Charged bodies can influence each other without contact through an electric field. A field that is created by static electrical particles, is called electrostatic.

Instructions

1. If another charge Q0 is placed in the electric field created by charge Q, then it will influence it with a certain force. This collation is called tension electric field E. It represents the ratio of the force F with which the field acts on a regular electric charge Q0 at a certain point in space to the value of this charge: E = F/Q0.

2. Depending on a certain point in space, the value of field strength E can change, which is expressed by the formula E = E (x, y, z, t). Consequently, the electric field strength refers to vector physical quantities.

3. Since the field strength depends on the force acting on a point charge, the electric field strength vector E is identical to 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 .

4. Michael Faraday proposed to visually depict the field strength of an electric charge with the support of tension lines. These lines coincide with the tension vector at all tangential points. In drawings they are usually designated by arrows.

5. If the electric field is uniform and its intensity vector is continuous in magnitude and direction, then the intensity lines are parallel to it. If the electric field is created by a properly charged body, the lines of tension are directed away from it, and in the case of a negatively charged particle, towards it.

Tip 2: How to detect electric field strength

In order to discover tension electrical fields, introduce a known test charge into it. Measure the force that acts on it from the side fields and calculate the tension value. If the electric field is created by a point charge or a capacitor, calculate it using special formulas.

You will need

• electrometer, dynamometer, voltmeter, ruler and protractor.

Instructions

1. Determination of the voltage of an arbitrary electrical fields Take a charged body, the dimensions of which are insignificant compared to the size of the body generating the electric field. A charged metal ball with low mass is ideal. Measure the amount of its charge with an electrometer and place it in an electric field. Balance the force acting on the charge from the electric fields dynamometer and take readings in Newtons. After this, divide the force value by the amount of charge in Coulombs (E=F/q). The result will be tension electrical fields in volts per meter.

2. fields point charge If the electric field is generated by a charge whose magnitude is known, to determine its intensity at a certain point in space distant from it, measure the distance between the selected point and the charge in meters. After this, divide the amount of charge in Coulombs by the measured distance raised to the second power (q/r?). Multiply the resulting total by 9*10^9.

3. Determination of electrical voltage fields capacitor Measure the potential difference (voltage) between the capacitor plates. To do this, connect a voltmeter parallel to it, record the result in volts. After this, measure the distance between these plates in meters. Divide the voltage value by the distance between the plates, the result will be tension electrical fields. If air is not placed between the plates, determine the dielectric constant of this medium and divide the total by its value.

4. Definition of electrical fields made by several fields mi If the field at a given point is the result of the superposition of several electric fields, find the vector sum of the values ​​of these fields, taking into account their direction (field superposition thesis). If you need to detect the electric field formed by two fields we, construct their vectors at a given point, measure the angle between them. After this, square each of their values ​​and find their sum. Calculate the product of the field strength values, multiply it by the cosine of the angle, the one that is equal to 180? minus the angle between the tension vectors, and multiply the total by 2. Then subtract the resulting number from the sum of the squares of the tensions (E=E1?+E2?-2E1E2*Cos(180?-?)). When plotting fields, consider that the field lines leave the correct charges and enter the negative ones.

Video on the topic

The objects of vector algebra are line segments having a direction and a length called a modulus. In order to determine module vector, should be removed square root from a quantity representing the sum of the squares of its projections onto the coordinate axes.

Instructions

1. Vectors are characterized by two basic properties: length and direction. Length vector called the module or norm and represents a scalar value, the distance from the start point to the end point. Both properties are used for graphic image different quantities or actions, say, physical strength, movements of elementary particles, etc.

2. 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 autonomy allows the use of vector algebra tools in different calculations, say, determining the angles between spatial lines and planes.

3. The entire vector can be specified by the coordinates of its ends. Let's first look at two-dimensional space: let's preface vector is located at point A (1, -3), and the end is at point B (4, -5). In order to detect their projections, lower the perpendiculars to the abscissa and ordinate axis.

4. Determine the projections of yourself vector, which can be calculated using the formula: ABx = (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.

5. In the graphical image you will see 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|? = ABx? +ABy? ? |AB| = ?((xb – xa)? + (yb – ya)?) = ?13.

6. Apparently, for three-dimensional space the formula is complicated by adding a third coordinate - applicate zb and za for the ends vector:|AB| = ?((xb – xa)? + (yb – ya)? + (zb – za)?).

7. Let in the example considered 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 identical magnitude, measure the force of their interaction and the distance between them and make a calculation. If it is necessary to detect 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.

You will need

• – torsion scales;
• - ruler;
• - calculator;
• – electrostatic field meter.

