Explanatory note

Cards in this series will help students better become familiar with new concepts of electrostatics. In addition, skills in problem solving, converting units of measurement, and calculations using a calculator are developed.

Methodology for working with cards

The card designs depict two metal balls carrying electrical charges. The values ​​of these charges are indicated on the cards. To find the sizes of the balls and the distance between them (their centers), a checkered grid is used. Each card indicates the length of the side of the cell of this grid. The mass of the ball on which the test charge is located at point B and the magnitude of this charge are also indicated on the cards.

After familiarizing students with Coulomb's law, it is recommended to do independent work with cards. The first two questions are suggested. Distances are calculated from the length of the cells on the appropriate scale using the Pythagorean theorem.

The second time it is useful to use the cards after studying the concept of electric field strength. Offering students questions 3, 4,5. Students should draw the location of all the charges in their notebook (lined in a square) and draw the vectors on the selected scale And and their total vector. It is interesting to ask students to draw the approximate location of the tension line passing through point B.

If you wish, you can ask questions 1-5 at the same time.

Questions for the cards “Interaction of Electric Charges”

1. What is the distance between the centers of the balls?
2. With what force do the charges on the balls interact with each other?
3. Calculate the field strength at point B created by each charge. Draw the location of the balls and test charge q in your notebook. On the selected scale, draw the intensity vectors created by each charge at point B. Find the magnitude and direction of the total intensity vector at this point in the field. Draw the approximate location of the tension line passing through point B.
4. What force does the electric field exert on a test charge q placed at point B?
5. What acceleration does a body with test charge q and mass m acquire?
6. Determine the radii of the balls by scale and calculate their potentials.
7. Determine the electric field potentials at points B and C.
8. How much work must be done by external forces to move a test charge q from point B to point C?

Example solution for card No. 8

1. Distance between ball centers:

10, r = 10 cm = 0.1 m

1. Modulus of interaction force between charges q 1 and q 2:

1. Electric field strength modulus at point B:

Let us depict the tension vectors And on the drawing to scale (see picture)

Let's build the tension vectorIts direction is indicated in the drawing, and the module is calculated:

Let's draw an approximate line of electric field strength through point B. This line should be tangent to the direction of the vectorand is perpendicular to the surface of the ball carrying charge q 2 .

1. The magnitude of the force with which the field acts on the test charge q at point B:

1. The acceleration module at point B will be:

1. Potentials on balls carrying charges q 1 and q 2:

1. Potentials at points B from charges q 1 and q 2 will be as many times less than the potentials on the balls as the distances from the centers of the balls to this point are greater than the radii of the balls. In this example, 8 and 6 times, respectively. Therefore, the total potential at point B is equal to:

The potential at point C from the same charges is determined by first finding the distances from the balls to this point.

13.6 cm = 0.136 m

8.06 cm = 0.081 m

1. The work of external forces required to move a test charge q from point B to point C:

J

Example of a programmed exercise

Questions:

1. Potential of a sphere with charge q 1 , V
2. Potential of a sphere with charge q 2 , V
3. Potential at point B, B
4. Potential at point C, B
5. Work to move charge q from point to point C, μJ

Answers to cards No. 1, 3, 5, 7, 9

	4 500	22 500	7 200	2 200
	5 400	7 200		2 800
	18 000	9 000
	3 200	18 000
	22 500	3 600	2 000

Code to check:

№1 – 25 431

№3 – 23 512

№5 – 34 125

№7 – 51 243

№9 – 12 354

Answers to cards No. 2, 4, 6, 8, 10

	9 000	54 000	12 000
	36 000	9 000		1 400
	36 000	18 000	1 700	8 200
	18 000	7 200	2 300	1 200
	27 000	45 000		2 300

Code to check:

