Field work. Tension. Potential

2101 ... Two electrons, which are at the initial moment far from each other, move towards meeting along one straight line with the same speed in absolute value v o \u003d 1000 km / s. What is the smallest distance they will approach? decision

2102 ... Two electrons are at a great distance from each other. Initially, one electron is motionless, while the other approaches it with an initial velocity v o \u003d 1000 km / s directed along the straight line connecting the electrons. What is the smallest distance they will approach? At what speeds will they fly apart? decision

2103 ... Four balls having the same charges are located along one straight line so that the distance between adjacent balls is a. What kind of job A you need to complete in order to place these balls: a) at the vertices of a square with a side a; b) at the vertices of a tetrahedron with an edge a? decision

2104 ... Two identical metal balls of radius R\u003d 1 mm connected with a long thin wire. One of them is located in rarefied air, and the other is in the middle of a large vacuum chamber. An electron flow with an initial velocity of v o \u003d 3000 km / s. What charge Q can you save up on balls in this way? What will be the answer if we increase the initial velocity of electrons to v o / \u003d 10000 km / s? Electrical breakdown of air occurs at tension electric field E o \u003d 3 × 10 4 V / m. decision

2105 ... On a thin metal ring of radius R evenly distributed charge q... determine field strength E and potential j at the point Alocated on the axis of the ring at a distance h from its center. decision

2106 ... The electron is on the axis of a thin ring of radius R on distance h from its center. The ring gets a positive charge q and begins to attract an electron. Does the electron have to fly through the center of the ring? How fast v can he fly near this point? decision

2107 ... What is the strength of the electric field on the surface of the conductor if the density surface charge s. decision

2108 ... Inside a ball of radius R there is a space charge of constant density r.

1) Find the dependence of the electric field strength on the distance to the center of the ball.

2) Find the dependence of the potential on the distance to the center of the ball. decision

2109 ... Find the electric field strength inside and outside an infinitely long cylinder charged with a bulk density r. Cylinder radius R. decision

2110 ... On two coaxial infinite cylinders with radii R and 2 R evenly distributed charges with surface densities s 1 and s 2. Required:

1) Using the Ostrogradsky-Gauss theorem: find the dependence E(r) electric field strength versus distance for three regions I, II, III... To accept s 1 = s, s 2 = –S;

2) Tension E at a point remote from the axis of the cylinders at a distance r, and indicate the direction of the vector E, to accept s\u003d 30 nC / m2, r= 4R;

3) Build a graph E(r). decision

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Lesson objectives:

• Educational:
  • continue the formation of ideas and knowledge about the strength of the electric field;
  • teach to analyze the condition of the problem and predict the type of graphical dependence;
  • teach to apply the studied patterns in a changed situation;
• Developing:
  • develop the ability to independently apply knowledge in a complex (informatics and physics);
  • develop skills in plotting function;
  • improving the skills of solving physical problems using computer technology;
• Educational:
  • education of a culture of mental work;
  • creating a positive motivation for learning.

Means of education:computers, multimedia projector, screen, textbook "Physics" Grade 10, author V.A.Kasyanov, M.: Drofa, 2002.

Lesson type: lesson in the complex application of knowledge.

Teaching methods: verbal, visual, research, practical.

Lesson summary

The lesson in solving problems with plotting the graphs of the intensity versus distance is carried out after studying the topics: "Principle of superposition of fields", "Conductors and dielectrics in an electric field" so that you can cover the options for problems containing mixed media (conductors, dielectrics). Then the graphs are more visual and it is easy to trace the differences between the magnitude of the field strength in different environments.
Functions graphing tasks often cause difficulties for students due to the large amount of processed numerical data. The use of computer applications (Excel) simplifies the construction of geometrically complex graphs and allows you to do it with a given interval of varying values. When making corrections to the numerical data, the result is quickly displayed on the monitor, which allows you to visually analyze the construction. When solving different problems, the results are easily compared. Using a multimedia projector allows you to quickly display the result on the screen, after which you can start a collective discussion. The integration of traditional teaching and innovative technologies in the study of this topic gives a sustainable positive result... The lesson is held in a computer class.

