Chapter 16 Lecture

Chapter 16 Lecture - Chapter 16 Electric Charge and...

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Unformatted text preview: Chapter 16 Electric Charge and Electric Field Units of Chapter 16 • Static Electricity; Electric Charge and Its Conservation • Electric Charge in the Atom • Insulators and Conductors • Induced Charge; the Electroscope Charge; the Electroscope • Coulomb’s Law • Solving Problems Involving Coulomb’s Law and Vectors • The Electric Field • Field Lines •Electric Fields and Conductors 16.1 Static Electricity; Electric Charge and Its Conservation and Its Conservation Objects can be charged by rubbing can be charged by rubbing 16.1 Static Electricity; Electric Charge and Its Electric Charge and Its Conservation Charge comes in two types positive and types, positive and negative; like charges repel and opposite repel and opposite charges attract 16.1 Static Electricity; Electric Charge and Its Conservation and Its Conservation Electric charge is conserved – the arithmetic sum of the total charge cannot change in any interaction. 16.2 Electric Charge in the Atom Atom: Nucleus (small, massive, positive charge) charge) Electron cloud (large, very low density, negative charge) 16.2 Electric Charge in the Atom Atom is electrically neutral. Rubbing charges objects by moving electrons from one to the other. 16.2 Electric Charge in the Atom Polar molecule: neutral overall, but charge not evenly distributed evenly distributed 16.3 Insulators and Conductors Conductor: Insulator: Charge flows freely Almost no charge flows Metals Most other materials Some materials are semiconductors. 16.4 Induced Charge; the Electroscope Metal objects can be charged by conduction: 16.4 Induced Charge; the Electroscope They can also be charged by induction: 16.4 Induced Charge; the Electroscope Nonconductors won’t become charged by conduction or induction but will experience conduction or induction, but will experience charge separation: 16.4 Induced Charge; the Electroscope The electroscope can be used for can be used for detecting charge: 16.4 Induced Charge; the Electroscope The electroscope can be charged either by conduction or by induction. 16.4 Induced Charge; the Electroscope The charged electroscope can then be used to determine the sign of an unknown charge. determine the sign of an unknown charge. ConcepTest ConcepTest 16.1a Electric Charge I Two charged balls are repelling each other as they hang from the ceiling. What can you say about their charges? 1) one is positive, the other is positive the other is negative 2) both are positive 3) both are negative 4) both are positive or both are negative ConcepTest ConcepTest 16.1a Electric Charge I Two charged balls are repelling each other as they hang from the ceiling. What can you say about their charges? 1) one is positive, the other is positive the other is negative 2) both are positive 3) both are negative 4) both are positive or both are negative The fact that the balls repel each ba eac other only can tell you that they have the same charge but you do charge, not know the sign So they can not know the sign. So they can be either both positive or both negative. FollowFollow-up: What does the picture look like if the two balls are oppositely charged? ConcepTest 16.1b Electric Charge II From the picture, what can you conclude about the charges? 1) 1) have opposite charges 2) have the same charge 3) all have the same charge 4) one ball must be neutral (no charge) ConcepTest 16.1b Electric Charge II From the picture, what can you conclude about the charges? 