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# UniversityPhysicsII PHY146

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This 166 page Class Notes was uploaded by Willard Grady on Monday October 5, 2015. The Class Notes belongs to PHY146 at Central Michigan University taught by Staff in Fall. Since its upload, it has received 98 views. For similar materials see /class/218952/phy146-central-michigan-university in Physics 2 at Central Michigan University.

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Date Created: 10/05/15

Lecture 23 Capacitance Chapter 25 Capacitors Capacitance Calculating the Capacitance ParallelPlate Capacitor Cylindrical Capacitor Spherical Capacitor Charged Isolated Conductor Excess charge on an isolated conductor will be located only on the surface Electric field is zero inside the conductor There is no charge on the cavity walls There is no component of the electric field tangent to the surface Capacitors Capacitors devices that can store energy as potential energy in an electric field Charge is accumulated in a capacitor and is released when the electric potential energy is to be used Capacitors very important for the electronic industry eg Random Access Memory RAM Capacitance A capacitor is a device whose purpose is to store electrical energy which can then be released in a controlled manner during a short period of time A capacitor consists of 2 spatially separated conductors which can be charged to 0 and Q respectively The capacitance is defined as the ratio of the charge on one conductor of the capacitor to the potential difference between the conductors C E g V The capacitance belongs only to the capacitor independent of the charge and voltage 4 Capacitance Typical capacitors two charged conductors isolated electrically Fig 252 Parallelplate capacitor Fig 253 f Capacitance jCV C q Units farad F 1farad1F1CV Charging a Capacitor Electric circuit a path through the charge can flow Battery a device that maintains a certain potential difference between it terminals Schematic diagram symbolic representation of an electric I i circuit Fig 254b Terminal b Checkpoint 25 1 NO Capacitance does 2g q a a 2 2C gt C not depend on charge or V potential Ca C Sample Problem 251 A storage capacitor on a random access memory RAM chip has a capacitance of 55 fF If the capacitor is charged to 53 V how many excess electrons are on its negative plate i1 gt qCV V gg 55x10 15F53V e e 16gtlt1019C 18 gtlt106 electrons Practical Application Microphone condenser 9 gt Current sensor T Battery Moveable plate Fixed plate Sound waves incident 9 pressure oscillations 9 oscillating plate separation d 9 oscillating capacitance Ci 9 oscillating charge on plate 9 oscillating current in wire 1 d Q 9 oscillating electrical signal 9 See this in action at httpmicromgqnetisueduelectromagiavamicrophone Calculating the Capacitance Strategy calculate the difference in potential and divide by the charge Calculating the electric field Gauss law gogg dzzi q For the parallelplate capacitor Fig 255 q 80 E A d A4 Gaussian q 39W V surface Path of integration Calculating the Potential Difference Knowing the field one can calculate the potential difference f Vf K jEd We will choose a path that follows the electric field line We i 9 Vf Ii V d Aquot V V V V Gaussian L q surface V j Eds r Path of integration Recall Two Infinite Sheets into screen Field outside the sheets is zero Gaussian surface encloses zero net charge Field inside sheets is not zero Gaussian surface encloses nonzero net charge Q 0A AEnside E 12 A ParallelPlate Capacitor Flg 255 w d 4 Gaussian EA L V V VHS V V W V surface 41 90 I l Path of d q integration szEdszEjdszEdz d 0 A 80 885gtlt1012Fm 2885pFm 13 P 253 A parallelplate capacitor has plates of 82 cm radius and 13 mm separation a calculate the capacitance and b what charge will appear on the plates if a potential difference of 120 V is applied 3 The tapacitante of a parallelplate capacitor is given by C 1314 where A is the area of each plate and d is the plate separation Since the plates are circular the plate area is A th3 where R is the radius of a plate Thus 12 2 2 Cg mei1335sl FfmImBEXIU ml s 13x1o393m 144gtlt10 1 s144ps h The rharge on the positive plate is given by q CV where V is the potential difference across the plates Thus e 144 1 11119 Flint V173 X m4 3173 14 A Cylindrical Capacitor Fig 256 ngOEAngE27rrL gt E q 2723980LI b VJEds q dr q luv Example Suppose we have 4 concentric cylinders of radii abcd and charges Q Q Q 39Q Question What is the capacitance between a and d Note Efield between b and c is zero WHY A cylinder of radius r1 b lt r1lt c encloses zero charge b d Vad ILdrOILdr a 2723980FL C 2723980FL Q 27z80L C E W Vad dr dr J Note This is just the result for 2 a r c r cylindrical capacitors in series 16 A Spherical Capacitor Fig 266 from a different perspective ngOEAngEmirrz gt E q 2 4722901 b b VZJEds q Cl 2 q i ijz q a 47Z80al 4750 a b 4750 ab Tom charge 1 39I mal charge q Checkpoint 252 l or capacitors charged by the same battery does the charge stored by the capacitor increase decrease or remain the same in each ot thc following situationsquot an The plate separation olia parallel plate capacitor is increased In The radius of the inner cylinder ofzi crluuhicalcapacuorisincreased cl rhetachuslt die outer spherical shell ofa spherical capacitor is Gaussian surfact increased a C 2 8014 C lt d39gtd 1717 d 11 d d bCW1 2750 CbzzngL ng a39gta lnb a 1nb a39 1nb a a a a c 7 4 C7224 4 b39 b 5 0 31 01 ab c gol ab39lt gol ab gt An Isolated Sphere In the spherical capacitor formula take b gtoo aR C4ngL l ab C47T80R 19 P 255 Assuming that a drop of mercury is an isolated sphere What is the capacitance of a drop that results when two drops each of radius R 2 mm merge 5 Assuming censervatien of volume we find the radius of the combined spheres then use C inlay tn i39nld the capacitance When the drops combine the volume is doubled It is then V 241I4 31g The new radius R39is given by gigsquot 4p 2 222 The new capacitance is Cquot 4DEUR392 4DEUEER 504DEUR With R 20 mm we awake 504st sewn1 3 Hm quot1 j scam3mquot 280x10 EF 20 Lecture 37 Review Ampere s law Magnetic field inside a wire Solenoids and Toroids Review id x I7 BiotSavart law dB 3 47239 r A Long Straight Wire 3 0i 272391 A Circular Arc of Wire 3 Z 0 W5 47rR Two Parallel