References:

1. Physics in grade 10. Lesson models. Yu.A. Saurov... - Moscow: Education, 2005 .-- pp. 183-194.
2. Physics. Zadachnik. 9-11 grades. Goldfarb N.I. - Moscow.: Bustard, 1998. - pp. 87-88.
3. Collection of problems for the general course of physics. Volkenshtein V.S. - St. Petersburg .: "Special Literature", 1997. - p. 106.
4. Electronic textbook "Open Physics" part 2, edited by SM Kozel.

Lesson plan:

Lesson steps	Time, min	Techniques and methods
Organizing time	1
Knowledge update. Reiteration.	7	Frontal poll.
Statement of the educational problem. The solution of the problem.	7	Explanation of the teacher. Media projector.
Skills formation. Collective problem solving.	15	The work of students at the computer.
Physical education for the eyes.	2
Improving knowledge and skills. Analysis of the solved problems.	10	Students' speech. Assessment of knowledge.
Summarizing	3	Highlighting the main thing. Teacher's message.

During the classes:

Questions are discussed: (there are slides on the screen, the material of which students can use in the answer)

1. What is the principle of superposition of fields?
2. How does a conductor behave in an electrostatic field? What about the field inside the conductor?
3. Is there electric field inside a dielectric in the absence of an external field; in the presence of an external field?
4. What is the difference between the processes occurring in a conductor and a dielectric placed in an electric field?
5. What formula can be used to calculate the strength of the field formed by a charged metal ball?
6. How to find the field strength inside the dielectric layer?

Picture 1

Picture 2

Consider the following problem: (Displayed using a multimedia projector).

Objective 1.

A charged metal ball is placed in the center of a thick spherical layer made of metal. Plot field strength versus distance from the center of the sphere.

Let's discuss the solution to the problem:

As you know, inside a charged ball, the electric field strength is zero. Therefore, in the area from 0 to R, the graph is a line that coincides with the r-axis (the graph "lies" on the axis).

On the surface of the ball, the field strength is (the graph shows an increase in the value of the strength E at r \u003d R).

When r changes from R to R 1 and from R 2 to infinity, the value of E decreases according to the law: (graph-hyperbola).

The "dip" of the graph in the section from R 1 to R 2 shows the decrease in tension to zero inside the metal layer.

Thus, on the screen we see an approximate graph of the dependence of the field strength on the distance r.

Now we will solve two problems (by options) and build graphs of the dependence E (r) using the Excel computer program, after which we can compare the graphs and analyze the results. You see the problem conditions on the screen. When building a spreadsheet, read the step of plotting a graph of 5 cm. (The task conditions are displayed using a multimedia projector).

Problem 2(for option 1)

Metal ballradius 20 cm, having a charge of 10 nC, placed in the center of the spherical layer with an inner radius of 50 cm and an outer radius of 80 cm, manufactured dielectric a permeability of 2. Plot the field strength versus distance from the center of the sphere.

Problem 3(for option 2)

Charge Q \u003d 20 nC uniformly distributed over the volume balls radius of 30 cm, made non-conductive material with a permeability equal to 2.5. The ball is placed in the center thick spherical metal layer 50 cm thick. The air gap between the ball and the sphere is 25 cm thick. Draw a graph of field strength versus distance from the center of the sphere.

Students get started on the computer (15 min.)
The results of solving problems (both options) are displayed on the screen (multimedia projector).
During this time, the students do some relaxing eye exercises.

Students at the screen explain the solution to the problem and describe the resulting graph.

Let's start analyzing the resulting graphs.