1) 1) have opposite charges 2) have the same charge 3) all have the same charge 4) one ball must be neutral (no charge) The GREEN and PINK balls must have the same charge, since they repel each other The YELLOW repel each other. The YELLOW ball also repels the GREEN, so it must also have the same charge as the GREEN (and the PINK). ConcepTest ConcepTest 16.2a Conductors I A metal ball hangs from the ceiling 1) positive by an insulating thread. The ball is 2) negative attracted attracted to a positive-charged rod positive rod 3) neutral neutral held near the ball. The charge of 4) positive or neutral the ball must be: the ball must be: 5) negative or neutral negative or neutral ConcepTest 16.2a Conductors I A metal ball hangs from the ceiling 1) 1) positive by an insulating thread. The ball is 2) negative attracted attracted to a positive-charged rod positive rod 3) neutral neutral held near the ball. The charge of 4) positive or neutral the ball must be: the ball must be: 5) negative or neutral negative or neutral Clearly, the ball will be attracted if its the ball will be attracted if its charge is negative However, even if negative. the ball is neutral the charges in the neutral, ball can be separated by induction (polarization), leading to a net attraction. tt remember the ball is the ball is a conductor! FollowFollow-up: What happens if the metal ball is replaced by a plastic ball? ball? ConcepTest 16.2b Conductors II Two neutral conductors are connected Two neutral conductors are connected 1) 0 0 by a wire and a charged rod is brought 2) + – 3) – + 4) + + 5) – – near, but does not touch. The wire is taken away, and then the charged rod is removed. What are the charges on the conductors? the conductors? 0 0 ? ? ConcepTest 16.2b Conductors II Two neutral conductors are connected Two neutral conductors are connected 1) 0 0 by a wire and a charged rod is brought 2) + – 3) – + 4) + + 5) – – near, but does not touch. The wire is taken away, and then the charged rod is removed. What are the charges on the conductors? conductors? While the conductors are connected, positive 0 0 ? ? charge will flow from the blue to the green ball due to polarization Once disconnected, polarization. th the charges will remain on the separate th conductors even when the rod is removed. FollowFollow-up: What will happen when the conductors are reconnected with a wire? 16.5 Coulomb’s Law Experiment shows that the electric force between two charges is proportional to the between two charges is proportional to the product of the charges and inversely proportional to the distance between them. 16.5 Coulomb’s Law Coulomb’s law: (16-1) This equation gives the magnitude of the force. the force. 16.5 Coulomb’s Law The force is along the line connecting the li charges, and is attractive if the charges are opposite and repulsive if they are the same opposite, and repulsive if they are the same. 16.5 Coulomb’s Law Unit of charge: coulomb, C The proportionality constant in Coulomb’s law is then: Charges produced by rubbing are typically around a microcoulomb: 16.5 Coulomb’s Law Charge on the electron: Electric charge is quantized in units charge is quantized in units of the electron charge. 16.5 Coulomb’s Law The proportionality constant k can also be written in terms of written in terms of , the permittivity of free the permittivity of free space: (16-2) ConcepTest 16.3a Coulomb’s Law I If F1 = 3N, what is the what magnitude of the force F2? 1) 1) 1.0 N 2) 1.5 N 3) 2.0 N F1 = 3N Q Q F2 = ? 4) 3.0 N 5) 6.0 N ConcepTest 16.3a Coulomb’s Law I If F1 = 3N, what is the what magnitude of the force F2? 1) 1) 1.0 N 2) 1.5 N 3) 2.0 N F1 = 3N Q Q F2 = ? 4) 3.0 N 5) 6.0 N The force F2 must have the same magnitude as F1. This is same due to the fact that the form of Coulomb’s Law is totally symmetric with respect to the two charges involved. The force of one on the other of a pair is the same as the reverse reverse. Note that this sounds suspiciously like Newton’s 3rd Law!! Note that this sounds suspiciously like Newton’s 3rd Law!! ConcepTest ConcepTest 16.3b Coulomb’s Law II F1 = 3N 3N Q Q F2 = ? 2) 3.0 N If we increase one charge to 4Q If we increase one charge to 4Q, what what is the magnitude of F1? F1 = ? 4Q Q 1) 3/4 N F2 = ? 3) 12 N 12 4) 16 N 5) 48 N 48 ConcepTest ConcepTest 16.3b Coulomb’s Law II F1 = 3N 3N Q Q F2 = ? 2) 3.0 N If we increase one charge to 4Q If we increase one charge to 4Q, what what is the magnitude of F1? F1 = ? 4Q Q 1) 3/4 N F2 = ? 3) 12 N 12 4) 16 N 5) 48 N 48 Originally we had: F1 = k(Q)(Q)/r2 = 3 N Now we have: F1 = k(4Q)(Q)/r2 which is 4 times bigger than before. is times bigger before FollowFollow-up: Now what is the magnitude of F2? MCAT MCAT Test 16-1 16The force between two charges force between two charges 1) 9 F separated by a distance d is F. If 2) 3 F the charges are pulled apart to a 3) 3) F distance 3d, what is the force on 4) 1/3 F each charge? 