General Physics II
General Physics II PHYS 408
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Date Created: 10/29/15
Two identical capacitors are placed in parallel in a circuit The resulting effective capacitor is how many times larger A B 12 C1D2 E4 Two identical capacitors are placed in series in a circuit The resulting effective capacitor is how many times larger A B 12 C1D2 E4 Four identical capacitors are arranged in the following circuit C1 C3 0 N O h The resulting effective capacitor is how many times larger A B 12 C1D2 E4 1 Which of these 4 0 l circuits gives a 1 1 B 1 different result A 1 E They are all 3 F a different 0 l i l 1 0 131 Cglig D 1 39 Glass Capacitor A parallel plate capacitor has a piece of glass between the 2 parallel plates If the capacitor is charged and then before discharging the glass is removed the resulting spark will be A Bigger spar than you d get if the glass was left B The same spark than you d get if the glass was left C Smaller spark than you d get if the glass was left D Don t know Single Current Loop For the most simple circuits AAAAA VVVVVV e11 Single Current Loop II A current loop is a circuit with any number of elements in it that are connected so that current can ow around the loop We compute such a loop by looking at the voltages across the devices the sum of all these voltages has to add up to zero Circuit Rules Loop Rule The sum of the changes in potential encountered going all the way around a circuit loop must be zero Resistance Rule When a resistor is encountered moving around the loop the change in potential voltage drop is iiR moving in the direction of the current iR moving opposite Where igt0 EMF Rule battery rule When an EMF device is encountered the change in potential is moving in the direction ofthe EMF arrow 7 moving in the opposite direction Example Loop I 1R1 4122 1123 0 Batteries in series R1 Om V V1V2V3 tot AAAAA vvvvvv R2 AAAtit vvvvvv Resistors in series R3 RmR1R2R3m AAAAA vvvvvv N Rtot i1 Mul riLoop Cir39cui rs 12 Junction Rule The total current into a junction ill must equal the total current out of a E E junction conservation of charge EER 5st Mul riIoop Circui rs II Multi loop circuits are solved by combining the loop rule and the junction rule Often you can signi cantly simplify the task by rst simplifying the circuit replacing parallel or series resistors with the equivalent resistor Lec rure 11 Thermo III Chap rer39 20 First Law of Thermodynamics AEth QW Between any 2 equilibrium states the change in internal energy is equal to the difference of the heat transfer into the system and the work done by the system Work by Ideal 605 Now that we have the relationship between the volume pressure and temperature of a gas we can look again at the work that it does when expanding or compressing A P Vf Vf T W pdV j Nk dV V V V V1 V2 V If we keep T constant W NkT I dV NkT 1n V V V If we keep Vconstant W0 i If we keepp constant W p dV 2 pAV The Roo r Mean Square RMS stands for Root Mean Square This is a recipe Take the root of the mean ie average of all the values squared vfv v vj 1 N 2 2 v Zv v rms N N 1 avg X xsquare 1 1 2 4 3 9 4 16 ave rms 25 273861 The RMS speed in a gas The pressure in a gas can be found to be p A Rearrange and use pVNkTnRT 2 3pV 3nRT 3RT 3kT vIms gt vIms nM nM M m At 300K the RMS speed of N2 is about 517 ms Velocity Pressure Temperature demo Translational Kinetic Energy Kinetic energy due to the movement of the gas molecules 2 2 Kavg mv avg 2 va 3kT an m kT Mean Free Path The mean free path answers the question How far does a gas particle go before it collides with another gas particle A 1 J57de NV Relevant This gives a measure of how well the velocities will distribute It is also relevant for the very complicated problem of diffusion Problem A cube of copper Cu of one cubic centimeter at a temperature of 100 C is added to one liter of air at 20 C and 1Atm The gas is confined in a fixed volume What is the final pressure of the gas after the cube is added Useful numbers 1 Atm 1013x105 Pascal R831 JmolK 1cal 4186 J Copper Specific heat c386 JkgK245 JmolK Melting point1063 C Density 896 gcm3 Molecular weight 6355 Heat of fusion207kJkg Air N2 Specific heat at constant volume 207 JmolK Molecular weight2801 Boiling point1958 C ReSI s ra nce How much push electric potential is needed to achieve a certain ow electric current R l X0hm Q l39 A A circuit element for