Instructions

1. If there are two charges identical in modulus, measure the force of their interaction using a Coulomb torsion balance, which is also an emotional dynamometer. Later, when the charges come into balance and the wire of the scales compensates for the force of electrical interaction, record the value of this force on the scale. Later, using a ruler, caliper, or a special scale on the scale, find the distance between these charges. Consider that unlike charges attract, and like charges repel. Measure force in Newtons and distance in meters.

2. Calculate the value of the modulus of one point charge q. To do this, divide the force F with which two charges interact by the exponent 9 10^9. Take the square root of the result. Multiply the result by the distance between charges r, q=r?(F/9 10^9). You will receive the charge in Coulombs.

3. If the charges are unequal, then one of them must be previously known. Determine the interaction force between the known and unknown charge and the distance between them using Coulomb torsion balances. Calculate the modulus of the unknown charge. To do this, divide the force of interaction of charges F by the product of the exponent 9 10^9 by the modulus of the famous charge q0. Take the square root of the resulting number and multiply the total by the distance between the charges r; q1=r ?(F/(9 10^9 q2)).

4. Determine the modulus of an unfamiliar point charge by introducing it into an electrostatic field. If its intensity at a given point is unknown in advance, insert an electrostatic field meter sensor into it. Measure voltage in volts per meter. Place a charge at a point of known tension and, with the support of an emotional dynamometer, measure the force in Newtons acting on it. Determine the charge modulus by dividing the value of the force F by the electric field strength E; q=F/E.

Video on the topic

Pay attention!
The tension vector has only one direction at any point in space, therefore the tension lines never intersect.

How do we detect any force or interaction? According to the result of the impact. We hit the ball and the speed of the ball changed. The earth attracts us; we cannot push off with our feet and fly away, but always land back. Unfortunately:)

Likewise, with an electric field, it is not enough just to know that it exists; it is necessary to find some of its characteristics that will describe the result of its influence.

We know that the field affects the charge. Actually, we can detect the electric field only by its effect on the charge. Accordingly, we must introduce a value characterizing the strength of this influence.

Tension as a characteristic of the electric field

When placing various charges in a constant electric field, it was possible to discover that the magnitude of the force acting on the charge is always directly proportional to the magnitude of this charge.

According to Coulomb's law, everything is correct. After all, the field is created by charge q_1, therefore, with a constant value of charge q_1, the field created by it will act on the charge q_2 placed in it with a Coulomb force proportional to the value of charge q_2.

Therefore, the ratio of the force of the field on a charge placed in it to this charge will be a value independent of the magnitude of the charge creating this field.

This value can be considered as a characteristic of the field. It was called the electric field strength:

where E is the electric field strength, F is the force acting on a point charge, q is a charge placed in the field.

Field strength the quantity is vector, the direction of the intensity vector at any point in the field is always along the straight line connecting this point and the charge placed in the field. The tension vector always coincides in direction with the force vector acting on the charge.

Principle of field superposition

We know that if several different forces act on a body, directed in different sides, then the resultant of these forces will be equal to their geometric sum: F =F_1+F_2+...+F_n.

The direction of influence of this force is determined by the rule of vector addition. In the case when we have a charge located in the zone of action of several electric fields, then several forces will act on it.

The magnitude and direction of each individual force will depend on the strength of each field separately. The resultant of these forces, as in the case of a body, will be equal to their geometric sum.

It is logical to assume that then the resulting field strength for our charge will be the sum of the strengths of all fields present at this point. This is the essence of the principle of field superposition.

This principle has been confirmed experimentally: if at a given point in space various charged particles create electric fields whose strengths are E_1, E_2,…, E_n, then the resulting field strength at this point is equal to the sum of the strengths of these fields.

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. The strength characteristic of the electric field is the intensity 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

To graphically depict electrostatic fields, the concept of lines of force is used.

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 $\rho$ 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 considered field point. 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 obtain:

\[(dE)_x=\frac(2\pi R\tau dRcos\varphi )(R^2)=\frac(2\pi \tau dRcos\varphi )(R)=\frac(\tau cos\varphi d\varphi )(R)\ \left(1.4\right),\]

where $2\pi dR=d\varphi $.

We'll find full projection$E_x$, by integrating expression (1.4) over $d\varphi $, where the angle changes $0\le \varphi \le 2\pi $.

Let us deal with the projection of the tension vector onto axis OY, by analogy without special explanations let's 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 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 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.

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. They call him

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, the algebraic sum of 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 of electrical interaction between 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 the vector E is determined in the indicated way. We can say that the electric field strength vector is equal to the force acting on a unit positive charge placed at a given point in space.

The definition of E G = F G /q can be generalized to the case of variable (time-dependent) fields.

2. Let us 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 intensity vector at an 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 is the charge of the ball, 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 dielectrics ε > 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 starting and ending 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 electric current and the capacitor will discharge. When current flows in the conductor, it will release a certain amount heat, 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 .