5) 1/9 F 1/9 F F Q Q d ? ? Q Q 3d ConcepTest 16.3c Coulomb’s Law III The force between two charges force between two charges 1) 1) 9 F separated by a distance d is F. If 2) 3 F the charges are pulled apart to a 3) 3) F distance 3d, what is the force on 4) 1/3 F each charge? 5) 1/9 F 1/9 F Originally we had: F Q Q Fbefore = k(Q)(Q)/d2 = F Now we have: Fafter = k(Q)(Q)/(3d)2 = 1/9 F d ? ? Q Q 3d FollowFollow-up: What is the force if the original distance is halved? 16.5 Coulomb’s Law Coulomb’s law strictly applies only to point charges. Superposition: for multiple point charges, the forces on each charge from every other charge can be on each charge from every other charge can be calculated and then added as vectors. 16.6 Solving Problems Involving Coulomb Law and Vectors Coulomb’s Law and Vectors The net force on a charge is the vector sum of all the forces acting on it. 16.6 Solving Problems Involving Coulomb’s Law and Vectors Vector addition review: Particles of charge +75, +48, and -85 μC are placed in a line (Fig. 16–49). The center one is 0.35 m from each of the others. Calculate the from each of the others Calculate the net force on each charge due to the other two. Let the right be the positive direction on the line of charges. Use the fact that like charges repel and unlike charges attract to determine the di direction of the forces. In the following th th expressions, k = 8.988 ×10 N ⋅ m C . 9 2 2 ( 75μ C ) ( 48μ C ) ( 75μ C ) ( 85μ C ) F+75 = −k +k = −147.2 N ≈ −1.5 ×102 N 2 2 0.35 m ) 0.70 m ) ( ( ( 75μ C ) ( 48μ C ) ( 48μ C ) ( 85μ C ) +k = 563.5 N ≈ 5.6 ×102 N F+48 = k 2 2 0.35 m ) 0.35 m ) ( ( ( 85μ C ) ( 75μ C ) ( 85μ C ) ( 48μ C ) −k = −416.3 N ≈ −4.2 ×102 N F−85 = −k ( 0.70 m )2 ( 0.35 m )2 1 2 ConcepTest 16.6 Forces in 2D 3 Which of the arrows best 4 represents the direction of the net force on charge of the net force on charge d +2Q +Q +Q due to the other two charges? d +4Q 5 1 2 ConcepTest 16.6 Forces in 2D 3 Which of the arrows best 4 represents the direction of the net force on charge of the net force on charge d +2Q +Q 5 +Q due to the other two d charges? +4Q The charge +2Q repels +Q towards charge +2 the right. The charge +4Q repels +Q upwards, but with a stronger force. but with stronger force. Therefore, the net force is up and to +2Q the right, but mostly up up. FollowFollow-up: What happens if the yellow charge would be +3Q? +4Q 16.7 The Electric Field The electric field is the force on small charge, force on a small charge, divided by the charge: (16-3) 16.7 The Electric Field For a point charge: (16-4a) (16-4b) 16.7 The Electric Field Force on a point charge in an electric field: (16-5) Superposition principle for electric fields: 16.7 The Electric Field Problem solving in electrostatics: electric forces and electric fields forces and electric fields 1. Draw a diagram; show all charges, with signs and electric fields and forces with signs, and electric fields and forces with directions 2. Calculate forces using Coulomb’s law forces using Coulomb’s law 3. Add forces vectorially to get result ConcepTest 16.7 Electric Field You are sitting certain distance from You are sitting a certain distance from a point charge, and you measure an point electric field of E0. If the charge is doubled and your distance from the di th charge is also doubled, what is the doubled, electric field strength now? (1) (1) 4 E0 (2) 2 E0 (3) E0 (4) 1/2 E0 (5) 1/4 E0 4 ConcepTest 16.7 Electric Field You are sitting certain distance from You are sitting a certain distance from a point charge, and you measure an point electric field of E0. If the charge is doubled and your distance from the di th charge is also doubled, what is the doubled, electric field strength now? (1) (1) 4 E0 (2) 2 E0 (3) E0 (4) 1/2 E0 (5) 1/4 E0 4 Remember that the electric field is: E = kQ/r2. that the electric field is: kQ/r Doubling the charge puts a factor of 2 in the numerator, but doubling the distance puts a factor of of 4 in the denominator, because it is distance squared!! Overall, that gives us a factor of 1/2 1/2. FollowFollow-up: If your distance is doubled, what must you do to the charge to maintain the same E field at your new position? ConcepTest 16.8a Field and Force I Between the red and the the red the blue charge, which of them experiences the greater electric field due to the green charge? +2 +1 d 1) 1) +1 2) +2 3) the same for both +1 +1 d ConcepTest 16.8a Field and Force I Between the red and the the red the blue charge, which of them experiences the greater electric field due to the green charge? 