1. How does the magnitude of tension depend on the distance at each section of the graph?
2. Where are the graphs different and why?
3. What explains the "dips" of the graphs when r is from 0.5 to 0.8 m? Why do they look different?
4. What value in condition 2 of the problem determines the "depth of failure"?
5. How will the appearance of the graphs change with a decrease (increase) in the magnitude of the electric charge?
6. How will the appearance of the graphs change with decreasing (increasing) the geometric dimensions of the sphere, the thickness of the layer?
7. Why does the function E have two meanings at some points?
8. What are the features of using Excel in this task?
9. Why does the table of values \u200b\u200bhave a "multistage" view?
10. What difficulties did you face when solving the problem?

The result of solving problem 2, obtained by students of option 1.

r	0	0,05	0,1	0,15	0,2	0,25	0,3	0,35	0,4	0,45	0,5	0,55	0,6	0,65	0,7	0,75	0,8	0,85	0,9	0,95	1
E	0	0	0	0	0
E					2250	1440	1000	734,69	562,5	444,44	360						140,62	124,56	111,11	99,72	90
E											180	149	125	107	92	80	70

The result of solving problem 3, obtained by students of 2 options.

r	0	0,05	0,1	0,15	0,2	0,25	0,3	0,35	0,4	0,45	0,5	0,55	0,6	0,65	0,7	0,75	0,8	0,85	0,9	0,95	1
E	0	133,3	266,6	399,9	533,2
E					4500	2880	2000	1469,38	1125	888,88	720						281,25	249,13	222,22	199,44	180
E											0	0	0	0	0	0	0

Summarize:

Solving problems on plotting the dependence E (r) allows you to visualize the geometry of the electric
fields and more accurately describe it. It is also interesting to carry out virtual experiments with the introduction of dissimilar bodies into the electric field and observe the change in the field pattern in these cases.

Homework

1. Answer the question: What is the difference: a conductor and a dielectric are placed in an electric field and cut in half; removed from the field?
2. Create and solve a problem similar to that solved in the class with modified conditions. Pass the result to the teacher
in the printed form.

1. On the r axis, it is impossible to display two values \u200b\u200bof one argument, the graph in this case is distorted ("stretched" horizontally and "lays down", see the graph below), therefore, it is necessary to draw up not one table, but a separate one for each section of the graph described by a separate function E (r).
2. When specifying a function in a table, the denominator must be enclosed in parentheses. 0

0 0 0 0 281,25 249,13

Aslamazov L. Tension, tension, potential // Kvant. -1978. - No. 5. - S. 38-43.

By special agreement with the editorial board and editors of the Kvant magazine

Each point of the electric field is characterized by a vector quantity - the field strength. Tension field at a given point is equal to the force acting on a positive test charge placed at this point, and referred to the unit of charge. It - power characteristic electric field.

When moving electric charge work is being done in the field. An electrostatic field has a very important property of potentiality: the work of moving a charge from one point of the field to another does not depend on the shape of the trajectory. This allows us to introduce the concept of voltage (or potential difference). Voltage Ubetween two points of the field (* Under the words "belt", "electric field" here and in what follows we mean an electrostatic field, that is, a field created by stationary charges.) is equal to the work done by the electric field to move the unit positive charge from one point to another.

Unlike the tension defined at a single point, tension characterizes two zero points. If you fix one point, choosing it as the origin, then any point of the field will have a certain voltage in relation to the selected point. This voltage is called potential φ. Obviously, the zero potential corresponds to the origin. Most often, the zero potential is attributed to a point infinitely distant from the charge that creates the field. In this case, the potential φ of some point of the field equal to workperformed by an electric field by moving a unit of positive charge from this point to infinity. It - energy characteristic electric field.

Sometimes ask at every point scalar - potential φ is more convenient than the vector value of the intensity . Naturally, these two quantities must be related to each other.

Let us first consider a uniform electric field. His tension is the same at all points; the lines of force of such a field are parallel straight lines (Fig. 1).