№2 – 53 241

№4 – 42 513

№6 – 31 425

№8 – 25 134

№10 – 14 352

Application

option
charge q 1, 10 -9 C	1,50	30,00	6,00	40,00	20,00	2000,00	50,00	40,00	5,00	50,00	40,00	500,00
charge q 2, 10 -9 C	1,00	20,00	10,00	20,00	20,00	3000,00	50,00	50,00	8,00	40,00	30,00	300,00
charge q, 10 -9 C	30,00	5,00	50,00	1,00	5,00	400,00	30,00	2,00	30,00	2,00	5,00	20,00
weight, kg	0,0020	0,0200	0,0001	0,0050	0,0020	0,0200	0,0050	0,0500	0,0100	0,0002	0,0002	0,0020
1. distance between charges, m	0,05	0,10	0,10	0,20	0,08	10,00	0,16	0,10	0,20	9,90	0,50	0,80
2. interaction force module, 10-5 N	0,54	54,00	5,40	18,00	56,25	54,00	87,89	180,00	0,90	0,02	4,32	210,94
	8,00	42,00	15,00	14,00	72,00	0,75	45,00	56,00	0,88	1,50	2,00	18,00
	10,00	50,00	14,00	12,50	72,00	0,28	45,00	125,00	0,26	2,00	3,00	10,80
	12,81	65,30	20,52	18,77	86,40	0,80	72,00	136,97	0,70	3,00	3,61	23,50
4. modulus of force acting on the charge, 10-5 N	38,43	32,65	102,59	1,88	43,20	32,00	216,00	27,39	2,10	0,60	1,80	47,00
5. charge acceleration module, 10-2 m/s 2	19,22	1,63	1025,90	0,38	21,60	1,60	43,20	0,55	0,21	3,00	9,01	23,50
1, kV	5,40	27,00	5,40	18,00	18,00	36,00	9,00	36,00	4,50	9,00	7,20	45,00
6. potential of a sphere with charge q 2, kV	3,60	18,00	9,00	9,00	18,00	54,00	9,00	45,00	7,20	7,20	5,40	27,00
7. potential at point B, kV	0,64	0,38	2,00	0,75	7,20	2,25	0,00	12,00	0,46	1,70	0,00	3,60
7. potential at point C, kV	0,35	1,20	2,20	0,25	2,85	1,90	0,26	8,23	0,06	2,30	0,44	4,80
8. work of external forces, 10-6 J	8,70	4,10	10,00	1,00	21,75	141,20	7,71	7,54	12,00	1,20	2,20	24,00

Interaction of electric charges

The figure shows two charged balls and a test charge B. The magnitude of the charges and the mass of the body are given on the card. Using this data, complete the tasks and answer the questions.

1 What is the distance between the centers of the balls?

2 With what force do the charges on the balls interact with each other?

3 Draw the location of the balls and test charge q in your notebook, calculate and draw the electric field strength vectors at point B from each charged ball on a selected scale, find the magnitude and direction of the total vector at this point in the field.

4 With what force does the electric field act on a test charge placed at point B?

5 What acceleration does a body with a test charge q receive at this point? (Body weight is indicated on the card.)?

6 Determine the radii of the balls using the scale and calculate the potentials on the balls in kilovolts.

7 Calculate the electric field potentials at points B and C.

8 How much work must be done by external forces to move a test charge q from point B to point C?

Option 1

Option 2

Option 3

Option 4

Option 5

Option 6

Option 7

Option 8

Option 9

Option 10

1 Distance between ball centers:

2 Modulus of the interaction force between charges q 1 and q 2:

3 Electric field strength modulus at point B:

Let us depict the tension vectors in the drawing to scale: the side of the cell is equal to . Let's construct a tension vector. Its direction is indicated in the drawing, and the module is calculated:

4 The magnitude of the force with which the field acts on the test charge q at point B:

5 The acceleration module at point B will be:

Let's draw an approximate line of electric field strength through point B. This line should be tangent to the direction of the vector and perpendicular to the surface of the ball carrying charge q 2. Since the test positive charge q approaches the negative charge q 2, the force and acceleration will increase as the charge q moves.

6 Potentials on balls carrying charges q 1 and q 2. In SI units, determined by the formula: Where units SI, then

The card shows a parallel plate capacitor. Its thickness is indicated. The shape of the capacitor plate is shown nearby. The plate dimensions are given in millimeters. Using the data on the card, complete the tasks and answer the questions.