Currents E oLiaib R Ci 1 de 7 a b a A BN l C 4 c 1 ltd a Fl 7 in39 EL 1 Ampere s Law 39 Closed IOOIO around currents 113 Amperian loop Fig 2912 org integration l od Bc089ds ol39enc Algebraic sum of currents enclosed Fig 2913 z391 21 f Direction of integration Bcos ds 0 I39m 001 i2 Magnetic Field Outside a Long Straight Wire with Current Check result obtained previously Fig 2914 Bcoslt9ds B ds 32727 B27r r uol39 139 B 0 27H Magnetic Field Inside a Long Straight Wire with Current That s new Fig 2915 41st 430036019 3de 32727 2 72 739 l l 2 enc R 2 72391 01 327239 l I B r 0 72122 272122 Magnetic Field of a Long Wire Inside the wire r lt a a 271561 B Outside the wire rgta I 34L 72139 J Checkpoint 293 The gure shows three equal currents 139 two parallel and one antiparallel and four Amperian loops Rank the loops according to the magnitude of t d along each greatest rst 210 ienc a I m z z z z b I39m z 0z 0 c 0 z dz39mz39z392z39 ii 8716 Rank 0 a and 1 tie then c Sample Problem 293 ngdEuoim Figure 315a shows the cross section of a long conducting cylinder with inner radius a 20 cm and outer radius 9 40 cm The cylinder carries a current out of the page And the current density in the cross section is given by J c r2 with c 30X106 Am4 and r in meters What is the magnetic eld B at a point that is 30 cm from the central aXis of the cylinder q r 2 r 3 jencszdAszdr 1m 2 IJ dA Icr 27rrdr 2 270 Ir dr I39m 27rc i r 2394 614 827N2Iuor4a4 4 a 2 C 4 4 erian B 1 10 4r r a g IAOIEFE b 47rx107TmA30gtlt106Am4 4003m b gtlt003 m4 002m4 2Ogtlt105 T Example Two cylindrical conductors each carry current I into the screen as shown The conductor on the left i solid and has radius R3a The conductor on the right has a hole in the middle and carries current 0 ly between Ra and R3a What is the relation between the magnetic field at R 6a for the two cases Left Rright a B L6alt B R6a b B L6a B R6a c B L6agt B R6a Use Ampere s Law in both cases by drawing a loop in the plane of the screen at R6a Both fields have cylindrical symmetry so they are tangent to the loop at all points thus the field at R6a only depends on current enclosed I I in both cases enclosed Example Two cylindrical conductors each carry current I into the screen as shown The conductor on the left is solid and has radius R3a The conductor on the right has a hole in the middle and carries a current only between Ra and R3a What is the relation between the 3 3 magnetic field at R 2a for the two cases Left Rright a BL2alt BR2a bBL2a BR2a cBL2agt BR2a l Again field only depends upon current enclosed LEFT cylinder RIGHT cylinder L 41411 74361 Magnetic Field of a Solenoid A constant magnetic field can in principle be produced by an 00 sheet of current In practice however a constant magnetic field is often produced by a solenoid A solenoid is defined by a current z flowing through a wire that is wrapped n turns per unit length on a cylinder of radius a and length L If a ltlt L the B field is to first order contained within the solenoid in the axial direction and of constant magnitude In this limit we can calculate the field using Ampere39s Law Magnetic Field of a Solenoid A solenoid a coil Fig 2917 Magnetic field lines in a real solenoid Fig 2918 Magnetic field lines in an ideal solenoid Fig 2919 Solenoids The magnetic field of a solenoid is essentially identical to that of a bar magnet The big difference is that we can turn the solenoid on and off It attractsrepels other permanent magnets it attracts ferromagnets etc Magnetic Field Inside an Ideal Solenoid o Fig 2920 Solenoid Applications Digital onoft I Doorbells nmmmun q Magnet off 9 plunger held in place by spring Magnet on 9 plunger expelled 9 strikes bell Power door locks Magnetic cranes Electronic Switch relay Close switch Advantage 9 current A small current can be used 9 magnet1c eldpulls 1n plunger to switch a much Iar er one 9 closes larger c1rcu1t 9 J Starter In washerdryer car ignition 15 Solenoid Applications Analog deflection 1 0c Variable AIC valves Speakers Permanent magnet Coil Electrical signals Dust cap 39 tau Cone vibrates Inner suspension spi clerjl Outer suspension surround Solenoids are everywhere In fact a typical car has over 20 solenoGids Sample Problem 294 A solenoid has length L 123 m and inner diameter d 355 cm and it carries a current i 557 A It consists of five closepacked layers each with 850 turns along length L What is B at its center L N B 39 uoln n L N 5gtltNl 5gtlt8504250 turns of 4250 N B i 2 47rx10 7TmA 557A 0 L L23m 242 gtlt10 2 T 242 mT Magnetic Field of a Toroid quotIIll Ilw 39 Fig 2921 l 0 ienc B27rruOiN a 7i 18 Magnetic Field of a Current Sheet Consider an 00 sheet of current described by n wireslength each carrying currenti into the screen as shown Calculate the B field What is the direction of the field Symmetry gt vertical direction wa o Z OdfBwOBwOZBw I nwi therefore 43617 2 0 2 Calculate using Ampere39s law for a square of B uom39 2 l 9quot gt constant gtltgtltgtltgtltgtltgtltgtltgtltgtlt g constant Lecture 40 Induction Review Inductors Se Jnduc on RL circuits Energy stored in a Magnetic Field Review 3 Z 372 3 L Induction end Energy 11 Transfers Z E EWLEEWLKJ mf v Lb4 Mechanical energy transform i i electric and then in thermal energy BZLZV2 2 1 PFvzR R PE Review Reformulation of Faraday s Law Electric eld lines Inductors and Inductance Inductor store energy in the magnetic field Inductance example a solenoid NCD i L inductance de ned Units 1 henry1 H lTmZA Inductance of a Solenoid Let s calculate it from the definition Nc1gtn1BA nz I Bzyom L NOCID nlBA nlLi0 inA 0 nZZA l l inductance of a solenoid Consider the loop at the right Rloop0 SelfInductance m XXX XXX switch closed gt current starts to flow in the XXXX 39oop is it infinite 1 V0 m waitgiSfu ztance a 3 Therefore magnetic field produced in the area enclosed by the loop B proportional to 1 Therefore flux through loop increases as the current increases Therefore emf induced in loop opposing initial direction of current flow because it opposes increasing ux Faraday s Law this emf causes Va gt Vb like a battery to oppose the real battery reduce current ow Fact the current turns on at such a rate to give VaVb battery voltage d dch oc V a V 9 dt dt SelfInduction changing current through a loop inducing an opposing emf in that same loop SelfInductance The magnetic field produced by the