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PSE Chapter 15 pg. 197 Textbook Chapter 32. ELECTROSTATICS – the study of electric charges, forces and fields Two types of charges exists, arbitrarily.

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Presentation on theme: "PSE Chapter 15 pg. 197 Textbook Chapter 32. ELECTROSTATICS – the study of electric charges, forces and fields Two types of charges exists, arbitrarily."— Presentation transcript:

1 PSE Chapter 15 pg. 197 Textbook Chapter 32

2 ELECTROSTATICS – the study of electric charges, forces and fields Two types of charges exists, arbitrarily named POSITIVE and NEGATIVE By Benjamin Franklin So what is positive and what is negative?

3 We know that charged particles exist in atoms Electrons are responsible for negative charge and Protons for positive charge Benjamin Franklin did not know about the existence of these particles but, he did investigate the behavior of static discharge and lightning.

4 Ben knew that if certain electrically neutral objects are rubbed, they can become charged. For example; when rubber is rubbed with a wool cloth, both become charged. The rubber scrapes electrons from fur atoms. So the rubber is negatively charged and the cloth is positively charged. or a comb through hair

5 Ben also knew that a charge separation occurs when a glass rod is rubbed with a silk cloth In the case of the glass and silk, the glass rod loses negative charge and becomes positively charged while the silk cloth gains negative charge and therefore becomes negatively charged.

6 Ben experimented with the interactions between the charge objects. He suspended one and brought other charged objects near… Ben observed that like charged object repel and unlike charges attract repel attract

7 CONSERVATION OF ELECTRIC CHARGE In the process of rubbing two solid objects together, electrical charges are not created. Instead, both objects contain both positive and negative charges. During the rubbing process, the negative charge is transferred from one object to the other leaving one object with an excess of positive charge and the other with an excess of negative charge. The quantity of excess charge on each object is exactly the same. A NEGATIVE charge is an EXCESS of electrons A POSITIVE charge is a SHORTAGE of electrons

8 An electroscope is a device that detects static charge. The metal leaves of the electroscope move apart if a charged object is brought near the knob. Benjamin Franklin used a similar device when he investigated charges. Positively charged Negatively charged

9 The SI unit of charge is the coulomb (C). 1 C = 6.25 x electrons or protons The charge carried by the electron is represented by the symbol -e, and the charge carried by the proton is +e. A third particle, which carries no electrical charge, is the neutron. The charge of a single electron = -1.6 x C The charge of a single proton = +1.6 x C

10 insulator An insulator is a material in which the electrons are tightly held by the nucleus and are not free to move through the material. There is no such thing as a perfect insulator, however examples of good insulators are: glass, rubber, plastic and dry wood. A material can be an electric … conductor A conductor is a material through which electrons are free to move through the material. Just as in the case of the insulators, there is no perfect conductor. Examples of good conductors include metals, such as silver, copper, gold and mercury. semiconductor A semiconductor is a poor conductor of electricity at normal temperatures. As the temperature rises, electrons break free and move through the material. As a result, the ability of a semiconductor to conduct improves with temperature.

11 Objects become charged by… Friction Induction Conduction Electrons are rubbed off one insulator onto another insulator Charging an object WITHOUT touching a charged object Charging by CONTACT with a charged object

12 Charging by Contact Some electrons leave rod and spread over sphere. Charging by conduction results in an object with the same charge

13 Neutral objects can be temporarily attracted to charged objects by a process called POLARIZATON. To understand the process of charging by Induction

14 Electrons are free to move in metals. Nuclei remain in place; electrons move to bottom Polarization occurs because the electrons are attracted or repelled by the charged object. This results in a polarization or temporary separation of the charge, and attraction results.

