Elastic springs. The spring pitch is determined by the dependence. First, general concepts
The electric field that surrounds the charge is a reality that is independent of our desire to change something and somehow influence it. From this we can conclude that the electric field is one of the forms of existence of matter, just like matter.
The electric field of charges at rest is called electrostatic. To detect the electrostatic field of a certain charge, you need to introduce another charge into its field, which will be acted upon by a certain force in . However, without the presence of a second charge, the electrostatic field of the first charge exists, but does not manifest itself in any way.
Tension E characterize the electrostatic field. Tension at a certain point in the electric field is a physical quantity that is equal to the force acting on a unit positive stationary charge placed at a certain point in the field, and directed in the direction of the force.
If a “test” positive point charge q is introduced into the electric field created by a charge q, then according to Coulomb’s law a force will act on it:
If different test charges q / pr, q // pr and so on are placed at one point in the field, then each of them will be affected various forces, proportional to the amount of charge. The ratio F/qpr for all charges introduced into the field will be identical, and will also depend only on q and r, which determine the electric field at a given point. This value can be expressed by the formula:
If we assume that q pr = 1, then E = F. From here we conclude that the electric field strength is its force characteristic. From formula (2) taking into account the expression Coulomb force(1) follows:
From formula (2) it is clear that the unit of intensity is taken to be the intensity at a certain point in the field, where a unit of force will act on a unit of charge. Therefore, in the CGS system the unit of tension is dyne/CGS q, and in the SI system it will be N/Cl. The relationship between the given units is called the absolute electrostatic tension unit (AGS E):
The intensity vector is directed from the charge along the radius when the field is formed by a positive charge q+, and when the field is negative – q – towards the charge along the radius.
If the electric field is formed by several charges, then the forces that will act on the test charge add up according to the rule of vector addition. Therefore, the strength of a system consisting of several charges at a given point in the field will be equal to the vector sum of the strengths of each charge separately:
This phenomenon is called the principle of superposition (imposition) of electric fields.
The intensity at any point in the electric field of two point charges - q 2 and +q 1 can be found using the superposition principle:
According to the parallelogram rule, the addition of vectors E 1 and E 2 will occur. The direction of the resulting vector E is determined by the construction, and its absolute value can be calculated using the formula below:
Where α is the angle between vectors E 1 and E 2.
Let's look at the electric field that a dipole creates. Electric dipole – this is a system of equal in magnitude (q = q 1 = q 2), but opposite in sign charges, the distance between which is very small when compared with the distance to the points of the electric field under consideration.
Electric dipole moment p, which is the main characteristic of a dipole and is defined as a vector directed from a negative charge to a positive one, and equal to the product of the dipole arm l by the charge q:
The vector is also the arm of the dipole l, directed from the negative charge to the positive one, and determines the distance between the charges. The line that passes through both charges is called - dipole axis.
Let's determine the electric field strength at a point that lies on the dipole axis in the middle (figure below a)):
At point B, the tension E will be equal to the vector sum of the tensions E / and E //, which are created by positive and negative charges separately. Between the charges –q and +q, the intensity vectors E / and E // are directed in the same direction, therefore, in absolute value, the resulting intensity E will be equal to their sum.
If we need to find E at point A lying on the extension of the dipole axis, then the vectors E / and E // will be directed in different directions, respectively, in absolute value the resulting intensity will be equal to their difference:
Where r is the distance between the point that lies on the dipole axis and at which the intensity is determined, and the midpoint of the dipole.
In the case of r>>l, the value (l/2) in the denominator can be neglected, then we obtain the following relation:
Where p is the electric dipole moment.
This formula in the GHS system will take the form:
Now you need to calculate the electric field strength at point C (Figure above b)), lying on the perpendicular restored from the midpoint of the dipole.
Since r 1 = r 2, then the equality will take place:
The dipole strength at an arbitrary point can be determined by the formula:
Where α is the angle between the dipole arm l and the radius vector r, r is the distance from the point at which the field strength is determined to the center of the dipole, p is the electric moment of the dipole.
Example
At a distance R = 0.06 m from each other there are two identical point charges q 1 = q 2 = 10 -6 C (figure below):
It is necessary to determine the electric field strength at point A, which is located on a perpendicular restored at the center of the segment that connects the charges, at a distance h = 4 cm from this segment. You also need to determine the voltage at point B, located in the middle of the segment that connects the charges.