current in the loop shown is proportional to that current BoCI The flux therefore is also proportional to the DEE IFS65 0C1 current We define this constant of proportionality between flux and current to be the inductance L Combining with Faraday s Law gives 0 the emf induced by a changing G current SelfInductance The inductance of an inductor a set of coils in some geometry eg solenoid then can be calculated from its geometry alone if the device is constructed from conductors and air similar to the capacitance of a capacitor If extra material eg iron core is added then we need to add some knowledge of materials as we did for capacitors dielectrics and resistors resistivity L C E KC R Archetypal inductor is a long solenoid just as a pair of parallel plates is the archetypal capacitor 1 rltlt d N turns 8 dl dt CEg V dltlt 4 SeHJnduc on i increasing Direction of the selfinduced emf Fig 3017 Lenz Law L1b F MfgQ 239 decreasing L J b Checkpoint 305 The gure shows an emf induced in a coil Which of the following can describe the current through the coil a constant and rightward b constant and leftward 0 increasing and rightward a decreasing and rightward 6 increasing and leftward 09 decreasing and leftward i 1ncreas1ng e i decreasing a Review RC circuits where dq q REE4E qt O O q C l e tRC C l e tTC charging C TC 2 RC t7C q q0 e discharging C RL Circuits 30 o I At t0 the switch is closed and R the current 1 starts to flow b Loop rule g IR L 20 dt Note that this equation is identical in form to that for the RC circuit with the following substitutions 61 RC e Q R Q0 gt RC RL RL Q gt1 C dt R 1 L E Therefore TRCRC gt TRL 2 RL Circuits To find the current Ias a function of time t we need to choose an exponential solution which satisfies the boundary condition 6139 8 dt R We therefore write M b 4W Rs L3 The voltage drop across the inductor is given by dI VL L gemL dt RL Circuit 2 on Current W x 3 I Z e RtL Max sR 63 Max at tLR W 1 lnl mgp nn I I VL 2 Ld dt RL 86 t Max sR 7 Max at tLR 8 UR 2LR RL Circuits After the switch has been in position a for a long time redefined to be t0 it is moved to position b Loop rule IRLd I0 dt The appropriate initial condition is 1r 0 I ie RtL The solution then R must have the form VL RL Circuit 8 off Current Rt L I 8 6 R Max aR 37 Max at tLR Voltage on L VL LdI 8e Rt dt Max 8 7 Max at tLR W SR LR 2LR Sketch curves 8 on gR LR 2LR W a I I1e RtL Rlt 0 I t s I V RtL L VL dt o a 8 Off SR LR 2LR I I 5e RtL I R o a t 0 fIF39Fquot VL L RtZ dt 398 Checkpoint 306 ego ere TLLR The gure shows three circuits with identical batteries inductors and resistors Rank the circuits according to the current through the battery a just after the switch is closed and b a long time later greatest rst a 1391 lim R i i 3 Rm 2R 2 3 I R IE REE E L L L lt1 2 L 3 E Q5 If b 1 e tTL 2 2 1 Hme 2R 2R l2 E R R2 R 1 5 2 3 I R 18 Energy of an Inductor How much energy is stored in I I an inductor when a current is 300 gt R flowing through it b Start with loop rule sIRL L Multiply this equation by I 81 12R LI 61 1 dt From this equation we can identify PL the rate at which energy is being stored in the inductor dU d We can integrate this equation to find an expression for U the energy stored in the inductor when the current I UTdUjLIdI gt U LI2 0 0 Where is the Energy Stored Claim without proof energy is stored in the magnetic eld itself just as in the capacitor l electric field case To calculate this energy density consider the uniform field generated by a long solenoid N 1 B 2 yo I 1 N2 The inductanceL is L 0 7272 Z Nturns 1 2 1 N2 2 2 1 2 32 E U U LI 7zr 2 7er nergy 2 2 10 z j 2 0 We can turn this into an energy density by dividing by the volume containing the field U 43 72721 22 u 1 Checkpomt 307 us g o n2 2 The table lists the number of turns per unit length current and cross sectional area for three solenoids Rank the solenoids according to the magnetic energy density within them greatest rst Turns per SUlCIlUlLl U n il Length Currcnl Area M 2m 1quot 2A 7 n 21 A I 61L r39 H l a 5 510 2770211Z 210 quot12112 1 a and b tie then c b 5 Euoniz211Z 210 quot12112 l l c 5 Euo quot02112 Euo quot12112 Sample Problem 307 UB Li2 A coil has an inductance of 53 mH and a resistance of 037 Q a If a 12 V emf is applied across the coil how much energy is stored in the magnetic field after the current has built up to its equilibrium value loo limt gtoo E 7 E E 1 quotML LR 1et L l e TL R R R U300 Lz 253x10 3H343A2 31 J b After how many time constants will half this equilibrium energy be stored in the magnetic field UBUBOO gt lukejlu 21421 gtz iz39w 2 2 2 2 J2 E T E r l e gt e R JER 1 i0293 gt i 1n0293123 5 TL 22 Sample Problem 308 uB zigz A long coaxial cable Fig 3022 consists of two thinwalled concentric conducting cylinders with radii a and b The inner cylinder carries a steady current 139 the outer cylinder providing the return path for the current The current sets up a magnetic eld between the two cylinders a Calculate the energy stored in the magnetic eld for a length l of the cable dU l uBE B dUBzquV UszquVz IBZdV dV 270 0i 2 luozi2 1 Bd uoz B27r01 gt B gtB 2 2 W 27H 47 r 2 2 b 2 b 2 OlllerT UB 1 012 27l 0 l Z 0 l r cylxntk 210 47 a r 47 a r 47 nn cylindrical shell 2 U3 2710 15112 47 a Mutual Inductance Demo current is induced in one coil when the current is changed in a neighboring coil b We can describe this effect quantitatively in 7 terms of the mutual inductance the ratio of flux through the loop to current in opposite loop H H Dab flux through coil a due to current in coil b ME 511 2 1M 1b 1 a Notthas this symmetry not obvious perhaps but true Change current in loop a 9 changes Ba 9 change flux through loop b 9 induces current in b to oppose this 9 loop b produces a field Bb in the opposite direction as Ba The same thing happens with just one loop This is called self inductance Applications of Mutual Inductance Transformers still to come Change one AC voltage into another V2 Airport Metal Detectors Pulsed current 9 pulsed magnetic field 9 Induces emf in metal p Ferromagnetic metals draw in more B 9 larger mutual inductance 9 larger emf l Emf 9 current how much how long it i lasts depends on the resistivity of the material Decaying current produces decaying magnetic field 9 induces current in receiver coils Magnitude amp duration of signal depends on the composition and geometry of