resistor is indicated with the symbol Resistivity and Conduc rivi ry Resistance is a property of a speci c object and so depends on the geometry of that object Resistivity is a property of a particular material E l p27 Qm 02 91ml Ezpj JzaE For simple geometry L R pA Temperature varia rions ppo p006TTo Ohm39s Law Ohm s Law when the current through a device is proportional to the voltage applied across the device When the resistance of the device is independent of the voltage across it A material obeys Ohm s law when the resistivity of the material is independent of the applied magnetic eld Power in Cir cui rs Remember that Power is the amount of energy transferred per time P The amount of energy transferred to a charge q dt traversing a potential difference U 301 dE d V P L IV d1 611 For resistive circuits P I 2R resistors Electromo rive Force Emf What really happens in a battery Charge is pumped from a low potential to a higher potential which requires work We define the EMF as the amount of work per amount of charge g Aq dq But note dW qu V do dq The correct way to interpret this last equation is The EMF device causes a potential difference of V7across it s terminals Single Cur r en r Loop For the most simple circuits L g V E 12 glir 55R R R l A negatively charged particle traveling in the X direction enters an area with a homogeneous magnetic eld in the 2 direction Which statement is incorrect A The particle accelerates B The particle experiences a force C The particles kinetic energy increases D The particle starts to follow a circular path E The particles speed remains constant There is a force so the particle accelerates this acceleration is perpendicular to its velocity so the path is a circle and the speed remains constant The kinetic energy remains the same though because the eld does no work since the force is perpendicular to the motion 2 A crossed E and B eld are set up with E in the 2 direction and B in the y direction A particle with VxEB enters this region The particle will A Go straight through B Will go straight only if it is charged C Will go straight only if it is charged D Not enough information This is exactly what crossed E and B fields are about To make sure about the answer you need to draw a careful picture to make sure the E and B elds do indeed create opposite forces Reread the relevant section in the book ifyou are confused 3 A crossed E and B eld are set up with E in the 2 direction and B in the y direction A charge particle with VxEB Vy5 ms enters this region The particle will A Go straight B Curve because F3 is stronger C Curve because FE is stronger D Not enough information The answer is the same as the previous because the forces did not change 4 A crossed E and B eld are set up with E in the 2 direction and B in the y direction A charge particle with VxEB VZ5 ms enters this region The particle will A Go straight B Curve because F3 is stronger C Curve because FEis stronger D Not enough information In this case the force FE stays the same but FB gets larger because the velocity perpendicularto the B eld is larger Dielectrics If the space between the plates of a capacitor is lled with something other than vacuum or air then the capacitance will increase by a constant factor which is called the dielectric constant Ck kCair Elec rr39ic Current A loop of conducting wire will have the same potential everywhere inside This approximation holds for 2 different cases 1 Electrostatics there is no battery 2 Superconducting wire Battery For normal wire with a battery attached the battery sets up a potential difference and then maintains the potential difference This is not electrostatics Electrons will ow continuously through the wire trying to cancel the potential difference the battery will continue to supply more electrons trying to maintain it The result is a constant current of electrons owing through the wire Electric Current II We define the electric current as the net positive ow of charge per unit time d i q A amperes dt S Current is conserved because charge is conserved What we stuff into the wire at one end must come out the other Except for that little bit of charge that may end up increasing the potential of the wire itself Electric Current Density The current density is the amount of current per unit area 1 J J dA For a uniform current parallel to the area vector we have Jzi A Charge carrier electron drift J 2 me if