Explanatory note

The flashcards in this series will help students become more familiar with the concepts of electrostatics that are new to them. In addition, the skills of solving problems, converting units of measurement, and calculating using a calculator are developing.

How to work with cards

The drawings of the cards show two metal balls carrying electric charges. The values ​​of these charges are indicated on the cards. A checkered grid is used to find the sizes of the balls and the distance between them (their centers). 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 value 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. Suggest the first two questions. Distances are calculated from the length of cells on an appropriate scale using the Pythagorean theorem.

The second time it is useful to apply the cards after learning the concept of electric field strength. By offering students questions 3, 4,5. Students should redraw the location of all charges in their notebook (lined in a cage) and draw vectors on the selected scale and and their sum vector. it is interesting to invite 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 to 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 values ​​at point B created by each charge. Redraw the location of the balls and test charge q in your notebook. On the selected scale, depict 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. With what force does the electric field act on a test charge q placed at point B?
5. What is the acceleration of a body with test charge q and mass m?
6. Determine the radii of the balls from the scale and calculate their potentials.
7. Determine the potentials of the electric field at points B and C.
8. What work must be done by external forces to move the test charge q from point B to point C?

Solution example for card #8

1. Distance between the centers of the balls:

10,r=10cm=0.1m

1. Modulus of the interaction force between charges q 1 and q2 :

1. Electric field strength module at point B:

Let's depict the tension vectors and on the drawing to scale (see figure)

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 the charge q 2 .

1. The modulus 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 q2 :

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, respectively, 8 and 6 times. Therefore, the total potential at point B is:

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 the test charge q from point B to point C:

J

Example of a programmed exercise

Questions:

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

Appendix

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. modulus of force of interaction, 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. module of the 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 ball with a 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 in the card. Use this information to 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 on the selected scale the vectors of the electric field strength at point B from each charged ball, find the magnitude and direction of the total vector at this point of the field.

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

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

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

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

8 What work must be done by external forces to move the 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 Ball center distance:

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

3 Electric field strength module at point B:

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

4 The modulus 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 the 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, it is determined by the formula: where units SI, then

The card shows a flat capacitor. Its thickness is indicated. Nearby is the shape of the capacitor plate. Plate dimensions are given in millimeters. Using the data on the card, complete the tasks, answer the questions.

1 Calculate the active area of ​​the capacitor.

2 Calculate the capacitance of the capacitor.

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

4 Find the amount of charge on the capacitor plate.

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

6 What electric capacitance 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 is laid between them of the same thickness. The dielectric constant of mica is assumed to be 6.

Did what I could

• 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. To study the physical essence of the gravitational constant.

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

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

What happens if...

• What happens if...

• We dropped the luggage from the 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 the straight line connecting the bodies; - are equal in size; are opposite in direction.

The formula applies:

• The formula applies:

• - if the dimensions 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 dimensions and mass of which are much larger than that of the second body

Task #1

• Task #1

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

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

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

• The force is so insignificant that it won't even break the thread.

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

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

GIVEN: Decision:

• GIVEN: Decision:

• m1=m2=100kgFrom the law of the world

• gravity:

• F= 0.01N F= G*m1m2/ R2

• _____________ let's express the distance:

• R-? R = (G*m1m2/ F) ½

• Let's calculate:

• R \u003d (6.67 * 10 -11Nm2 / kg2 * 100kg * 100 kg / 0.01N) 1/2

• R = 8.2 * 10-3 m

• Answer : R=8.2*10-3m

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

GIVEN: Decision:

• GIVEN: Decision:

• m1=m2 = mFrom the law of the universal

• 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:

• motion of bodies under the influence of gravitational forces near the surface of the Earth;

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

• trajectories of comets and meteors;

• the phenomenon of ebbs and flows;

• possible trajectories of celestial bodies were explained;

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

Let's summarize:

• Let's summarize:

• Newton set

• what all bodies in the universe mutually attract each other.

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

§ 15, exercise 15 (3; 5)

• § 15, exercise 15 (3; 5)