the metal object 25 Pacemakers 4quot Applications of Mutual Inductance It s not easy to change the battery Instead use an external AC supply Alternating current 9 alternating B 9 alternating DB inside wearer 9 induces AC current to power pacemaker Applications of Mutual Inductance Upon leaving the hospital with his new pacemaken Herman rea Zes that bananas are NOT Included Lecture 46 Magnetism of matter Review Magnets Magnetism and electrons Magnetic materials Review Gauss Law for the Magnetic Field ch d21 0 Ampere Maxwell law d y050ampy0 I39m dt dQDE I D tC t l 5 ISp acemen urren d 0 dt n M II axwe S ua IONS Gauss law for electricity IA hm8 Rcl39alu nct electric ux to net enclosed electric change Gauss law for magnetism 99 E zIT L Relates net magnetic llux to net enclosed magnetic charge a MFR Faraday s lnw I Iv 7 I Relates induced electric held tn changing magnetic ux 4 7 tl l E Alnpet ciMuchll lttw B ls Mae H MM Relules Induced magnetic eld l0 changmg electric 11le and lo Clll39l Cl ll Written on the nasumptiun that no diclccnic or magnetic material are prcxcnll M 39 A qur r eld mum me msmmun caused by passmg charged Females rm the Sun 39 x wwlm wunugnmu quHII Mp 3 3 39fquotiquot with Earth s rmauunal an R The sunlh p l uf39he mm 15 m Earth s Nunth Hmspham Furthe maimed eld anlgure 114 the magnlude nfL39158XlUZZJT and the mrecuun U 139 makes anangle nf 11 5 w h the 39 gmm y The eld the eld id x aulu mm M w w 77 U Age tlmllmn mm The Magnetism of Earth Fig The earth toxins a giant magnet The direcuun ufthe magma eld pruduced by the cure reverses abuut every mlllmn ymrs 39n n I I n v 0 u e A u at Ihe earth39s magnetic eld is not aligned with the earth39s axis The average xtrength of the ealth39s magnetic eld is about half a Gauss 50 IT The gquot 39 quot 39 quot quot g 39 39 39 39 39 to point on the earth The eld 39 quot 39 39 the de ation 39 39 39 39 39 39 39 39 the deviation 0 TL e u 39 39 39 Bay in anada l The quotuonhquot pole of quot 39 39 39 39 a avulii mugm l 39 5 de ned 0The eanh39s magnetic eld gets stranger and weaker over a peu39od of centuries and has reverle itself many times over millions of years 39Ihe earth s lnagie e eld extends into space and traps tharged partitles from the sun in the Van Allen belts 39This protects the earth from higher radiation eqyoslu39e Source of Magnetism What is the source of magnetism if not magnetic charge Answer electric charge in motion eg current in wire surrounding cylinder solenoid produces very similar field to that of bar magnet Therefore understanding source of field generated by bar magnet lies in understanding currents at atomic level within bulk matter Orbits of electrons about nuclei Intrinsic spin of electrons more important effect Magnetism and Electrons An election has an intliusie angular momentum called spin S 39Assotiated m th this angular momentum is an intn nsit spin magnetic dipole moment it The magnetic moment in is related to S by S Elev us where e 16x1019 C and m 911x1031 kg and the minus sign means that S and u are oppositel directed Fig 32710 Thehsem E5 39 339 mmmpm 5p ere S is different from a mau oseopic anguhu momentum in that it cannot be directly lneasiu39ed its component along any axis can be meastn39ed OS is also quantized which means it is limited to certain values in this case the same value either positive 0139 negative 51 mg with ms 1 where m is the spin magnetic quantum number and h is Planck39s Constant h 663 x 1034 Js Orbital Magnetic Dipole Moment 1 Ha I aroluul the nucleus Tle orb angular momentlun is symbolized by L Tlle orbital magnetic dipole moment in is a e gt um Lnrn Tle component along the z axis is Lnrhz me 2quot with rn 0 1 1 i 2 10mm 39quot quot 39 39 nilunrmuumu a 39 139 does eh unrnz me m Loop hiotlel for Electron Orbits The colnplete explanation for the behavior of electrons in atoms is given by quantlnn mechanics But semi classical models yield sinlilar results in lnost cases oThe semimlassical model of an electron orbiling about the nucleus gives a magnetic moment purl LA The culrent i is that of the electron circling the nucleus i charge 6 e time T 211rv Where the period is the circulnference divided by the speed Since A 139 and substituting for current evr Horn IA 7 i i an Eurv The angular momentum for a circular orbit is L anl enclusesanarai u dlpule mumem 7m evr S L e L mauunufthe m 2 2m 2m 050 including the Sign due to opposing directions of L and n e Hun 39239 Lorb IThe spin magnetic moment 11 also cannot be directly can measured but Its component 3 7 quot5 27 m Z a quot5392 mm52n 4nm 4152 is called the Bohr magneto and has the value eh Mg m 927X1024JT l39 quotI 39 H quot 39 elementaryll 39 quot HIB 1 t t t I e i U 39Mg39Bext 115259 thi can39panem fths arb a angular namenhvm by min W Ln m unpmen m the mth argular nummum xs gvm by Lm5 r m Wm m w w mm mm Phme aw l n w w mm msz Whm eh 1 7 im mms m Wl39a39EME eAan 9 2 m 11 sthEEuhr magmnn 3 m 1 the magma Earnpanmtu Mr 7 7M 7 is a A 107241 139 mm m 72113mmsuredcunpmmlxs W 2 1 a x 10 Loop Model in a NonUmfoim Field 39It has previously been shown that the w force exerted by a lnlifoim magnetic field on a elment loop is zero 3980 how are the magnetic forces between atoms created 0h should be noted that the magnetic dipoles nithin atoms do not produce lmifolm magnetic fields ehllact l quot 39 quot quot 39 uumd39 I 39 and hop off rapidly vsith distance Ulltlt l39 these cii39clnnstances clm39ent loops near atoms are subject to a net magnetic force Magnetic Materials Materials can be classified by how they respond to an applied magnetic field Balm Paramagnetic aluminum tungsten oxygen Atomic magnetic dipoles atomic bar magnets tend to line up with the eld increasing it But thermal motion randomizes their directions so only a small effect persists Bind Bapp 105 Diamagnetic gold copper water The applied field induces an opposing field again this is usually very weak Bind Bapp o10395 Exception Superconductors exhibit perfect diamagnetism 9 they exclude all magnetic fields Ferromagnetic iron cobalt nickel Somewhat like paramagnetic the dipoles prefer to line up with the applied field But there is a complicated collective effect due to strong interactions between neighboring dipoles they tend to all line up the same way Very strong enhancement Bind Bapp o10t5 12 DIagneIizzlrion Vector and DIngne r Fidll Strength