2 ELECTROSTATICS – the study of electric charges, forces and fields Two types of charges exists, arbitrarily namedPOSITIVE and NEGATIVEBy Benjamin FranklinSo what is positive and what is negative?
3 We know that charged particles exist in atoms Electrons are responsible for negative chargeand Protons for positive chargeBenjamin 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.or a comb through hairThe rubber scrapes electrons from fur atoms. So the rubber is negatively charged and the cloth is positively charged.
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 like charged object repel and unlike charges attract Ben experimented with the interactions between the charge objects. He suspended one and brought other charged objects near…repelrepelattractBen observed thatlike charged object repeland unlike charges attract
7 A NEGATIVE charge is an EXCESS of electrons 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 electronsA POSITIVE charge is a SHORTAGE of electrons
8 An electroscope is a device that detects static charge. Negatively chargedPositively chargedThe 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.
9 The charge of a single electron = -1.6 x 10-19 C The SI unit of charge is the coulomb (C).1 C = 6.25 x 1018 electrons or protonsThe 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 singleelectron = -1.6 x CThe charge of a singleproton = +1.6 x C
10 A material can be an electric … 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.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… Electrons are rubbed off one insulator onto another insulatorFrictionCharging an object WITHOUT touching a charged objectInductionCharging by CONTACT with a charged objectConduction
12 Charging by conduction results in an object with the same charge Charging by Contact Some electrons leave rod and spread over sphere.Charging by conduction results in an object with the same charge
13 To understand the process of charging by Induction Neutral objects can be temporarily attracted to charged objects by a process called POLARIZATON.
14 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.Electrons are free to move in metals.Nuclei remain in place; electrons move to bottom
15 The charge on the object is opposite Charging by Induction permanent chargepolarizationgroundingThe 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
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 electroscopeHere a negatively charge rod repels electrons into the ground from the electroscope
17 To review…Induction results in an OPPOSITE CHARGEConduction 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.
20 F = electrostatic force (N) q = charge (C) k = 9x109 N m2/C2 COULOMB’S LAWCoulomb’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 = 9x109 N m2/C2r or d = separation between charges (m)
21 Two negatively charged balloons are 0.70m apart. If the charge of each is 2.0 x 10-6C, What is the electric force between the two balloons?q1 = q2 = 2.0 x 10-6 Cd = r = 0.70 mF = 9.0 x 10 9 N m2/C2 (2.0 x 10-6 C)2(0.70m)2F = NAn attracting or repelling force?
22 Two equally charged balloons repel each other with a force of 4 Two equally charged balloons repel each other with a force of 4.0 x 10-3 N. If they are m apart, what is the charge of the each balloon?F = 4.0 x 10-3 Nd = mq2 = Fd2kq2 = (4x10-3N)(0.015m)2(9x109Nm2/C2)q1 = q2 = 1.0 x 10-8C
23 Static Electricity WORK on PSE problems # 1-5 for next 15 mins. Due EOC today!!
24 Electric FieldsChapter 33 in TextbookPSE 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.-Q+Electric field line flow Out of positive charges and into Negative charges.
27 The electric field is strongest in regions where the lines are close together and weak when the lines are further apart.
28 These fields can be detected in lab… Threads floating on oil bath become polarized and align themselves with the electric field.
29 The electric field intensity E at a distance r from a single charge q can be found as follows: Units: N/C
30 = 2.7x104 N/C, towards q or to the left Example What is the electric field intensity at a distance of 2 m from a charge of -12 μC?r = 2 mq = -12 μCq = -12μC= 2.7x104 N/C, towards q or to the left
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/CNote we will look at direction of the field to know whether fields add or subtract at a point.
32 r2 r2 X = kq1 + kq2 = (9 x 109)(8 x 10-6) + (9 x 109)(12 x 10-6) Two charges q1=-8 μC and q2=+12 μC are placed 120 mm apart in the air. What is the electric field at the midpoint between them?The fields have the same directionso they addETq1 = -8 μCq2 = +12 μCr = mE1E2-q1X+q2= kq1 + kq2r r2= (9 x 109)(8 x 10-6) + (9 x 109)(12 x 10-6)(0.06) (0.06)2E= 2.0 x x 107 = 5.0 x 107 N/Cto the left
33 r2 r2 X = kq1 - kq2 = (9 x 109)(8 x 10-6) - (9 x 109)(12 x 10-6) Two charges q1=+8 μC and q2=+12 μC are placed 120 mm apart in the air. What is the electric field at the midpoint between them?The fields are in oppositedirections so they subtractETq1 = + 8 μCq2 = +12 μCr = mE2E1+q1X+q2= kq1 - kq2r r2= (9 x 109)(8 x 10-6) - (9 x 109)(12 x 10-6)(0.06) (0.06)2E= 2.0 x x 107 = -1.0 x 107 N/CE = 1.0 x 107 N/C to the left
34 Electric Fields Work on the PSE Problems # 6-9 for the remainder of class.Due BOC W.26/R.27!!