1 Calculate the active area of ​​the capacitor.

2 Calculate the electrical capacity of the capacitor.

3 What is the field strength between the plates of a capacitor?

4 Find the amount of charge on the capacitor plate.

5 With what force does the capacitor field act on the charge q 1, the value of which is indicated on the card?

6 What electrical capacity in microfarads will 100 of the same capacitors connected in parallel have if the distance between the plates is reduced to 0.1 mm and mica of the same thickness is placed between them. The dielectric constant of mica is considered equal to 6.

I did what I could

• I did what I could

• let others do better.

• I. Newton.

• . Formulate the law of universal gravitation and write down a formula expressing the relationship between quantities.

• 2. Study the physical essence of the gravitational constant.

• 3. Limits of applicability of the law of universal gravitation

• 4. Learn to solve problems using the law of universal gravitation.

What will happen if...?

• What will happen if...?

• We dropped the luggage from our hands...

• We threw the ball up...

• We threw a stick horizontally...

M. Lomonosov

• M. Lomonosov

• English scientist Isaac Newton was the first to formulate the law of universal gravitation

• - long-range; - there are no barriers for them; - directed along a straight line connecting the bodies; - equal in size; - opposite in direction.

The formula applies:

• The formula applies:

• - if the sizes of the bodies are negligibly small compared to the distance between them;

• - if both bodies are homogeneous and have a spherical shape;

The formula applies:

• The formula applies:

• - if one of the interacting bodies is a ball, the size and mass of which is significantly greater than that of the second body

Task No. 1

• Task No. 1

• Calculate the force of universal gravity between two students sitting at the same desk.

• The masses of the students are 50 kilograms, the distance is one meter.

• We get a force equal to 1.67*10 -7 N .

• The force is so insignificant that even the thread will not break.

• With what force is Aunt Masha’s goat attracted to the cabbage in Baba Glasha’s garden if he is grazing at a distance of 10 meters from her? The weight of the goat Grishka is 20 kg, and this year the cabbage has grown large and juicy, its weight is 5 kg.

• What is the distance between the balls, each weighing 100 kg, if they are attracted to each other with a force of 0.01 N?

GIVEN: Solution:

• GIVEN: Solution:

• m1=m2 =100kgFrom the universal law

• gravity:

• F= 0.01N F= G*m1m2/ R2

• _____________ Let's express the distance:

• R -? R = (G*m1m2/ F) ½

• Let's calculate:

• R = (6.67*10 -11Nm2/kg2 *100kg*100kg/0.01N)1/2

• R = 8.2*10-3 m

• Answer : R = 8.2*10-3 m

• Two identical balls are located at a distance of 0.1 m from each other and attract with a force of 6.67 * 10 -15 N. What is the mass of each ball?

GIVEN: Solution:

• GIVEN: Solution:

• m1=m2 = mFrom the universal law

• R=0.1 m gravity:

• F= 6.67*10 -15N F= G*m1m2/ R2

• _____________ Let's express the mass of bodies:

• m-? m= (F*R2/G) ½

• Let's calculate:

• m= (6.67*10 -15 N *0.01m2/6.67*10 -11Nm2/kg2)1/2

• m= 0.001 kg

• Answer: m= 0.001 kg

• The discovery of the law of universal gravitation made it possible to explain a wide range of terrestrial and celestial phenomena:

• the movement of bodies under the influence of gravitational forces near the Earth's surface;

• movements of the planets of the solar system and their natural and artificial satellites;

• trajectories of comets and meteors;

• the phenomenon of ebb and flow;

• the possible trajectories of celestial bodies were explained;

• solar and lunar eclipses were calculated, masses and densities of planets were calculated

Let's summarize:

• Let's summarize:

• Newton established

• What all bodies in the Universe mutually attract each other.

• The mutual attraction between all bodies is called universal gravity – gravitational force.

§ 15, exercise 15 (3; 5)

• § 15, exercise 15 (3; 5)