Tn describe Illa magnet eld 1 materials 1 new symbol BI is used 4 pl39ar To describe the manual magnetic h 1 presence of materials a new symbol lL i used m39oidcmll llsionuitll V 7 m Hamll39 39 Bio 39 39 r Alnpsmetu39 B pugI M 39 39 g r 39 I consulta 39 39 I M r c H 1m1m1mg the magnetic msceplnb ity in 1 B equation BpnI1IpmII CHpmlIl 1 And the mgmm permeability m a mumm is plmpn1cIgtBpmH Diamagnetism J amngjlelic materials do not have penmmem dipole Inolnelm the externle Imlgnelic eld lull created tllem Dimnngerit material llll Weaken lie external eld Paramagnetism r I I permanent magnetic dipole moment 1 IIII I lIIv I I eld I lung u 7 J 39 Iquot d 39 the aliglnnent 39 39 quot 39 compete with the random thenml Inolion of the dipole IThe average translational kinetic energy from theor is 15 kT where k is Boltzman39s Constant k 133 X 10723 Jr t l quota I I II I ivofEUBr I 39 39 M 39 39 II llll external 39 39 39 l Ililmles are mostly aligned quot quot Inie39sLaw uyxlrur Iniein39 quot 39r39l39 Hex M C T where is at constant for each maltelial Ferromagnetism Fen39omng1Ietixlnix a much stronger effect than pam 0139 Iliamagnetism m I II I I I I I I I II I II a IIIIIIIIII 39 IIIIIII m I 39 39 I I H 39 l quot39 l 39 39 mrl39rl oreobalt TlIe efftttI 39 quot 5 Iquot 39 39 I 39 I 39 539 quot39 39 39 39 quotquot quot 39 IIIIIIIII I I My 39KquotAI Vl Ivr II I II I I Ilomains U1ultr these quot r I quot 39 II IIIIIIIIIIIIVIIIII39II NannallyIlIowevelu 39 quotquot 39 39 V 39 39 739 39 r 39 I39IIIIIIIIIII oAIIyI39 39 39 39 39 39 39 39 IIIIII IIIIII IIIIIIIIII39IIIIIIII 39 quot to melt other quot1 39 39 Iquot 39 39I 39 I 39I IIIIL39 IlIrII quot 39 39 oftlIe extemalfleltl I I II I I I I III I I I nou feu olnag uelic matm als 739 Iquot quot 39 I 39 39 I39 39 nttlIe aligmneuttlteylteltlcreatingapennzmeutlmgletie moment Le a pennaneut magnet I I I II 7 II I I1 I p III I39Vl I I I I Almugll temporary magnetic field Ferromagnets Even in the absence of an applied B the dipoles tend to strongly align over small patches domains Applying an external field the domains align to produce a large net magnetization Magnetic Domains Soft ferromagnets The domains rerandomize when the field is removed Hard ferromagnets The domains persist even when the field is removed Permanent magnets Domains may be aligned in a different direction by applying a new field Domains may be rerandomized by sudden physical shock If the temperature is raised above the Curie point 770 for iron the domains will also randomize gt paramagnet Example Which kind of material would you use in a video tape a diamagnetic c sof ferromagnetic b paramagnetic d hard ferromagnetic How does a magnet attract screws paper clips refrigerators etc when they are not magnetic 18 Example Which kind of material would you use in a video tape a diama netic c soft ferromagnetic g b paramagnetic d hard ferromagnetic Diamagnetism and paramagnetism are far too weak to be used for a video tape Since we want the information to remain on the tape after recording it we need a hard ferromagnet These are the key to the information age cassette tapes hard drives ZIP disks credit card strips 19 Example How does a magnet attract screws paper clips refrigerators etc when they are not magnetic The materials are all soft ferromagnets The external field temporarily aligns the domains so there is a net dipole which is then attracted to the bar magnet The effect vanishes with no applied B field It does not matter which pole is used S N End of paper clip 20 Lecture 41 Electromagnetic oscillations and Alternating current Chapter 31 LC oscillations Damped oscillations in RLC circuits Alternating current Simple circuits Resonance and Power in Alternating current circuits Transformers Review of Voltage Drops Across Circuit Elements I l M Q J Idt Voltage determined by V Z integral of current and l C C capacitance l l m 2 V L d L d Q Voltage determined by 2 derivative of current and df df inductance Where are we going Oscillating circuits radio TV cell phone ultrasound clocks computers GPS What s Next Why and how do oscillations occur in circuits containing capacitors and inductors naturally occurring not driven for now stored energy capacitive ltgt inductive LC Circuits Consider the RC and LC I series Cll39CUltS shown HH HH CR CL Suppose that the circuits are i I formed at t0 with the capacitor charged to value Q There is a qualitative difference in the time development of the currents produced in these two cases Why Consider from point of view of energy In the RC circuit any current developed will cause energy to be dissipated in the resistor In the LC circuit there is NO mechanism for energy dissipation energy can be stored both in the capacitor and the inductor LC Circuits Qualitativer We discussed RC and RL circuits In an LC circuit energy oscillates between the capacitor E field and inductor B field LC Oscillations qualitative QZQ0 TT U I0 0 L 111i L QZO l Q Q0 I Oscillations UB UE RCLC Circuits I gt1 Q fc L C R i RC 1 LC 1 current decays exponentially L current oscillatesJ o LC Oscillations L with finite R lfL has finite R then energy will be dissipated in R the oscillations will become clamped 0 R0 R 0 The Electrical Mechanical 2 a 11 c One can make the analogy with mechanical oscillations Table 311 q corresponds to X 1C corresponds to k icorresponds to v L corresponds to m k 1 a block sprmg a LC czrcult m R TABLE 35 The Energy in Two Oscillating Systems Compared Block Spring System 1quot Oscillator Elcmcnl Energy Elemenl Energy S pi39ing Polonlial x Cupuciim39 Electric lC39lqj Block Kinclic m inductor i lugneiic ill 139 Irll i T dqr39ll LC Oscillations Quantitativer BlockSpring Oscillator 1 1 UUbUs va kx2 2 2 O dUm E 2 dx t2 m x X cosa t displacement a 2 E m dzx clt2 clv V clt kx0 kx dt dxdzx m clt a t2 dx x clt block spring oscillations a1 d2x di di di clt2 2 dltv gt2V di alt 02X 0080 1 5 a2x t Cdt dzq L d12 q 0 LC oscillations q Q cosa t charge a 12 LC Oscillations Quantitativer See also Fig 314 14 dt 2 2 q Q 2 cos wt E 2C 2C 5 Li2 2 U3 szQZ sin2at UB Q Zsin2al 2C 2 aQsinat I sinat q Qcosat sin2 05cos2 a 1 LC Oscillations Energy Check 1 OscHlatIon frequency a has been found from the loop equation 0 VLC The other unknowns Q0 are found from the initial conditions Eg in our original example we assumed initial values for the charge Qi and current 0 For