Solution
According to the principle of superposition (superposition of fields), the field strength E is determined. Thus, the vector (geometric) sum is determined by the E created by each charge separately: E = E 1 + E 2.
The electric field strength of the first point charge is equal to:
Where q 1 and q 2 are charges forming an electric field; r is the distance from the point at which the voltage is calculated to the charge; ε 0 – electrical constant; ε is the relative dielectric constant of the medium.
To determine the intensity at point B, you first need to construct the vectors of the electric field strength from each charge. Since the charges are positive, the vectors E / and E // will be directed from point B in different directions. By condition q 1 = q 2:
This means that in the middle of the segment the field strength is zero.
At point A it is necessary to perform a geometric addition of vectors E 1 and E 2. At point A the tension will be equal to:
Forces acting at a distance are sometimes called field forces. If you charge an object, it will create an electric field - an area with changed characteristics surrounding it. An arbitrary charge that enters the electric field zone will be subject to the action of its forces. These forces are influenced by the degree of charge of the object and the distance to it.
Forces and charges
Let's say there is some initial electric charge Q that creates an electric field. The strength of this field is measured by the electrical charge in the immediate vicinity. This electric charge is called a test charge, since it serves as a test charge when determining the voltage and is too small to influence the generated electric shock.
The control electric charge will be called q and have some quantitative value. When it is placed in an electric field, it is subject to attractive or repulsive forces F.
As a formula for electric field strength, denoted by the Latin letterE, serves as a mathematical notation:
Force is measured in newtons (N), charge - in coulombs (C). Accordingly, the unit used for tension is N/C.
Another unit often used in practice for homogeneous EP is V/m. This is a consequence of the formula:
That is, E depends on the electrical voltage (the potential difference between its two points) and the distance.
Does tension depend on the quantitative value of electric charge? From the formula you can see that an increase in q entails a decrease in E. But according to Coulomb's law, a greater charge also means a greater electrical force. For example, a twofold increase in electric charge will cause a twofold increase in F. Therefore, there will be no change in voltage.
Important! The electric current voltage is not affected by the quantitative indicator of the test charge.
How is the electric field vector directed?
For a vector quantity, two characteristics are necessarily used: quantitative value and direction. The initial charge is acted upon by a force directed towards it or towards the opposite side. The choice of a reliable direction is determined by the charging sign. To resolve the question in which direction the tension lines are directed, the direction of the force F acting on the positive electric charge was accepted.
Important! The field strength lines created by the electric charge are directed from the charge with the “plus” sign to the charge with the “minus” sign. If you imagine an arbitrary positive initial charge, then lines will come out of it in all directions. For a negative charge, on the contrary, the entry of field lines from all surrounding sides is observed.
A visual display of vector quantities of EF is carried out using power lines. A simulated ES sample can consist of an infinite number of lines, which are arranged according to certain rules that give the maximum possible more information about the nature of the EP.
Rules for drawing power lines:
- Electric charges of larger magnitude have the strongest electric field. In a schematic diagram this can be shown by increasing the frequency of the lines;
- In areas connected to the surface of an object, the lines are always perpendicular to it. On the surface of objects regular and irregular shape there is never an electrical force parallel to it. If such a force existed, any excess charge on the surface would begin to move, and a electricity inside an object, which is never the case with static electricity;
- When leaving the surface of an object, the force can change direction due to the influence of electron beams of other charges;
- Electrical lines must not cross. If they intersect at some point in space, then at this point there should be two EPs with their own individual direction. This is an impossible condition, since each place of the EP has its own tension and direction associated with it.
The power lines for the capacitor will run perpendicular to the plates, but will become convex at the edges. This indicates a violation of the homogeneity of the EP.
Taking into account the condition of a positive electric charge, we can determine the direction of the electric field strength vector. This vector is directed towards the force acting on the electric charge with a plus sign. In situations where the electric shock is created by several electric charges, the vector is found as a result of the geometric summation of all forces to which the test charge is exposed.
At the same time, the lines of electric field strength are understood as a set of lines in the area of action of the electric field, tangent to which the vectors E will be at any arbitrary point.
If an electric shock is created from two or more charges, lines appear surrounding their configuration. Such constructions are cumbersome and are performed using computer graphics. When solving practical problems, the resulting vector of electric field strength for given points is used.
