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 -- 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”.
Tension electrostatic field does not depend on time. If the intensity at all points of the field is the same, then the field is called homogeneous. IN otherwise the field is not uniform.
Power lines
To graphically represent electrostatic fields, the concept is used power lines.
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.
The electrostatic field lines 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 $ -- bulk density charge distribution.
Field strength
The field strength in the dielectric is equal to vector sum 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, and for it we will write an expression for the intensity (we will use GHS 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 much explanation, we’ll 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 surface density charge, we get:
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 $E=\frac(\sigma)(4(\varepsilon )_0).$
5. Electrostatics
Coulomb's law
1. Charged bodies interact. There are two types of charges in nature, they are conventionally called positive and negative. Charges of the same sign (like) repel, charges of opposite signs (opposite) attract. The SI unit of measurement for charges is the coulomb (denoted
2. In nature, there is a minimum possible charge. He is called
elementary and denoted by e. Numerical value elementary chargee ≈ 1.6 10–19 C, Electron chargeq electron = –e, proton chargeq proton = +e. All charges
V nature are multiples of the elementary charge.
3. In an electrically isolated system algebraic sum charges remains unchanged. For example, if you connect two identical metal balls with charges q 1 = 5 nC = 5 10–9 C and q 2 = – 1 nC, then the charges will be distributed
between the balls equally and the charge q of each of the balls will become equal
q = (q 1 + q 2 ) / 2 = 2 nC.
4. A charge is called a point charge if its geometric dimensions are significantly smaller than the distances at which the effect of this charge on other charges is studied.
5. Coulomb's law determines the magnitude of the force electrical interaction two stationary point charges q 1 and q 2 located at a distance from each other (Fig. 1)
k |q | |q | ||||||
F = | F | |= |F | |||||
Here F 12 is the force acting on the first charge from the second, F 21 is the force
acting on the second charge from the first, k ≈ 9 10 9 N m2 / Cl2 – a constant in Coulomb’s law. In the SI system, this constant is usually written in the form
k = 4 πε 1 0 ,
where ε 0 ≈ 8.85 10 − 12 F/m is the electrical constant.
6. The force of interaction between two point charges does not depend on the presence of other charged bodies near these charges. This statement is called the principle of superposition.
Electric field strength vector
1. Place a point charge q near a stationary charged body (or several bodies). We will assume that the magnitude of the charge q is so small that it does not cause the movement of charges in other bodies (such a charge is called a test charge).
From the side of the charged body, a force F will act on a stationary test charge q. In accordance with Coulomb's law and the principle of superposition, the force F will be proportional to the amount of charge q. This means that if the magnitude of the test charge is increased, for example, by 2 times, then the magnitude of the force F will also increase by 2 times; if the sign of the charge q is changed to the opposite, then the force will change direction to the opposite. This proportionality can be expressed by the formula
F = qE.
Vector E is called the electric field strength vector. This vector depends on the distribution of charges in bodies creating an electric field, and
from the position of the point at which in the specified way vector E is defined. We can say that the electric field strength vector equal to force, acting on a unit positive charge placed in this point space.
The definition of E G = F G /q can be generalized to the case of variable (time-dependent) fields.
2. Let's calculate the vector of the electric field strength created by a stationary point charge Q. Let us choose some point A located at a distance from the point charge Q. To determine the voltage vector at this point, let’s mentally place a positive test chargeq at it. On
test charge from the side of a point charge Q, there will be an attractive or repulsive force depending on the sign of the charge Q. The magnitude of this force is equal to
F = k| Q| q. r2
Consequently, the magnitude of the electric field strength vector created by a stationary point charge Q at point A, distant from it at a distance r, is equal to
E = k r |Q 2 |.
Vector E G starts at point A and is directed from charge Q, if Q > 0, and towards charge Q,
if Q< 0 .
3. If the electric field is created by several point charges, then the voltage vector in arbitrary point can be found using the principle of field superposition.
4. Line of force (vector line E) is called a geometric line,
the tangent to which at each point coincides with the vector E at that point.
In other words, the vector E is directed tangentially to the field line at each of its points. The direction of the force line is assigned - along the vector E. The picture of the power lines is a visual representation force field, gives an idea of the spatial structure of the field, its sources, and allows you to determine the direction of the intensity vector at any point.
5. A uniform electric field is a field, vector E of which is the same (in magnitude and direction) at all points. Such a field is created, for example, by a uniformly charged plane at points located fairly close to this plane.
6. The field of a uniformly charged ball over the surface is zero inside the ball,
A outside the ball coincides with the field of a point charge Q located in the center of the ball:
k | Q| | for r > R | ||
E = r2 | |||
at r< R | |||
where Q – ball charge, R is its radius, r is the distance from the center of the ball to the point, in
which defines the vector E.