1) 1) +1 2) +2 3) the same for both +2 +1 d Both charges feel the same electric field due to the green charge because they are at the same point in space space! +1 +1 d Q E=k 2 r ConcepTest 16.8b Field and Force II Between the red and the the red the blue charge, which of them experiences the greater electric force due to the green charge? +2 +1 d 1) 1) +1 2) +2 3) the same for both +1 +1 d ConcepTest 16.8b Field and Force II Field Between the red and the the red the blue charge, which of them experiences the greater electric force due to the green charge? +2 +1 1) 1) +1 2) +2 3) the same for both +1 d The electric field is the same for both charges, but the force on a given charge also depends on the magnitude of that specific charge charge. +1 d F = qE ConcepTest 16.9a Superposition I -2 C 2 What is the electric field at 1 the center of the square? 3 5) E = 0 -2 C 4 ConcepTest 16.9a Superposition I -2 C 2 White dot = -1 C 1 What is the electric field at the 3 center of the square? 5) E = 0 -2 C For the upper charge, the E field vector at the center the upper charge the vector at the center of the square points towards that charge. For the lower charge, the same thing is true. Then the vector sum of these two E field vectors points to the left left. FollowFollow-up: What if the lower charge was +2 C? What if both charges were +2 C? 4 ConcepTest 16.9b Superposition II -2 C -2 C 2 What is the electric field at 1 the center of the square? 3 5) E = 0 -2 C 4 -2 C ConcepTest 16.9b Superposition II -2 C -2 C 2 What is the electric field at 1 the center of the square? 3 5) E = 0 -2 C The four E field vectors all point outwards from the center of the square toward their respective charges. Because they are all respective charges. Because they are all equal, the net E field is zero at the center center!! FollowFollow-up: What if the upper two charges were +2 C? What if the right-hand charges were +2 C? right- 4 -2 C ConcepTest 16.11 Uniform Electric Field In a uniform electric field in empty 1) 12 N space, a 4 C charge is placed and it 2) 8 N feels an electrical force of 12 N. If 3) 24 3) 24 N this charge is removed and a 6 C charge is placed at that point instead, what force will it feel? it Q 4) no force 5) 18 N 18 ConcepTest 16.11 Uniform Electric Field In a uniform electric field in empty 1) 1) 12 N space, a 4 C charge is placed and it 2) 8 N feels an electrical force of 12 N. If 3) 24 N 24 this charge is removed and a 6 C charge is placed at that point instead, what force will it feel? it 4) no force 5) 18 N 18 Since the 4 C charge feels a force, there must the charge feels force, there must be an electric field present, with magnitude: E = F / q = 12 N / 4 C = 3 N/C 12 Once the 4 C charge is replaced with a 6 C Q charge, this new charge will feel a force of: F = q E = (6 C)(3 N/C) = 18 N FollowFollow-up: What if the charge is placed at a different position in the field? different 16.8 Field Lines The electric field can be represented by field lines. These lines start on a positive charge and end on a negative charge. 16.8 Field Lines The number of field lines starting (ending) on a positive (negative) charge is proportional to the magnitude of the charge proportional to the magnitude of the charge. The electric field is stronger where the field lines are closer together. 