15 Charging by Induction The rod does not touch the sphere. It pushes electrons out of the back side of the sphere and down the wire to ground. The ground wire is disconnected to prevent the return of the electrons from ground, then the rod is removed. The charge on the object is opposite polarizationgrounding permanent charge

16 Grounding is allowing charges to move freely along a connection between a conductor and the ground. The Earth (the ground) is a practically infinite reservoir of electric charge. Here a positively charge rod attracts electrons from the ground into the electroscope Here a negatively charge rod repels electrons into the ground from the electroscope

17 To review… Induction results in an OPPOSITE CHARGE Conduction results in the SAME CHARGE

18 Applications of Electrostatic Charging Fine mist of negatively charged gold particles adhere to positively charged protein on fingerprint. Negatively charged paint adheres to positively charged metal.

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20 COULOMB’S LAW Coulomb’s Law states that two point charges exert a force (F) on one another that is directly proportional to the product of the magnitudes of the charges (q) and inversely proportional to the square of the distance (r) or (d) between their centers. The equation is: F = electrostatic force (N) q = charge (C) k = 9x10 9 N m 2 /C 2 r or d = separation between charges (m)

21 Two negatively charged balloons are 0.70m apart. If the charge of each is 2.0 x C, What is the electric force between the two balloons? q 1 = q 2 = 2.0 x C d = r = 0.70 m F = 9.0 x 10 9 N m 2 /C 2 (2.0 x C) 2 (0.70m) 2 F = N An attracting or repelling force?

22 Two equally charged balloons repel each other with a force of 4.0 x N. If they are m apart, what is the charge of the each balloon? F = 4.0 x N d = m q 2 = (4x10 -3 N)(0.015m) 2 (9x10 9 Nm 2 /C 2 ) q 2 = Fd 2 k q 1 = q 2 = 1.0 x C

23 WORK on PSE problems # 1-5 for next 15 mins. Due EOC today!!

24 Chapter 33 in Textbook PSE Chapter 15 page 200

25 ELECTRIC FIELD An electric field is said to exist in a region of space in which an electric charge will experience an electric force. The magnitude of the electric field intensity is given by: Units: N/C

26 The direction of the electric field intensity at a point in space is the same as the direction in which a positive charge would move if it were placed at that point. The electric field lines or lines of force indicate the direction.electric field intensity Electric field line flow Out of positive charges and into Negative charges. + -Q-Q

27 The electric field is strongest in regions where the lines are close together and weak when the lines are further apart.electric field

28 Threads floating on oil bath become polarized and align themselves with the electric field. These fields can be detected in lab…

29 The electric field intensity E at a distance r from a single charge q can be found as follows: Units: N/C

30 Example What is the electric field intensity at a distance of 2 m from a charge of -12 μC? r = 2 m q = -12 μC = 2.7x10 4 N/C, towards q or to the left q = -12μC

31 When more than one charge contributes to the field, the resultant field is the vector sum of the contributions from each charge. Units: N/C Note we will look at direction of the field to know whether fields add or subtract at a point.

32 Two charges q 1 =-8 μC and q 2 =+12 μC are placed 120 mm apart in the air. What is the electric field at the midpoint between them? q 1 = -8 μC q 2 = +12 μC r = m -q1-q1 +q2+q2 E1E1 E2E2 ETET X = k q 1 + kq 2 r 2 r 2 = (9 x 10 9 )(8 x ) + (9 x 10 9 )(12 x ) (0.06) 2 (0.06) 2 The fields have the same direction so they add E= 2.0 x x 10 7 = 5.0 x 10 7 N/C to the left

33 Two charges q 1 =+8 μC and q 2 =+12 μC are placed 120 mm apart in the air. What is the electric field at the midpoint between them? q 1 = + 8 μC q 2 = +12 μC r = m +q1+q1 +q2+q2 E1E1 E2E2 ETET X = k q 1 - kq 2 r 2 r 2 = (9 x 10 9 )(8 x ) - (9 x 10 9 )(12 x ) (0.06) 2 (0.06) 2 The fields are in opposite directions so they subtract E= 2.0 x x 10 7 = -1.0 x 10 7 N/C E = 1.0 x 10 7 N/C to the left

34 Work on the PSE Problems # 6-9 for the remainder of class. Due BOC W.26/R.27!!


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