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Physics 208 Exam Review I 03/10/2006 1 Chapters 23-28 • • • • • • • • • • Coulomb force between charges Work done to move a charge Electric potential as U/q Electric field as F/q Motion of charged particle in electric field Relation between electric field and electric potential Electric field lines and equipotentials Electric fields in and near conductors Electric flux and Gauss' law Electric field and potential of a point charge, line charge, sheet charge, spherical shell • Capacitance of an isolated conductor, and of a pair of conductors • Calculating capacitance, parallel plate, cylinder, sphere • Energy in a capacitor, energy density, energy in E-field • Dielectrics • Current and resistance, Ohms law • Power dissipation in a resistor •03/10/2006 Simple circuits (parallel and series) and RC circuits 2 Charging processes •Triboelectric: -glass rod rubbed with silk positive -rubber rod rubbed with fur negative •Induction •Conduction +- + +++ + ++ + +++++ - -+ +- + electron flow + + + + ++ + + + + + + Less positively charged rod 03/10/2006 Positively charged metal 3 Quiz on Charge Conservation If you rub an inflated balloon against your hair, the two materials attract each other, as shown in this figure. Fill in the blank: the amount of charge present in the system of the balloon and your hair after rubbing is _____ the amount of charge present before rubbing. (a) less than (b) the same as (c) more than 03/10/2006 4 Quiz on charge sign 3 pithballs suspended from thin threads are charged by touching them with a charged object. It is found that pithballs 1 and 2 repel each other and pithballs 2 and 3 repel each other. Which of the following statements are correct? 1) 1 and 3 carry opposite charge 2) All of them carry charges of the same sign 3) We need to do more experiments to determine the sign of the charges 1 + 03/10/2006 2 3 + + 1 - 2 3 - - 1 + 2 3 + 5 Quiz on charge sign cont. 3 objects are brought close to each other, 2 at a time. When objects A and B are brought together, they attract. When objects B and C are brought together, they repel. From this, we conclude that: (a) objects A and C possess charges of opposite sign. (b) all three of the objects possess charges of the same sign. (c) we need to perform additional experiments to determine information about the charges on the objects. A B C A B C + - - - + + 03/10/2006 6 Answer Answer: (c). In the first experiment, objects A and B may have charges with opposite signs, or one of the objects may be neutral since they can attract one the other even due to induction. The second experiment shows that B and C have charges with the same signs, so that B must be charged. But we still do not know if A is charged or neutral. 03/10/2006 7 Insulators and conductors One of these isolated charged spheres is copper and the other is rubber. Label which is which and support your answer with an explanation. QuickTime™ and a TIFF (Uncompressed) decompressor are needed to see this picture. a) A is rubber and B is copper. b) A is copper and B rubber c) They are both copper Charged rubber rods are placed near a neutral conducting sphere, causing a redistribution of charge on the spheres. Which of the diagrams below depict the proper distribution of charge on the spheres? List all that apply. QuickTime™ and a TIFF (Uncompressed) decompressor are needed to see this picture. a) A and D b) C and B c) E 03/10/2006 8 Coulomb’s Law • Electrical force between two stationary charged particles • The SI unit of charge is the coulomb (C ), µC = 10-6 C • 1 C corresponds to 6.24 x 1018 electrons or protons • ke = Coulomb constant ≈ 9 x 109 N.m2/C2 = 1/(4πeo) eo permittivity of free space = 8.854 x 10-12 C2 / N.m2 03/10/2006 9 Electric Force 03/10/2006 q1q2 Fe k 2 r m1m2 Fg G 2 r 10 Electric Force magnitude The electric force between 2 like charged objects is F=0.02 N. If the charge of one of the 2 objects is halved and the distance separating the 2 objects is doubled, what is the new force? Q1Q2 Q1Q2 F 3 F k 2 F' k 2.5 10 N 2 r 2(2r) 8 03/10/2006 11 Electric Force: Newton’s 3rd law 03/10/2006 12 Forces as vectors 2 small beads with positive charges +3q and +q are fixed at the opposite ends of an horizontal rod. A 3rd small charged bead is free to slide on the rod. a) Does the result depend on the sign of the charge on the bead? b) Find its equilibrium position. QuickTime™ and a TIFF (Uncompressed) decompressor are needed to see this picture. F2 QuickTime™ and a TIFF (Uncompressed) decompressor are needed to see this picture. + F1 d2 d1 1 2 F1 QuickTime™ and a TIFF (Uncompressed) decompressor are needed to see this picture. 03/10/2006 a) The result is independent on the sign of the charge of the bead b) Equations to solve the problem (q’ charge of the bead): - F2 3qq' qq' F1 F2 k 2 k 2 3 3 d d 0.63d d1 d2 1 2 d d1 d2 13 Electrostatic Potential of 2 like charges Electrostatic potential V (r) k 03/10/2006 Q (point charge Q) r 14 E field, potential difference and energy conservation An electron initially travels horizontally in a uniform electric field pointing upwards with 3 possible directions of the initial velocity of the same magnitude. Which of the following relation is correct? a) vb > v > va b) va > v > vb c) vb = va > v d) vb = v = va Answer: a) the greatest change x --------------------- L in potential V occurs for the case + with the max vertical deflection from the initial position. This occurs when the electron is projected at an angle below the horizontal. K = -U=-qV 03/10/2006 b + + + + + + + + + + + + + + ++ a dV E dx b b dV E dx V V a a Vb EL a 15 Example E Fields and Potential What is the electric field at point B? q qd1 cos 2k d2 d3 12 106 9 18 10 2 3 5.18 10 3 N /C 2 3/2 (3 4 ) E 2k What is the electric field at point A? y B x d2=4 m d +12 mC -12 mC - d1=3 m 3m + A 3m 1 1 3 q E kq 2 k 2 2 d1 (2d1) 4 d1 Calculate the force on a Q=-3mC charge placed at point A: F=QE 03/10/2006 16 Example E Potential B Calculate the electric potential at B q q VB k 0 d d Calculate the electric potential at A q q q VA k k d1 2d1 2d1 y x d -12 mC d2=4 m - +12 mC + d1=3 m 3 m A 3m Calculate the work YOU must do to move a Q=+5 mC charge from A to B. 03/10/2006 qQ W UA UB QVA VB k 2d1 17 Gauss’ Law 03/10/2006 18 Electric Flux 03/10/2006 19 Plane charge density The electric charge density for plate 1 is - and for plate 2 is +The magnitude of the electric field associated with plate 1 is: /(2e0) and the filed lines are shown in the figure. When they are placed parallel one to the other, the Magnitude of the field is: 1. /e0 between and ± /(2e0) outside 2. /e0 between and 0 outside 3. None of the above 03/10/2006 20 Insulator and Conductor sphere A solid nonconductive spherical ball of diameter d = 2R = 1.0 cm is uniformly charged such that is has an electric field of 20 kV/m at its surface. 1. How much charge is on the ball? Chose gauss surface with r = R E E dA e Q E 4 R 2 0 3 ER 2 20 10 0.5 10 Q k 9 10 9 Q e0 R 2 2 5.55 1011C 2. What is the magnitude of the electric field halfway to the center of the ball? kQ' inside E(r) 2 rR r Q Q' r3 Q' Q 3 4 R 3 4 r 3 R 3 3 the insulating shell E(r) k uniform charge distribution Qr R3 R kQ E E' 2 10kV /m 2 2R 2 3. If the ball were made of a conductive material, how would the answers to 1) and 2) change? 