7. In dielectrics, the field is weakened. For example, a point charge or a sphere uniformly charged over the surface, immersed in oil, creates an electric field
E = k ε |r Q 2 |,
where r is the distance from the point charge or the center of the ball to the point at which the voltage vector is determined, ε is the dielectric constant of the oil. Dielectric constant depends on the properties of the substance. The dielectric constant of vacuum is ε = 1, the dielectric constant of air is very close to unity (when solving problems it is usually considered equal to 1), for other gaseous, liquid and solid dielectricsε > 1.
8. When the charges are in equilibrium (if there is no ordered movement), the electric field strength inside the conductors is zero.
Working in an electric field. Potential difference.
1. The field of stationary charges (electrostatic field) has important property: the work of the electrostatic field forces to move a test charge from some point 1 to point 2 does not depend on the shape of the trajectory, but is determined only by the positions of the initial and end points. Fields with this property are called conservative. The property of conservatism allows us to determine the so-called potential difference for any two points of the field.
Potential differenceϕ 1 −ϕ 2 at points 1 and 2 is equal to the ratio of the work A 12 field forces to move a test charge q from point 1 to point 2 to the magnitude of this charge:
ϕ1 - ϕ2 =A q 12.
This definition of the potential difference makes sense only because the work does not depend on the shape of the trajectory, but is determined by the positions of the starting and ending points of the trajectories. In the SI system, potential difference is measured in volts: 1V = J/C.
Capacitors
1. The capacitor consists of two conductors (they are called plates), separated from one another by a layer of dielectric (Fig. 2), and the charge of one
facing Q, and the other –Q. The charge on the positive plate Q is called the charge on the capacitor.
2. It can be shown that the potential difference ϕ 1 −ϕ 2 between the plates is proportional to the amount of chargeQ, that is, if, for example, the chargeQ is increased by 2 times, then the potential difference will increase by 2 times.
ε S
ϕ 1ϕ 2
Fig.2 Fig.3
This proportionality can be expressed by the formula
Q = C (ϕ 1 -ϕ 2),
where C is the proportionality coefficient between the charge of the capacitor and the potential difference between its plates. This coefficient is called electrical capacity or simply capacitance of the capacitor. The capacity depends on the geometric dimensions of the plates, their relative position And dielectric constant environment. The potential difference is also called voltage, which is denoted by U. Then
Q = CU.
3. A flat capacitor consists of two flat conducting plates located parallel to each other at a distance d (Fig. 3). This distance is assumed to be small compared to the linear dimensions of the plates. The area of each plate (capacitor plate) is S, the charge of one plate is Q, and the charge of the other is Q.
At a certain distance from the edges, the field between the plates can be considered uniform. Therefore ϕ 1 -ϕ 2 = Ed, or
U = Ed.
The capacitance of a parallel-plate capacitor is determined by the formula
C = εε d 0 S ,
where ε 0 =8.85 10–12 F/m is the electrical constant, ε is the dielectric constant of the dielectric between the plates. From this formula it can be seen that to obtain a large capacitor, you need to increase the area of the plates and reduce the distance between them. The presence of a dielectric with a high dielectric constant ε between the plates also leads to an increase in capacitance. The role of the dielectric between the plates is not only to increase the dielectric constant. It is also important that good dielectrics can withstand high electric fields without causing breakdown between the plates.
In the SI system, capacitance is measured in farads. A flat-plate capacitor of one farad would have gigantic dimensions. The area of each plate would be approximately 100 km2 with a distance of 1 mm between them. Capacitors are widely used in technology, in particular, for storing charges.
4. If the plates of a charged capacitor are short-circuited with a metal conductor, then a electricity and the capacitor will discharge. When current flows in a conductor, a certain amount of heat will be released, which means that a charged capacitor has energy. It can be shown that the energy of any charged capacitor (not necessarily flat) is determined by the formula
W = 1 2 CU2 .
Considering that Q = CU, the formula for energy can also be rewritten in the form
W = Q 2 =QU .
If in the space surrounding electric charge, introduce another charge, then the Coulomb force will act on it; This means that in the space surrounding electric charges, there is force field. According to ideas modern physics, the field really exists and, along with matter, is one of the forms of existence of matter, through which certain interactions are carried out between macroscopic bodies or particles that make up the substance. IN in this case talk about an electric field - a field through which electric charges interact. We consider electric fields that are created by stationary electric charges and are called electrostatic.
To detect and experimentally study the electrostatic field, it is used test point positive charge - such a charge that does not distort the field under study (does not cause a redistribution of charges creating the field). If in the field created by the charge Q, place a test charge Q 0, then a force acts on it F, different in different points field, which, according to Coulomb's law, is proportional to the test charge Q 0 . Therefore the ratio F/ Q 0 does not depend on Q 0 and characterizes the electrostatic field at the point where the test charge is located. This quantity is called tension and is force characteristic of the electrostatic field.
Electrostatic field strength at a given point there is a physical quantity determined by the force acting on a test unit positive charge placed at this point in the field:
Field strength of a point charge in vacuum
The direction of vector E coincides with the direction of the force acting on the positive charge. If the field is created by a positive charge, then vector E is directed along the radius vector from the charge into external space (repulsion of the test positive charge); if the field is created negative charge, then vector E is directed towards the charge (Fig.).
