field strength at a point is determined A, created by two point charges q 1 And q 2 , q 1 >q 2
???
1. What is the electric field strength called?
2. What is the field strength of a point charge?
3. How is the charge field strength q 0 directed if q 0>0
? If q 0<0
?
4. How is the principle of field superposition formulated?
G.Ya.Myakishev, B.B.Bukhovtsev, N.N.Sotsky, Physics 10th grade
Lesson content lesson notes supporting frame lesson presentation acceleration methods interactive technologies Practice tasks and exercises self-test workshops, trainings, cases, quests homework discussion questions rhetorical questions from students Illustrations audio, video clips and multimedia photographs, pictures, graphics, tables, diagrams, humor, anecdotes, jokes, comics, parables, sayings, crosswords, quotes Add-ons abstracts articles tricks for the curious cribs textbooks basic and additional dictionary of terms other Improving textbooks and lessonscorrecting errors in the textbook updating a fragment in a textbook, elements of innovation in the lesson, replacing outdated knowledge with new ones Only for teachers perfect lessons calendar plan for the year; methodological recommendations; discussion program Integrated LessonsIf you have corrections or suggestions for this lesson,
Instructions
If another charge Q0 is placed in the electric field created by charge Q, then it will act on it with a certain force. This is called the electric field strength E. It is the ratio of the force F with which the field acts on a positive electric charge Q0 at a certain point in space to the value of this charge: E = F/Q0.
Depending on a specific point in space, the value of field strength E can change, which is expressed by the formula E = E (x, y, z, t). Therefore, the electric field strength is a vector physical quantity.
Since the field strength depends on the force acting on a point charge, the electric field strength vector E is the same as the force vector F. According to Coulomb’s law, the force with which two charged particles interact in a vacuum is directed along the direction that connects these charges.
Video on the topic
The objects of vector algebra are line segments having a direction and a length called a modulus. To determine module vector, you should extract the square root of a quantity that is the sum of the squares of its projections onto the coordinate axes.
Instructions
Vectors are characterized by two basic properties: length and direction. Length vector or norm and represents a scalar value, the distance from the start point to the end point. Both are used to graphically depict various actions, for example, physical forces, the movement of elementary particles, etc.
Location vector in two-dimensional or three-dimensional space does not affect its properties. If you move it to another place, then only the coordinates of its ends will change, however module and the direction will remain the same. This independence allows the use of vector algebra in various calculations, for example, angles between spatial lines and planes.
Each vector can be specified by the coordinates of its ends. Let us first consider two-dimensional space: let the beginning vector is at point A (1, -3), and is at point B (4, -5). To find their projections, drop perpendiculars to the x-axis and ordinate.
Determine the projections of yourself vector, which can be calculated using the formula: АВх = (xb - xa) = 3; ABy = (yb - ya) = -2, where: ABx and ABy are projections vector on the Ox and Oy axis; xa and xb are the abscissas of points A and B; ya and yb are the corresponding ordinates.
In the graphic image you will see a right triangle formed by legs with lengths equal to the projections vector. The hypotenuse of a triangle is the quantity that needs to be calculated, i.e. module vector. Apply the Pythagorean theorem: |AB|² = ABx² + ABy² → |AB| = √((xb - xa)² + (yb – ya)²) = √13.
Let in the considered example za = 3, zb = 8, then: zb – za = 5;|AB| = √(9 + 4 + 25) = √38.
Video on the topic
In order to determine the modulus of point charges of the same size, measure the force of their interaction and the distance between them and make a calculation. If you need to find the charge modulus of individual point bodies, introduce them into an electric field with a known intensity and measure the force with which the field acts on these charges.
12. Dielectrics in an electric field. Molecules of polar and non-polar dielectrics in an electric field. Polarization of dielectrics. Types of polarization.
1. Polar dielectrics.
In the absence of a field, each of the dipoles has an electric moment, but the vectors of the electric moments of the molecules are randomly located in space and the sum of the projections of the electric moments to any direction is zero:
If the dielectric is now placed in an electric field (Fig. 18), then a pair of forces will begin to act on each dipole, which will create a moment under the influence of which the dipole will rotate around an axis perpendicular to the arm, tending to the final position when the vector of the electric moment is parallel to the voltage vector electric field. The latter will be hampered by the thermal movement of molecules, internal friction, etc. and therefore 
the electric moments of the dipoles will make some angles with the direction of the external field vector, but now a larger number of molecules will have components of the projection of the electric moments in the direction that coincides, for example, with the field strength and the sum of the projections of all electric moments will already be different from zero.
A value indicating the ability of a dielectric to create greater or less polarization, that is, characterizing the compliance of the dielectric to polarization called dielectric susceptibility or dielectric polarizability ().
16. Flow of the electric induction vector (uniform and inhomogeneous induction). Flow through a closed surface. T.Gauss for el. Fields in the environment.
Similar to the flow of the tension vector, we can introduce the concept induction vector flow , leaving the same property as for tension - the induction vector is proportional to the number of lines passing through a unit surface area. You can specify the following properties:
1. Flux through a flat surface in a uniform field (Fig. 22). In this case, the induction vector is directed along the field and the induction line flux can be expressed as follows: 
2. The flux of the induction vector through a surface in a non-uniform field is calculated by dividing the surface into elements so small that they can be considered flat, and the field near each element is uniform. The total flux of the induction vector will be equal to: 
3. Flow of the induction vector through a closed surface.
Let us consider the flow of the induction vector crossing a closed surface (Fig. 23). Let us agree to consider the direction of the external normals to be positive. Then at those points of the surface where the induction vector is directed tangentially to the induction line outward, the angle 
and the flux of induction lines will be positive, and where the induction vector D will be positive, and where the vector D is directed inside the surface, the flux of induction lines will be negative, because and . Thus, the total flux of induction lines penetrating the closed surface through and through is zero.
Based on Gauss's theorem, we find that there are no uncompensated electric charges inside a closed surface conducted in a conductor. This property remains the same when the conductor is given an excess charge.
An equal but positive charge will appear on the opposite side. As a result, inside the conductor there will be induced electric field E ind , directed towards the external field, which will grow until it becomes equal to the external field and thus the resulting field inside the conductor becomes zero. This process occurs within a very short time.
The induced charges are located on the surface of the conductor in a very thin layer.
The potential at all points of the conductor remains the same, i.e. the outer surface of the conductor is equipotential.
A closed hollow conductor screens only the field of external charges. If electric charges are located inside the cavity, then inductive charges will arise not only on the outer surface of the conductor, but also on the inner one, and a closed conducting cavity no longer screens the field of electric charges placed inside it.
. The field strength near a conductor is directly proportional to the surface charge density on it.
