Presentation on theme: "ELECTROSTATICS: The study of the behavior of stationary charges"— Presentation transcript:
1 ELECTROSTATICS:The study of the behavior of stationary chargesELECTRIC CHARGEThere are two types of electric charge, arbitrarily calledpositive and negative. Rubbing certain electrically neutral objects together (e.g., a glass rod and a silk cloth) tends tocause the electric charges to separate. 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. After separation, the negative charges and positive charges are found to attract one another.
2 If the glass rod is suspended from a string and a second positively charged glass rod is brought near, a force of electrical repulsion results. Negatively charged objects also exert a repulsive force on one another.These results can be summarized as follows: unlike charges attract and like charges repel.
3 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.Electrons are free to move in metals.Nuclei remain in place; electrons move to bottom
4 The law of conservation of electric charge: "The net amount of electric charge produced in any process is zero." Another way of saying this is that in any process electric charge cannot be created or destroyed, however, it can be transferred from one object to another.Charged comb attracts neutral water molecules.Charged comb attracts neutral bits of paper.
5 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.e = 1.6 x Cmelectron = 9.11 x kgmproton = x kgmneutron = x kg.
6 Experiments performed early in this century have led to the conclusion that protons and neutrons are confined to the nucleus of the atom while the electrons exist outside of the nucleus. When solids are rubbed together, it is the electrons that are transferred from one object to the other. The positive charges, which are located in the nucleus, do not move.Rubber scrapes electrons from fur atoms.
7 INSULATORS AND CONDUCTORS 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 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.
8 A few materials, such as silicon, germanium and carbon, are called semiconductors. At ordinary temperature, there are a few free electrons and the material is a poor conductor of electricity. 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.
9 Charging by Contact Some electrons leave rod and spread over sphere.
10 Charging by Induction Rod does not touch 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.
11 INDUCED ELECTRIC CHARGE If a negatively charged rod is brought near an uncharged electrical conductor, the negative charges in the conductor travel to the far end of the conductor. The positive chargesare not free to move and a charge is temporarily induced atthe two ends of the conductor. Overall, the conductor is still electrically neutral and if the rod is removed a redistributionof the negative charge will occur.
12 If the metal conductor is touched by a person’s finger or a wire connected to ground, it is said to be grounded. The negative charges would flow from the conductor to ground. If the ground is removed and then the rod is removed, a permanent positive charge would be left on the conductor. The electrons would move until the excess positive charge was uniformly distributed over the conductor.
13 CHARGE DISTRIBUTIONS Charge on Metals Charge on Insulators Metal BallExcess charge on the surface of a uniform metal spreads out.Charge on InsulatorsPlastic Ball Charge on insulating materials doesn't move easily.Charge on Metal PointsExcess charge on a metal accumulates at points. Example: lightning rods.
14 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.
15 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) between their centers. The equation is:F = electrostatic force (N)q = charge (C)k = 9x109 N. m2/C2r = separation between charges (m)
16 The value of k can also be expressed in terms of the permittivity of free space (εo): 9x109 N. m2/C2The proportionality constant (k) can only be used if the medium that separates the charges is a vacuum. If the region between the point charges is not a vacuum then the value of the proportionality constant to be used is determined by dividing k by the dielectric constant (K).For a vacuum K = 1, for distilled water K = 80, and forwax paper K = 2.25
17 = 80 N = 120 N FR = 80 + 120 = 200 N, to the right 16.1 Two charges q1 = -8 μC and q2= +12 μC are placed 120mm apart in the air. What is the resultant force on a third charge q3 = -4 μC placed midway between the other charges?FRF2q1 = -8 μCq2= +12 μCq3 = -4 μCr = mF1-q1-q3+q2= 80 N= 120 NFR == 200 N, to the right
18 16.2 Three charges q1 = +4 nC, q2 = -6 nC and q3 = -8 nC are arranged as shown. Find the resultant force on q3due to the other two charges.F1FRq1 = +4 nCq2= -6 nCq3 = -8 nCF237˚θ= 2.88x10-5 N= 6.75x10-5 N
19 From the FBD: Σ Fy = F1 sin 37˚ Σ Fx = F2 - F1 cos 37˚ θΣ Fy = F1 sin 37˚= (2.88x10-5)(sin 37˚)= 1.73x10-5 NΣ Fx = F2 - F1 cos 37˚= (6.75x10-5) - (2.88x10-5)(cos 37˚)= 4.45x10-5 N= 4.8x10-5 NFR (4.8x10-5 N, 21˚)θ = 21˚
20 ELECTRIC FIELDAn electric field is said to exit 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
21 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. The electric field is strongest in regions where the lines are close together and weak when the lines are further apart.
23 F = qE = 1.6x10-19 (6x104) = 9.6x10-15 N, upward 16.3 The electric field intensity between two plates is constant and directed downward. The magnitude of the electric field intensity is 6x104 N/C. What are the magnitude and direction of the electric force exerted on an electron projected horizontally between the two plates?E = 6x104 N/Cqe = 1.6x10-19 CF = qE= 1.6x10-19 (6x104)= 9.6x10-15 N, upward
24 16.4 Show that the gravitational force on the electron of example 16-3 may be neglected. me = 9.1x10-31 kgFg = mg= 9.1x10-31 (9.8)= 8.92x10-30 NThe electric force is larger than the gravitational force by a factor of 1.08x1015!
25 The electric field intensity E at a distance r from a single charge q can be found as follows: Units: N/C
26 16.5 What is the electric field intensity at a distance of 2 m from a charge of -12 μC? r = 2 mq = -12 μC= 27x103 N/C, towards q
27 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
28 16.6 Two point charges q1 = -6 nC and q2 = +6 nC, are 12 cm apart, as shown in the figure. Determine the electric field a. At point Aq1 = -6 nCq2 = +6 nCE1ERE2= 3.38x104 N/C, left= 8.44x103 N/C, left
30 16.6 Two point charges q1 = -6 nC and q2 = +6 nC, are 12 cm apart, as shown in the figure. Determine the electric field b. At point BE2q1 = -6 nCq2 = +6 nC37ºθERE1= 6.67x103 N/C, down= 2.4x103 N/C at 37˚
31 From FBD Σ Ex = - E2cos 37˚ = - (2.4x103)(cos 37˚) = -1916.7 N/C 16.6 Two point charges q1 = -6 nC and q2 = +6 nC, are 12 cm apart, as shown in the figure. Determine the electric field b. At point BFrom FBDE237ºΣ Ex = - E2cos 37˚= - (2.4x103)(cos 37˚)= N/CθERE1Σ Fy = E2 sin 37˚- E1= (2.4x103)(sin 37˚) - (6.67x103)= N/C= N/C
32 16.6 Two point charges q1 = -6 nC and q2 = +6 nC, are 12 cm apart, as shown in the figure. Determine the electric field b. At point BE237ºθERE1= 70˚180˚ + 70˚ = 250˚ER (5566 N/C, 250˚)