Find the potential difference between the points Band D.Potential φ B points Bis equal to the work of moving a unit of charge from this point to infinity. The shape of the trajectory when calculating the work does not matter, so we will move the charge first along the segment BC then along the segment CD and then from point D into infinity. The force acting on a unit of charge from the electric field is equal to the strength. On the segment Sunthe work of this force is Elwhere E -the projection of the tension vector onto the field line, a l- segment length Sun... On the segment CDthe force does not work, since it is perpendicular to the displacement. Finally, the work of moving a unit of charge from a point Dto infinity is equal to the potential φ D... Therefore: or for the potential difference:

(1)

In order for formula (1) to give the correct sign of the potential difference, the value l it is necessary to assign a certain sign depending on the location of the points Band Con the power line. We will assume that l Is the projection of the vector BDon the direction of the field line. Then the sign is positive if the point Clies "lower" along the line of force than the point B and negative otherwise. For the case shown in Figure 1, l \u003e 0, and the potential difference, which corresponds to a decrease in the potential along the line of force.

So, in a homogeneous electric field between the intensity and the potential difference there is a simple connection, given by formula (1).

What is the relationship between potential and strength in the case of an inhomogeneous electric field? In such a field, the tension varies from point to point. Let, for simplicity of reasoning, the change in tension occurs only in one direction, which we take as the axis OH(fig. 2).

Then the field strength depends only on the coordinate x:. It is clear that in small areas of space, the intensity changes little, and the electric field there can be approximately considered uniform. Let's take close points Band Dand find the potential difference between them. Let's use formula (1). The potential, just like the tension, depends only on the coordinate x (*Plane x \u003d const is equipotential, since when the unit of charge moves in this plane, the electric field does not work.):

Vector projection per axis OHis equal to the difference of coordinates of points Dand B:

Thus, for close points Band Dwe get:

(2)

For formula (2) to become accurate, it is necessary to direct the point Bto the point Dand find the limit to which the right-hand side tends with unlimited convergence of points:

(3)

It is easy to see that the right-hand side of formula (3) is the derivative of the potential, taken with the opposite sign. Thus, in an inhomogeneous electric field, the relationship between the potential and the strength at each point is as follows:

The minus sign in formula (4) means that the potential decreases along the line of force: since the projection of the tension on the line of force , which means a decrease in the potential.

If we draw a graph of the dependence of φ on x, then the tangent of the angle of inclination α of the tangent to the graph at each of its points is equal to the derivative at this point (fig. 3) . Therefore, we can say that the electric field strength determines the slope of the tangent to the potential graph.

Let us now consider several specific tasks.

Problem 1... Sphere of radius R has charge Q... Find the dependence of tension and potential on distance r from the center of the sphere. Draw graphs.

Let's find the field strength first. There is no electric field inside the sphere: at r< RE\u003d 0. Outside the sphere, the field strength is the same as that of a point charge Q placed in the center of the sphere: at r> Rthe projection of the intensity onto the selected direction from the center, where ε 0 is the electrical constant. On the surface of the sphere, at r = R the electric field experiences a jump ... Dependence Efrom ris graphically shown in Figure 4, a.

The magnitude of the jump Δ E can be expressed in terms of the surface charge density (equal to the charge per unit surface area of \u200b\u200bthe sphere):

Note that this is a general property electrostatic field: on a charged surface, its projection onto the normal direction always experiences a jump regardless of the surface shape.

Let us now find out how the potential φ changes depending on r.We already know that at any point the tangent of the angle of inclination of the tangent to the potential graph must coincide with the value of the intensity projection (taken with the opposite sign). At 0< r < RE \u003d0, and therefore, at all these points, the tangent to the potential graph must be horizontal. This means that on the segment 0< r< Rpotential does not change: φ \u003d const.