16.8 Field Lines Electric dipole: two equal charges, opposite in sign: sign: 16.8 Field Lines The electric field between two closely spaced, oppositely charged parallel plates is constant plates is constant. MCAT Test 16-1 16An electron is located between An electron is located between a pair of oppositely charged pair parallel plates. As the electron parallel plates. As the electron approaches approaches the positively charged plate, the kinetic energy energy of the electron? - A: Increases B: Decreases C: Remains the same + + + + + + 16.8 Field Lines Summary of field lines: 1. Field lines indicate the direction of the lines indicate the direction of the field; the field is tangent to the line. 2. The magnitude of the field is proportional magnitude of the field is proportional to the density of the lines. 3. Field lines start on positive charges and li end on negative charges; the number is proportional to the magnitude of the proportional to the magnitude of the charge. 16.9 Electric Fields and Conductors The static electric field inside a conductor is zero – if it were not, the charges would move. The net charge on a conductor is on its surface surface. 16.9 Electric Fields and Conductors The electric field is perpendicular to the perpendicular to the surface of a conductor – again, if it were not, charges would move. What are the magnitude and direction of the electric field at a point midway between a -8.0 μC and a +7.0 μC charge 8.0 cm apart? Assume cm apart? Assume no other charges are nearby. +7.0 -8.0 8cm What are the magnitude and direction of the electric field at a point midway between a -8.0 μC and a +7.0 μC charge 8.0 cm apart? Assume cm apart? Assume no other charges are nearby. +7.0 -8.0 8cm The electric field due to the negative charge will point toward the negative charge, and the electric field due to the positive charge will point away from the positive charge. Thus both fields point in the same direction, towards the negative charge and so can be added negative charge, and so can be added. The direction is towards the negative charge . E = E1 + E2 = k = ( Q1 2 1 r +k Q2 r 4 8.988 ×109 N ⋅ m2 C2 (8.0 ×10 m) −2 2 =k 2 2 ) ( Q1 ( d / 2) 2 +k Q2 ( d / 2) 2 = ) 4k d 2 (Q 1 + Q2 ) 8.0 ×10−6 C + 7.0 ×10−6 C = 8.4 ×107 N C Summary of Chapter 16 • Two kinds of electric charge – positive and negative ti • Charge is conserved • Charge on electron: • Conductors: electrons free to move • Insulators: nonconductors Summary of Chapter 16 • Charge is quantized in units of e • Objects can be charged by conduction or induction • Coulomb’s law: • Electric field is force per unit charge: Summary of Chapter 16 • Electric field of a point charge: • Electric field can be represented by electric field lines field lines • Static electric field inside conductor is zero; surface field is perpendicular to surface surface field is perpendicular to surface Demonstration 16.1 One person should touch the ball on top of the electroscope with their finger and hold their finger th there, while another person brings a charged rubber rod near, but not touching the leaves of the electroscope. While the rubber rod is near the electroscope, the first person should remove their finger from the top of the electroscope. Then remove the rubber rod from near the electroscope the rubber rod from near the electroscope. Demonstration 16.1 One person should touch the ball on top of the electroscope with their finger and hold their finger there, while another person brings charged rubber there, while another person brings a charged rubber rod near, but not touching the leaves of the electroscope. Demonstration 16.1 One person should touch the ball on top of the electroscope with their finger and hold their finger th there, while another person brings a charged rubber rod near, but not touching the leaves of the electroscope. 1. While the rubber rod is near the electroscope, the first person should remove their finger from the top of the electroscope. Then remove the rubber rod from near the electroscope. 2. Remove the rubber rod first and then remove the th fi th th first person’s finger. ...
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This note was uploaded on 09/08/2011 for the course BIOL 1403 taught by Professor Dini during the Fall '09 term at Texas Tech.

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