rte is the carrier charge density in Cm3 ReSI s ra nce How much push electric potential is needed to achieve a certain ow electric current R l X0hm Q l39 A A circuit element for resistor is indicated with the symbol Magnetic Force on a Wire The current in a wire is essentially a large number of charged particle moving with vd In a magnetic eld they all contribute to a force on the wire For a wire segment with a volume of Vand length L a a e A FB zq xB qNeVneFm Vnet7dgtltB a V J nevd gt1 2 AJ 2 Anevd nevd gt Vnevd 21L L FM 2 1L X B So the force on a wire in a magnetic eld is proportional to the current the length of the wire the strength of the field and depends like sine on the angle between the wire and the eld Torque on a Current Loop A square current loop consists of 4 Wire segments With a current i in them What is the force and torque on this loop Q G lt3 3 IE r9 lt gt G e e 0 Fe 2 o G G a u G G l 9 67 E F 13391 i x iaB1339Z istianB ibBcoanB 2 2 2 2 2 3 FF Fpr th FneFO H Np 0 gtlt u H l Np 0 gtlt 9 H u wen u H NF 0 gtlt l Nle c 0 gtlt LU H w 11 B 3 Q Torque on a current loop II The torques from F1 and F3 are the only relevant ones they sum to Where A is the area of the loop Notice the vector component If we had N windings instead of a single loop we get N times the torque so then we nd NE A bar magnet acts like a big current loop The reason 2 bar magnets attract each other is that the field they produce is not uniform so Fmet is not equal to zero Magnetic Dipole Moment We can replace the term iNA in the torque equation with the dipole moment p a a u NA This is independent of the shape of A Compare with an electric We can now write for the torque dil o39ev p in an E eld 1 u x B r p X E The potential energy stored in a dipole is then given by Ul UD Fm Magnetic Field due To a Wire From experiment it was discovered that B uol for a long straight wire the B eld is 27239R Forjust a little segment of wire the eld is d ampid xffz idssin6 2 2 47rr 47 r So now you can just integrate and eld the eld of any current LC Oscillations Analyzed L LO d2 1 VC 1 2 I 2 1 q0 C dq dz dzq dz LC 2 2 VLdl dz dz dz 1 qQsinazt a E vVsinazt V i lcoswt lwQwVC t Radios At the resonant frequency the effective resistance of the LC combination is maximum At all other frequencies the resistance of the LC combination is small So all signals with frequencies other then the resonant frequency are The resonant frequency is found from 1 ZzVLC a LDf xLC 27 lfyou ever try this you will need a coil with low resistance use thick wire a big antenna and a very sensitive set ofhead phones 500 meters long You We and 1600 kHz You l4 quarter wave length antenna should be Lll4cl4l3x108ms39l4 105s39l75 uld want to tune to the Amplitude Modulationquot AM band between Gauss39 Law for39 Magnetism Gauss Law for electric fields was dA39 qenc 80 Because there are no magnetic monopoles Get the equivalent statement for magnetic fields is From this law you can conclude that the magnetic field lines must always close upon themselves forming complete loops Induced Magnetic Fields Except for the absence of magnetic charge the equations for electric and magnetic fields are quite similar You may then expect from symmetry that a changing electric field would induce a magnetic field This is indeed the case and the formula is almost the same as for induced electric fields a dCDE J33 d 080 dz Notice that there is no minus sign and that there now is a constant There are thus 2 ways to create a magnetic field one is with a current the other with a changing electric field Combining this law with Ampere s Law gives the AmpereMaxwell law 0800ienc Displacement Cur r en rquot a old 3 080 OZenc dd In this equation the term so d E takes the same role as a current I We can find What this current is compared to the current that charges a parallel plate capacitor Using Gauss Law we find the E field dA qgo 2 q EOEl Differentiating both sides we find the current charging the capacitor z ngdEg quE 139 dt 0 dt 0 dt So the charging current and the electric flux term are the same For this reason the term I is called the displacement current though nothing is dt being displaced Maxwell39s Equations Putting all our Electricity and Magnetism equations on one page we have Fortne mathematically Sophisticated you can rewrite the equations on the left l lhleOrm Gauss39LawE lSEdAqm80 VEpgo Gauss39LawB g1210 VE 0 Faraday s Law lSEd dq3 