Outside the sphere, at r\u003eRthe derivative is negative and its value decreases with distance r.Therefore, the potential should also decrease with distance, tending to zero at. Indeed, the further the point at which we are looking for the potential is located, the less work must be done when moving a unit of charge from this point to infinity. The value of the potential φ at r> Ris the same as that of a point charge placed in the center of a sphere:

Can the potential experience a jump on the surface of a sphere, that is, at r \u003dR? Obviously not. A potential jump would mean that when a unit charge moves between two very close points 1 and 2 the electric field would do the final work:

must remain finite, which is impossible. Thus, the potential does not jump.

The plot of φ versus rshown in Figure 4, b .

Problem 2... Ball radius R evenly charged throughout the volume. Full charge of container Q... Draw graphs of tension and potential versus distance r from the center of the ball.

Such a ball can be imagined as consisting of a large number of thin charged spheres, embedded one into another. Each sphere inside itself does not create a field, but outside it creates a field the same as a point charge placed in its center. Therefore, outside the ball, at r\u003eR the strength is the same as the field strength of the point charge Q placed in the center of the ball:

Inside the ball, in the distance R the field is created only by spheres with radii from 0 to r (for spheres with a larger radius, the point in question is inside them). Hence the tension at a distance sfrom the center of the ball is the same as the field strength of a point charge Q r... placed in the center of the ball, where Q r– total charge all spheres with radii from 0 to r, that is, the charge of the ball of radius r.If the ball of radius Rthere is a charge Q, then on a ball of radius rthere will be a charge

Thus, inside the ball, the field strength - it grows linearly with distance.

On the surface of the ball, at the point r \u003dR the tension of the leap is not experienced. This is in line with general rule, since the surface charge density in this case is equal to zero: the ball is charged uniformly, and an infinitely small charge falls on an infinitely thin surface layer.

Dependency graph E from r shown in Figure 5, a.

Let us now draw a graph of the potential. Derived from potential

is always negative ( E ≥ 0). Therefore, with increasing rthe potential should decrease monotonically. At the point r \u003d0, the derivative of the potential is zero. Therefore, the tangent to the graph at. this point is horizontal: at the point r \u003d0 has a maximum potential. At the point r = R neither the potential nor its derivative experience jumps. The first follows from general rule for potential, we have already spoken about the second above. Therefore, the curves depicting the dependence of the potential on the distance at r < R and r > R at the point r= Rshould be mated - smoothly, without kink, go one into another. With potential. The plot of φ versus ris presented in Figure 5, b.

Problem 3... Two planes are parallel to each other at a distance d and charged with surface density charge σ 1 and σ 2, respectively. Draw graphs of field strength and potential versus coordinate x (axis OH perpendicular to the plates). Consider the cases of the same (Fig. 6, a) and opposite (Fig. 7, a) charges on the plates.

Figure: 6 Fig. 7

Each plane creates on both sides of itself a uniform electric field, the strength of which

Using the principle of superposition, for the case of like charges, we come to the graph shown in Figure 6, b, and for opposite charges, to the graph in Figure 7, b. Tension surges again correspond to the general rule:

The corresponding plots for the potentials are shown in Figures 6, c and 7, c . In some areas, the dependence of the potential on the coordinate is linear, since the field strength is constant. Kinks occur in places where the field strength experiences a jump.

Note that in this problem the potential does not tend to zero at. This is obviously due to the fact that the plane is infinite. In reality, the dimensions of real plates are always limited; this leads to the fact that the potential decreases with increasing distance from the plates.

Task 4... Two identical parallel plates have charges + q and - q... How the potential difference changes U between plates with increasing distance d between them? Draw a dependency graph U from d.

As long as the distance between the plates is much less than their dimensions, such a system can be considered a flat capacitor. Then - voltage increases linearly with distance (initial section in Figure 8).