1. Same (all charge on surface) 2. E= 0 (field inside a conductor is 0) 03/10/2006 r 21 Capacitors • C Q/V • Parallel Plate C = e0A/d • Adding – Series • Veq=V1+V2 • Qeq = Q1=Q2 1 1 1 Ceq C1 C2 – Parallel • Veq=V1=V2 • Qeq = Q1+Q2 03/10/2006 Ceq C1 C2 22 Energy stored in a parallel plate capacitor The parallel plates of a capacitor with vacuum between them are given equal and opposite charge. If the plates are pushed closer from a distance d to D < d, does the energy stored in the capacitor 1) Increase 2) Decrease 3) Remain equal Q2 U 2C e0 A e0 A Dd Ci Cf C f Ci U f U i d D 03/10/2006 23 Metal slab in a capacitor In the slab E = 0 and we have 2 equivalent capacitors in series 1 1 1 d 2 Ceq C C 3e0 A In the other case the capacitance between the 2 bottom plates is (for the other 3e0 A 2 plates at the same potential 1/C 0) C 03/10/2006 eq d 24 Capacitor with slab of dielectric A dielectric slab is introduced between the plates of a capacitor. The capacitor is charged and then the dielectric is removed. Which of the following statements are correct: 1) The stored energy increases 2) The capacitance increases 3) The potential difference decreases C0 03/10/2006 C V0 V 0 = quantities in vacuum >1 Q2 U 2C 25 Calculate the equivalent Capacitance C1 = 10 mF C2 = 20 mF C3 = 30 mF C4 = 40 mF V = 50 Volts 1 1 1 1 Ceq 6.9mF Ceq C1 C2 C3 C4 C1 Compare V1 with V4 V V1 V23 V4 V23 V2 V3 Q Q1 Q23 Q4 Q23 Q2 Q3 Q Q Q Q V Ceq C1 C2 C3 C4 C1<C4 V1>V4 03/10/2006 C2 V C3 C4 parallel C1, C23, C4 in series 26 Resistors • • • • R = V/I R= L/A P = RI2 Adding – Series • Veq=V1+V2 • Ieq = I1=I2 – Parallel • Veq=V1=V2 • Ieq = I1+I2 03/10/2006 Req R1 R2 1 1 1 Req R1 R2 27 Resistors in Series 03/10/2006 28 Resistors in Parallel 1 1 Req R 03/10/2006 29 Light bulbs in Series Consider the resistance of the wire negligible Since the wire has R =0 all current will flow through it and bulb B will fade. Instead A will look brigther since the current in A increaseas from I = V/(RA+RB) to I = V/RA 03/10/2006 30 Light Bulbs of Different Power in Parallel A 100 W B 60 W Which of the following statements are correct? 1. A looks dimmer than B 2. A looks as bright as B 3. A looks brighter than B RA < RB more current flows in A: IA>IB RAIA2 > RBIB2 What happens if A and B are connected in series? 03/10/2006 31 Resistors in series and parallel 03/10/2006 32 Kirchhoff’s Rules I1 = I2 + I3 • Junction Rule: SIin = SIout • A statement of Conservation of Charge • Loop Rule: e closed loop k V i closed loop • A statement of Conservation of Energy 03/10/2006 - 33 Kirchhoff’s laws R1 = 10 W R2 = 20 W R3 = 30 W V1 = 50 Volts V2 = 10 Volts 2 loops Chose 2 currents and their verse V1 R1I1 R2 (I1 I2 ) R3 I2 V2 R3 I2 R2 (I2 I1 ) (R2 R3 )I2 V2 I1 2.45A R2 V2 I1 + - V1 -+ R1 I2 R2 R3 R2V1 (R1 R2 )V2 I2 1.18A R1(R2 R3 ) R2 R3 Calculate total power delivered by the batteries Ptot V1I1 V2 I2 134.5W PR,tot R1I12 R2 (I1 I2 ) 2 R3 I32 134.5W 03/10/2006 34 RC Circuits: CHARGE OF C Calculate current immediately after switch is closed. I = e/R = 2 A R C e S1 Calculate current after switch has been closed for 2 time constants e R=10W, C=30 mF =RC = 0.3 s I(t 2 ) e 0.27A and e=20 Volts R dq q Calculate current after switch has been closed eR dt C for a very long time I(t ) 0 2 charge on capacitor after switch Calculate has been closed for a long time 03/10/2006 q(t ) Ce 0.6C q(t) Ce(1 et / RC ) I(t) e R et / RC 35 RC Circuits You are given a 5-pack of R=2W resistors and a 5-pack of C=2mF capacitors. How do you configure them to produce a circuit of 21 ms time constant RC circuit? Time constant, = RC = 21 ms = 7 * 3 ms = 7 W * 3 mF 03/10/2006 R 7 3R R 3W 2 2 C 3 C C 2 2 36