The unit of electrostatic field strength is newton per coulomb (N/C): 1 N/C is the intensity of a field that acts on a point charge of 1 C with a force of 1 N; 1 N/C = 1 V/m, where V (volt) is the unit of electrostatic field potential. Graphically, the electrostatic field is represented using tension lines - lines, the tangents to which at each point coincide with the direction of vector E (Fig.).
Since at any given point in space the tension vector has only one direction, 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 emerging from the charge if it is positive (Fig. A), and included in it if the charge is negative (Fig. b). Due to the great visibility graphic method representation of the electrostatic field is widely used in electrical engineering.
In order to use tension lines to characterize not only the direction, but also the value of the intensity of the electrostatic field, it was agreed to draw them with a certain density: the number of tension lines penetrating a unit surface area perpendicular to the tension lines must be equal to the modulus of the vector E. Then the number of tension lines , penetrating the elementary area d S, normal n which forms an angle a with the vector E, equals E d Scos a = E n d S, Where E p-vector projection E to normal n to site d S(rice.).
Value dФ E =E n dS= E dS is called tension vector flow through platform d S. Here d S=d Sn- a vector whose modulus is d S, and the direction coincides with the direction of the normal n to the site. Selecting the vector direction n(and therefore d S) is conditional, since it can be directed in any direction. The unit of flux of the electrostatic field strength vector is 1 V×m.
For an arbitrary closed surface S vector flow E through this surface
,
where the integral is taken over the closed surface S. Flow vector E is algebraic quantity: depends not only on the field configuration E, but also on the choice of direction n. For closed surfaces, the positive direction of the normal is taken to be outer normal, that is, the normal pointing outward to the area covered by the surface.
TO Coulomb forces Let us apply the principle of independence of the action of forces, i.e. the resulting force F acting from the field on the test charge Q 0 is equal to the vector sum of the forces Fi applied to it by each of the charges Q i: . 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 strength of the field created by the charge Q i . Substituting this into the expression above, we get . This formula expresses the principle of superposition (imposition) of electrostatic fields, according to which the strength E of the resulting field created by a 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. An electric dipole is a system of two opposite point charges of equal magnitude (+Q, –Q), the distance l between which is significantly less than the distance to the field points under consideration. According to the principle of superposition, the strength E of the dipole field at an arbitrary point 
, where E+ and E– are the field strengths created by positive and negative charges, respectively.
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 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 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, should be removed Square root from a quantity representing the sum of the squares of its projections onto 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 represent various or actions, e.g. physical strength, movements elementary particles etc.
Location vector in two dimensions 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 graphic representation you will see right triangle, formed by legs with lengths, equal to 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 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.
Using tension lines or lines of force, you can visually depict an electrostatic field. Lines of force – curves, tangent at each point, which coincide with the direction of the tension vector E.
Lines of force are a relative concept and in reality they do not exist.
The field lines of positive and negative single charges are shown in the figure below:
Since a positive charge was used as a test charge, when introducing another one into its field positive charge their forces will be directed away from the charge. Therefore, it is believed that the lines of force “emanate” from the positive and “enter” the negative.
If we consider an electrostatic field formed by several stationary charges, then the lines of force can have the most different configuration. Based on the set of field lines, one can judge the change in the magnitude of the vector E in space and its direction, which characterizes the configuration (structure) of the electric field.
An electrostatic field is considered homogeneous in the case when the direction and density of the lines of force throughout the entire volume of the field are unchanged. Graphically, this is represented by straight parallel lines equidistant from each other.
Inside an area that doesn't have singular points(in which the voltage is zero) and does not have a boundary between two dielectrics, electric field lines are represented by smooth curves that do not have branches or kinks, do not intersect, and through each point of the field it is possible to draw no more than one field line.
If the number of field lines is numerically equal to the intensity E, they will characterize not only the direction of the field, but also its intensity. The number of lines is counted on a surface located perpendicular to each field line. This area will be part of the spherical surface in the case of a single charge.
Tension vector flow electrostatic field is the number of field lines N E that penetrate the area S, perpendicular to them.
IN general case the flux of the intensity vector through the area S is equal to:
Where E n is the projection of the vector E onto the normal n to the surface.
In the case of a flat surface and a uniform field, the flux of the vector E through the area S or its projection S / will be equal to:
Where α is the angle between the normal n and the vectors E to the surface S.
For example, it is necessary to determine the tension at a point lying on the boundary of two media: water (ε = 81) and air (ε ≈ 1). At this point (the point of transition from air to water), the electrostatic field strength decreases by 81 times. The flux of the tension vector will also decrease by a similar amount. When solving problems of calculating fields at junctions different environments the discontinuity of vector E causes certain inconveniences. To simplify the calculations, introduce new vector D, which is called electric displacement vector (induction vector). Numerically it is equal.