these values Q0 Qi 0 Question Does this solution conserve energy U130 QC id coszw gt 1 1 UB t 2 EU2 t EngQo2 s1n2a0t Energy Check Energy in Capacitor Energy in Inductor UBtLw Q Sm2wot 0 1 a 0 W U U3 UBt Q3 sinzw o W Therefore UEtUB1 o Problem At t0 the current flowing through the circuit is I 12 of its maximum value Q L Which of the following plots best 39 represents UB the energy stored in the inductor as a function of timbe time time time C Which of the following is a possible value for the phase 25 when the charge on the capacitor is described by Qt Q0cosat 15 8 30 b lt15 45 0 60 Problem cont At t0 the current flowing through the circuit is 12 of its maximum value Q L Which of the following plots best represents UB the energy stored in the inductor as a function of time a I bi c The key here is to realize that the energy stored in the inductor is proportional to the CURRENT SQUARED Therefore if the current at t0 is 1l2 its maximum value the energy stored in the inductor will be 1I4 of its maximum value Problem cont At t0 the current flowing through the circuit is 12 of its maximum value Which of the following is a possible value for the Q L phase when the charge on the capacitor is described by Qt Q0c0swt a 30 b 45 6 60 We are given a form for the charge on the capacitor as a function of time but we need to know the current as a function of time Ht 2 62 a0Q0 sin wot go Att o the current is given by 0 aOQO sin go 2 I COOQO max 1 o Therefore the phase angle must be given by sun go 2 i3 2 go I i 30 qu qQcoswt c C Checkpomt 312 iwQsmw A capacitor in an LC oscillator has a maximum potential difference of 17 V and a maximum energy of 160 u When the capacitor has a potential difference of 5 V and energy of 10 u What are a the emf across the inductor and b the energy stored in the magnetic eld di 1 a E L 2 EL L a 2Q cosat szq VC rJsttCzsr UBUE b U3 UEmX UE 160M 10M2150M QCVC Sample Problem Q51nm I sina t A 15 uF capacitor is charged to 57 V The charging battery is then disconnected and a 12 mH coil is connected in series With the capacitor so that LC oscillations occur a Assuming that the circuit contains no resistance What is the maximum current in the coil 1 C 1 61 0Q mCVC E C JLC a W 6 W57V0637A 12x10 H b What is the maXimum rate didzjmax at which the current 139 changes d 1 5mm t m I a cosat b d t d t 216021 0637A 4750As dtmaX ch J12x10 3H15x10 6F 20 Lecture 50 Images Chapter 34 Review Lenses Optical instruments Review Brewster s Law 93 6r 900 n Qthanl Zztanln r5 Plane mirrors wlw III a lt L Incident unpolarized Re ected ray ray 0 Component perpendicular to page 4D Component parallel to page F P n i 4 Virtual Images from Spherical Mirrors image Review Focal Point F Focal distance f WHJLH f r Spherical Mirrors Formula L Li p in f Lateral Magnification 39 k lt Lenses A lens is a piece of transparent material shaped such that parallel light rays are refracted towards a point a focus Convergent Lens POSitiVef light moving from air into glass will move toward the normal light moving from glass back into air will move away from the normal real focus Negativef Divergent Lens light moving from air into glass will move toward the normal light moving from glass back into air will move away from the normal virtual focus 4 The Lens Equation We now derive the lens equation which determines the image distance in terms of the object distance and the focal length Convergent Lens Ray Trace Ray through the center of the lens light blue passes through undeflected Ray parallel to axis white passes through focal pointf These are principal two sets of similar triangles h 3 1quot 2 L rays h s 5 f f eliminating h lh39 S39 S39 f 1 1 1 same as quot111101 eqn 39 F f gt gty T gt Ifwe define gt0 fgt0 magnification also same as mirror eqnll M Z Sr Mlt 0 for inverted image s 5 Thin Lenses Formula of thin lenses H 1 1 I Iq p i f JEH 139 Focal distance real focal pint virtual focal point n 1i i r1 r2 ChecprI nt 344 m i I9 A thin symmetric lens provides an image of a fingerprint with a magni cation of 02 when the ngerprint is 10 cm farther from the lens than the focal point of the lens What are the type and orientation of the image and What is the type of the lens mgt0 gt ilt0 Virtualimage 1 1 1 same orientation as the object E li 139 f flcm pzlflcm 1 fflcm mi gt0 onlyl39fflt0 lfl flcm p ifflcm divergent lens CL F1 F1 2 I Ffquot i 17 k rg gtlt 71 gti Hf f7 gtlt i gt Thin Lenses cont Lateral magnification Images from lenses O 1 I 0N The Lensmaker s Formula So far we have treated lenses in terms of their focal lengths How do you make a lens with focal lengthf Snell s Law at the convex surface nsin sina Assuming small angles 0 n6 The bendangle 8 is just given by 8 205 49 z n 16 The bendangle 8 also defines the focal lengthf z gt 71 1 The angle 6can be written in terms of R the radius of curvature of the lens h z R9 Look h and 6drop out Putting these last equations together This holds for any incident parallel ray 1 1 Lensmaker s Formula The thin lens equation for a lens with two curved sides in11 1 Rgt0ifconvex f R1 R2 when light ray hits it 10 Thin Lenses cont The twolens system write the lens equation for lens 1 write the lens equation for lens 2 taking p as the image distance of lens 1 adjusted for the distance between lenses Mzmlm2 Draw Rays Multiple Lenses We determine the effect of a system of lenses by considering the image of one lens to be the object for the next lens lt f 1 f 394 1 1 1 1 For the first lens s1 15 f1 1 Sample Problem 344 11 Figure 354a shows a jalapeno seed 01 that is p l f 9 placed in front of two thin symmetrical coaxial W lenses 1 and 2 with focal lengths f1 240 cm and 39139 f2 90 cm respectively and with lens separation L 10 cm The seed is 60 cm in front of lens 1 W ova Where does the system of two lenses produce an m 39 image of the seed r 039 l l l 1 1 1 I 6cm 1391 24cm 191 11 f1 e l 1 1 gt i1 24cm6cm 8cm JR17 1391 24cm 6cm 6cm 24cm 1 p22Lilllecm86m2186m 0 1 1 1 9cm18cm 180m 1392 9cm 2 2180m9cm 2 1 1801 omen MzmlmziLiXLZ 80mgtlt180m 1 w p1 p2 91 p2 6cm 180m 3 Simple Magnifying Lens Near point Fig 3417 6 2 6h Angular magnification 6 Z 25 cm 639 lw i i 4i Example A lens is used to image an object on a screen The right half of the lens is covered What is the nature of the image on the screen a left half of image disappears b right half of image disappears screen