This is consistent with the field strength. As soon as the distance between the plates becomes comparable to the dimensions of the plates, the electric field appears outside the space between the plates. Then the so-called edge effects become significant, and the dependence of the potential on the distance is rather complicated. However, it is qualitatively clear that, due to the weakening of the field in the region between the plates, the voltage will grow more slowly than linearly (middle section in Figure 8). With a further increase in the distance between the plates, it will become much larger than their dimensions. Then each plate can already be considered an isolated body, and its potential is where C 0 is the capacity of the solitary plate. Thus, at very large distances, the potential difference ceases to depend on the distance between the plates (plot of dependence Ufrom d.8 has a horizontal asymptote).

Boundary effects often turn out to be significant in solving electrostatic problems associated with the law of conservation of energy; consider, for example, such a version of an electron accelerator.

Problem 5... In the plates of a flat capacitor charged to a potential difference U a through hole is made. The capacitor is placed in a constant magnetic field directed perpendicular to the electric field in the capacitor (Fig. 9). An electron flies into the space between the capacitor plates, accelerates, acquiring energy eU flies out through the hole and. moving in a magnetic field in a circle, it returns to the capacitor. Then it accelerates again, moves in a circle with a larger radius, enters the capacitor again, etc. At first glance, it seems that in this way it is possible to accelerate an electron to high energies, that is, to create an accelerator. Is it so?

It turns out that such an accelerator will not work - the edge effect is not taken into account. Outside the capacitor, there is always a weak electric field, which slows down the electron as it moves in a circle. Negative work field in this case is exactly equal to the positive work during acceleration of an electron in a capacitor: work in an electrostatic field does not depend on the shape of the trajectory. The magnetic field does not work (the Lorentz force is perpendicular to the speed of the electron). Therefore, the total work of all the forces acting on the electron, when it returns to the starting point, will be zero, and the kinetic energy of the electron will not change. The accelerator will not work.

Exercises

1. Can there be an electrostatic field in which the lines of force are parallel straight lines, and the absolute value of the intensity changes only in the direction perpendicular to the lines of force (Fig. 10)?

2. Two concentric metal spheres of radii R 1 and R 2 have charges Q 1 and Q 2 respectively. Find the strength and potential of the electric field at an arbitrary distance r from the center of the spheres. Draw dependency graphs E from r and φ from r... Consider the cases of like and unlike charges. What the graphs for the case look like Q 1 = –Q 2 (spherical capacitor)?

3. Point charge qsurrounded by a metal sphere of radius Rwith charge Q.Find the field strength and potential at an arbitrary distance r from charge q if it is in the center of the sphere; draw dependency graphs Efrom rand φ from r.How will the graphs change if the charge is displaced from the center of the sphere? Solve the same problem for the case where the metal sphere is grounded.

4. An electron flies into the space between the plates of a flat capacitor so that its velocity makes an acute angle with the direction ley lines... Then, when moving in the capacitor, it will decelerate and fly out at a lower speed; his kinetic energy will decrease. Will this increase the energy of the capacitor?

5. Two identical capacitors with a capacity C each one of which is charged to voltage U and the second is not charged, connected in parallel. Find the energy of the system before and after connecting the capacitors. Why are these energies not equal?

6. Point charge qis outside an uncharged metal sphere of radius Ron distance d from its center. Find the potential of the sphere.

Answers.

1. It cannot, otherwise the work of moving the charge along a closed loop would be different from zero.

2. When R 1 > r \u003e 0 tension E \u003d 0 and ; at R 2 > r > R and ; at r > R 2 and (fig. 11).

3. When R > r \u003e 0 tension and ; at r > R and (fig. 12).

4. The energy of the capacitor does not change; the energy of interaction between the electron and the capacitor changes (the work of moving the electron to infinity from the starting and ending points is not the same).

5. exactly half of the energy went into heat (regardless of the resistance of the supply wires).

6. (the potential of the sphere is the same as in its center, and there the total potential of the field of the charges induced on the sphere is equal to zero).