Coulomb's law defines electric force:
F = (K x q x Q)/r², where:
- F – electric force directed along the line between two electric charges;
- K – constant of proportionality;
- q and Q – quantitative values of charges (C);
- r is the distance between them.
Constant proportionality is found from the relation:
K = 1/(4π x ε).
The value of the constant depends on the medium in which the charges are located (dielectric constant).
Then F =1/(4π x ε) x (q x Q)/r².
The law is valid in natural environment. For theoretical calculations, it is initially assumed that electric charges are in free space (vacuum). Then the value of ε = 8.85 x 10 (to the -12th power), and K = 1/(4π x ε) = 9 x 10 (to the 9th power).
Important! Formulas describing situations where there is spherical symmetry (most cases) include 4π. If there is cylindrical symmetry, 2π appears.
To calculate the tension modulus, you need to substitute the mathematical expression of Coulomb’s law into the formula for E:
E = F/q = 1/(4π x ε) x (q x Q)/(r² x q) = 1/(4π x ε) x Q/r²,
where Q is the initial charge that creates the electron beam.
To find the electric shock intensity at a specific point, you need to place a test charge at this point, determine the distance to it and calculate E using the formula.
Inverse square law
In the formulaic representation of Coulomb's law, the distance between electric charges appears in the equation as 1/r². This means that the application of the inverse square law will be fair. Another well-known such law is Newton's law of gravity.
This expression illustrates how changing one variable can affect another. Mathematical notation of the law:
E1/E2 = r2²/r1².
The value of the field strength depends on the location of the selected point; its value decreases with distance from the charge. If we take the electric field strengths at two different points, then the ratio of their quantitative values will be inversely proportional to the squares of the distance.
To measure EF voltage in practical conditions, there are special devices, for example, tester VX 0100.
Video
As you know, electrical voltage must have its own measure, which initially corresponds to the value that is calculated to power a particular electrical device. Exceeding or decreasing the value of this supply voltage negatively affects electrical equipment, up to its complete failure. What is tension? This is the electrical potential difference. That is, if, for ease of understanding, it is compared with water, then this will approximately correspond to pressure. According to science, electrical voltage is a physical quantity that shows how much work the current does in a given area when a unit charge moves through this area.
The most common voltage-current formula is one in which there are three basic electrical quantities, namely voltage itself, current and resistance. Well, this formula is known as Ohm’s law (finding electrical voltage, potential difference).
This formula sounds like this - electrical voltage is equal to the product of current and resistance. Let me remind you that in electrical engineering there are different units of measurement for various physical quantities. The unit of measurement for voltage is “Volt” (in honor of the scientist Alessandro Volta, who discovered this phenomenon). The unit of current is “Ampere” and resistance is “Ohm”. As a result, we have - an electrical voltage of 1 volt will be equal to 1 ampere multiplied by 1 ohm.
In addition, the second most used voltage formula is the one in which this same voltage can be found knowing the electrical power and current strength.
This formula sounds like this - electrical voltage is equal to the ratio of power to current (to find voltage you need to divide power by current). The power itself is found by multiplying the current by the voltage. Well, to find the current you need to divide the power by the voltage. Everything is extremely simple. The unit of measurement for electrical power is “Watt”. Therefore, 1 volt is equal to 1 watt divided by 1 ampere.
Well, now I’ll give a more scientific formula for electrical voltage, which contains “work” and “charges”.
This formula shows the ratio of the work performed to move electric charge. In practice this formula you're unlikely to need it. The most common one will be the one that contains current, resistance and power (that is, the first two formulas). But, I want to warn you that it will only be true for the case of using active resistances. That is, when calculations are made for electrical circuit, which has resistance in the form of ordinary resistors, heaters (with a nichrome spiral), incandescent light bulbs, and so on, then the above formula will work. In the case of using reactance (the presence of inductance or capacitance in the circuit), you will need a different current voltage formula, which also takes into account the frequency of the voltage, inductance, and capacitance.
P.S. The formula of Ohm's law is fundamental, and it is by it that one can always find one unknown quantity out of two known ones (current, voltage, resistance). In practice, Ohm's law will be used very often, so it is simply necessary for every electrician and electronics engineer to know it by heart.