I C entire image reduced in intensity Example What happens to the focal length of a lens when it is submerged in water ln air the focal length of a glass lens n15 air air fair is determined to bef When the lens is submersed in watern133 its focal length is measured to be fw What is the relation ater39 between fair andfw 9 ater 39 afwater lt air fwater fair c fwater gtfair The refraction is determined by the difference in refraction The smaller the difference the less the bend he two indices of the longer the focal length Think of the case where it is glass in glass f gt infinity Quantitatively look at the Lensmaker s formula n N glass 1 f water n water 1 iltLz R fair nair lit The Eye The Normal Eye Far Point 5 distance that relaxed eye can focus onto retina 00 Near Point 5 closest distance that can be focused on to the retina 250m 1 1 1 250m 1 1 1 1 v f23cm 25cm Therefore the normal eye acts as a lens with a focal length which ca vary from 25 cm the eye diameter to 23 cm which allows objects from 25 cm gt co to be focused on the retina This is called accommodation 17 Diopter 1f Eye 40 diopters accommodates by about 10 or 4 diopters Optical Aberrations Why really good lenses cost a lot Chromatic Aberration Chromatic aberration Due to dispersion index of refraction depends on frequency focal length can be different for different colors Spherical Aberration Spherical aberration But spherical Outside the paraxial limit lenses are optimal focusing occurs only for a mUCh Qheaper parabolic lens Spherical lenses to grlnd amp look parabolic for narrow field of POIiSh view Astigmatism Curvature of the lens not symmetric in transverse directions gt slightly cylindrical gt different focal lengths 8 Why use two lenses instead of one No choice gtvision correction Improve aberrations Spherical Aberration 19 Apparent Magnification Our sense of the size of an object is determined by the size of image on the retina Consequently the apparent magnification factor of a lens is just the ratio of the angular size with the lens to the angular size without the lens Use lens to make close up image fall in focus range of eye itive f lens ZJHL TX c h I h f gt Object at Near Point can t get nearer ObjeCtju t inSide Fo al Pomquot of Simple magnifier a z h an 7 Define Angular Magnification M E 7 2o Microscope Larger magnifications than are possible with a single lens can be obtained by combining lenses For example the compound microscope consists of two lenses a short focal length objective lens and an eyepiece The object placed just beyond I the focal point of the objective 2 produces a real inverted image 11 3911 at a position which isjust inside the focal length of the eyepiece f0 3 o This image then produces a virtual objective L eyepiece image 12 which is seen as magnified as in the previous slide The magnification of the objective is The magnification of the eyepiece is The total magnification M Maile is 21 Compound Microscope Eyepiece Objective 2T F2 EN S gtgt 0b ra s I y lzs To distant Virtual image le b eLb s fey 1 fob l S m P fab s 25 cm M mmt9 feb fey 22 Telescopes The purpose of a telescope is to gather light from distant objects and produce a magnified image Refracting telescopes use lenses so that the objects can be viewed directly As in the microscope we have two lenses an objective and an eyepiece Since the object is at a large distance the objective lens should have a long focal length to obtain a large magnification Reflecting telescopes use mirrors to create the image Most astronomical telescopes are reflectors since the most important feature for these telescopes is the light gathering ability and it is easier to make a large mirror than it is to make large lenses detector 23 Refracting Telescope Eyepiece Objective 6 90b 151 m6 6 I 0b Parallel rays To ima e I quot Parallel from g x rays distant h39 h39 object 6 b m 6 z 0 9y a Lb fob f ey h3939f 3y fob m6 2 h39f0b fey Hub 66gt39 EEJI L b fey 24 Hubble Space Telescope The HST was launched in 1990 it was discovered that a lens had been ground incorrectly so all images were blurry A replacement contact lens COSTAR was installed in 1993 Now Hubble s instruments have builtin corrective optics so COSTAR is no longer needed a Before COSTAR After COSTAR 1 Aperture of primary mirror 24 m 8 ft Mass of primary mirror 828 kg 1800 lbs Lecture 51 Interference and Diffraction Chapter 35 amp 36 Light as a wave The law of refraction Diffraction Yang s interference experiment Coherence Light as a Wave o Wave wavefront Fig 352 W Huygens Principle All points on a wavefront serve as a point sources of spherical secondary wavelets After a time t the new position of the wavefront will be that of a surface tangent to those secondary wavelets A1 Incident wave The Law of Refraction 91 Air Glass See Fig 353 W A 12 51116 Sll lg 1 he 2 hC 11061 vlT sin62 va v2 C C n12 V12 b V12 quot12 sm6 2 v2 cn2 n1 51m1 cnl n2 Refracth wave A2 gt V2 nlsln 61 n2 sm 62 Checkpoint 351 The gure shows a monochromatic ray of light traveling across parallel interfaces from an original material a through layers of materials b and c and back into material a Rank the materials according to the speed of light in them greatest rst v gt v nlsin lznzsin 62 c c quot12 V12 V12 quot12 sin 91 v1 v gt v sm 92 v2 b a Vb gt vc Vb gtVcgtVa Plane waves In mathematical form Maxwell s Eqns x t E Em Slnkx wt E Em sm27zZ 27zj C E a k wave Speed A 27rk 27rca 2 CT wave length B Bm ave ronts 1 a W f Ray 299792458 mS Vluogo E C B C y I E m B m Component component Wavelength and Index of Refraction C Wavelength inamedium 11 A 126T A ZVT 3 l lz 7 CT 0 c 71 Number of wavelengths Fig 354 gt gt Nzi Nl LL71 nz 2139 in 2 4 Ln Ln L N27N172 71Z 27quot1 Sample Problem 351 In Fig 354 the two light waves that are represented by rays have wavelength 550 nm before entering media 1 and 2 They also have equal amplitudes and they are in phase Medium 1 is now just air and medium 2 is a plastic layer of indeX of refraction 16 and thickness 26 um a What is the phase difference of the emerging waves in wavelengths radians and degrees b What is their effective phase difference in wavelengths L 26x10396m ANN N n 16 10 284 2 1 2 2 quot1 55x10397m E1Emsin 27239 x 2er E2Emsin 27239 x 2ni xL m T in T L 2 My 27239 27239 27239 AN p in in 4 27r284 178rad 210200 ANe AN IntAN284 2084 L Diffraction of Water Waves Fig 564 The diffraction of water waves in a ripple tank The waves are pro duced by an oscillating paddle at the left As they move from left to right they are out through an opening in a barrier along the water surface Diffraction Fig 357 If a wave encounters a barrier that has an opening dimension similar with the wavelength that part of the wave that passes through the opening will diffract spread out into the region beyond the barrier a A Ht it lit iii lit a b c 39 Young s Interference Experiment Two coherent waves intersecting in a point interfere Young s Interference experiment Fig 35 8 Interference pattern L SE bright bands maxima 1 dark bands minima Locating the Fringes The phase difference between two waves can change if the waves travel paths of different lengths D t E Em sinkx wt 2 Em sin27r x xi wt AL 2 d sin6 integeri d sin 9 mi m 012 maxima dsin 6 m 31 m 012 minima J a sin 91d 2 0 51 91d 2 sinf rst dark fringe 9 Path length difference AL 5 Sample Problem 352 What is the distance on the screen C in Fig 35lOa between adjacent maxima near the center of the interference pattern The wave length of the light is 546 nm the slit separation d is 012 mm and the slitscreen separation D is 55 cm Assume that G is small enough to permit the approximation Sin 9 tan 9 9 d sin 9 m2 m 012 maxima 6 m2 i s1n d ym m1 sin6 ztan6 z6 zy m D d D D ym mg ym m 1M quot D D D A 1 1 i u y ym ym m d m d d 55gtlt1072m 546gtlt 10 7m 012gtlt10 3m 25 gtlt lO Sm 25mm I nu AL Diffraction Diffraction pattern Light passing through a small hole exhibit a broad intense very bright central spot on a screen a number of narrower less intense secondaryside maxima interference effect Fresnel diffraction pattern by a circular aperture Fig 363 Diffraction by a Single Slit Fig 364 use Huygens principle and split the slit in two sources lHSHleSQI Vu 5 I 1 Incldcm I B d sin l m EM interference minirna a a 1 e Esm 6 E a sm 6 A rst rmnlmum I l3 iPalh length difference 12 Ir Diffraction by a Single Slit Four sources Fig 36 5 1 d sm 6 m EVL mterference m1n1ma sin 6 g a sin 6 21 second minimum asin6 m1 m123 minima a FEQIHEAREQI quot quotquot quot3 quotZ I b Checkpoint 371 We produce a diffraction pattern on a Viewing screen by means of a narrow slit illuminated by blue light Does the pattern expand away from the bright center or contract toward it if we a switch to yellow light or 9 decrease the slit width a sin 61 2 xi rst minimum sin 6 i 1 a yellow Abbie 1 sin y 2 y gt sin 61b 6 expands a 1 s1n6391 39 gt s1n 61 b expands a a Intensity Qualitativer Try to find the intensity at each point not only the location of the dark fringes Fig 366 phase difference path length difference A sin 9 ZA 2772sin6lZAx 2772sin6 a l 19 lt Innquot Intensity Quantitativer Try to calculate the intensity at each point not only the location of the dark fringes Fig 367 7Z39Cl m7ra sm9 m123 l l lll lll l la a sin6 2 m1 m 123 minima Checkpoint 363 Two wavelengths 650 nm and 430 nm are used separately in a single slit experiment The figure shows the results as graphs of intensity 1 versus angle 9 for the two diffraction patterns If both wavelengths are then used simultaneously what color will be seen in the combined diffraction pattern at a angle A and 1 angle B 7 650nm gt red 7 430 nm gt violet a sin 6 l first minimum megl lAltlB a A red visible violet dark B violet visible red dark Lecture 31 Circuits Review The Ammeter and Voltmeter RC circuits Review Multiloop Circuits Kirchhoff s junction rule 11 13 12 Resistances in Parallel Two Resistances in Parallel R Reg 1 RZ lt 11212 11212 l 1 lt Resistors I I I a In SerIes ReVIew 39 The Voltage drops R1 Va VHR Vb VisZ quot n KMaampgt R2 a c Whenever devices are in SERIES the current is the same through both This reduces the circuit to R effective Hence Re ective R1 R2 Ammeter and Voltmeter Ammeter instrument that measures currents inserted in series in a circuit Fig 2714 has a very small internal resistance Ammeter and Voltmeter Voltmeter instrument that measures potential differences inserted in parallel in a circuit Fig 2714 has a very large internal resistance RC Circuits Charging a capacitor Fig 2715 E iR cho chi led 1 C dt E R izo dt C dq q R rE t0 0 dt C q Charging a Capacitor Fig 2715 S gt a qC 1 e RC See Fig 2716 0 2 4 6 8 10 Timems a 2 4 6 8 10 Time ms 17 Derivation of Eq 2733 dt RC R CZq hq h0gtqhKeal dt RC RC al Kae Ke 0 RC 0 qquKe at 371720 quCE qC Kem q00 3 K C IRC ql C93 Case RC Cqsa e The time Constant Product RC has dimension of time time constant A12 Cg s u n T R C W 4 I C E e ZT 0 2 flk ime6m 10 a 101 CE1 e 1063CE Charging Capacitor Charge on C W Ca RC 2RC Q C81e tRC Max C3 63 Max at tRC quot W o Current W a R I d Q ie lRC dt R I Max aR 37 Max at tRC 10 Problem At i0 the switch is thrown from position b to position a in the circuit shown The capacitor is initially uncharged At time tt139r the charge Q1 on the capacitor is 8 11Ie of its asymptotic charge QfCe What is the relation between Q1 and Q2 the charge on the capacitor at time tt22 r 8 Q2 lt 2Q1 b Q2 2Q1 C Q2 gt 2Q1 The point of this ACT is to test your understanding of the exact time dependence of the charging of the capacitor Q Cg1 eRC So the question is how does this charge increase differ from a linear increase Charge increases according to From the graph at the right it is clear that the Q2 charge increase is not as fast as linear Q1 In fact the rate of increase is just proportional to Q the current dQdt which decreases with time 1 Therefore Q2 lt 2Q1 1 Discharging a Capacitor Back to Fig 2813 S gt b dq q R 0 t0 ZOE dt C q Io itRC qzqoe iz amp gitRC dt RC VC Ziz eitRC C C Discharging Capacitor The capacitor is initially fully charged Q Q0 At t 0 the switch is thrown from position a to position b in the circuit shown FromKVL R Q 20 dz C dQ 1 dz E39Q Qz must be an exponential function of the form 1 t 2 eat Therefore dQ t where a Q dt ae aQ RC From initial condition Q0 Q0 we get W QM 13 Discharging Capacitor Ck Charge on C W Max C8 Q 37 Max at tRC Current W I eitRC I dt R Max 8R 37 Max at tRC 39ER RC 2RC 14 Behavior of Capacitors Charging Initially the capacitor behaves like a wire After a long time the capacitor behaves like an open switch Discharging Initially the capacitor behaves like a battery After a long time the capacitor behaves like a wire 15

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