VXE a B dt 6 quE uoiem V X B Smut 6 H J dt 6 quot AmpereMaxwell d 080 Entropy John Von Neumann no one knows What entropy ls so In a debate you will always have the advantage Many processes are or at least appear to be irreversible even though no conservation law energy momentum etc would be violated if such a process were reversed Why is this so Examples Your coffee on the table heating up instead ofcooling le heat owing from a colder to a hotter object Two gasses say N2 and O2 separating spontaneously after they were mixed Statistical Mechanics tells us that these things do not happen because they are incredibly un 39 e y It would then be useful to have a measure that tells us how likely a specific configuration of a system is This measure is Entropy Entropy is thus a property ofthe system or entropy is a state function Entr39opy Statistical View Lets say we have a system which can be in a well defined number ofpossible configurations each with the same energy momentum etc Each configuration can be achieved by a set of specific microstates The number ofmicrostates is called the multiplicity of that configuration denoted by W Statistical Mechanics tells us that All microstates are equally probable The most likely configuration is then that configuration with the highest number of microstatesl Or the configuration with the largest W Example Ten uncoupled spin 12 particles Each particle can be in either an up or a down state The multiplicity of All 10 spins are up is then 1 The multiplicity of 5 up 5 down is Thatof6 up 4down W 210 Boltzmann s Entropy Eq So W is a measure of the likeliness of a configuration but W getvery large very rapidly with large numbers of microstates Entropy is more useful The two are connected through the Boltzmann s entropy equation 1896 Skan In a sense this defines entropy The entropy of our previous example the 5 up 5 down state was thus 10 Sk1nWkln 5l5 k1n10l721n5l 7610 23 Big numbers Note that 10 is not such a big number yet 10 is already fairly big Try 100 You will get 93E1 57 on your calculator and most calculators wi not do 101 I But for particles even 100 is not that big Trick Use the Stirling s approximation 1nNN1nN N Second Law of Thermodynamics A520 For a closed system entropy can only increase it can never decrease For an irreversible process the entropy increases For a reversible process the change in entropy is zero Entropy Thermodynamic View Rudolf Clausius 1865 Entropy is the amount of thermal energy put into a system that can not be extracted as work f dQ AS S f S j 7 139 In a cyclic process you can convert heat in to work Le a steam engine However you can never convert all the heat fully into work some of it is lost 59 At constant temperature we have T Entropy as State Function If entropy is a state function then the entropy of a system isthe same whenever it is in the same state Thus a cyclic process must haveAS0 For an ideal gas we can write down the change in entropy between 2 states as Vf Tf ASann nCV1n V T 1 1 Entropy Sr then some Other statements about entropy and the second law of thermodynamics Entropy is associated with the amount of disorder The world only goes from a more ordered state to a disordered state chaos Claude Shannon 1948 finds a relationship between entropy and information Information Theory The amount of information a system contains is directly related to its entropy 1 Pipe A with length L and pipe B with length 3Lhave one open and one closed end Which harmonic ie ha1monic number of pipe B has the same frequency as the fundamental of pipe A Answer nAl fundamental mode so fA 14 vL 3IIB4 v3L so nB3 2 Pipe A with length L and pipe B with length 2Lhave one open and one closed end Which harmonic ie ha1monic number of pipe B has the same frequency as the fundamental of pipe A Answer Using the same formula as above you would find nB2 however in a pipe with one open end only odd harmonics can exist So the answer is It cannot exits 3 I am 10 times as far from the loudspeaker as my friend The sound intensity where I am compared to where my friend is is Answer llOOth since the intensity goes like IP411r2 4 I am 10 times as far from the loudspeaker as my friend The sound level where I am compared to where my friend is is Answer 20 dB since 3 10 dB logIIo and in part 3 we found IIol100 o EXAM II Electrostatics Chapters 22232425 Thursday March 27 at 7pm Iddles 101 Rules Calculator OK One page ofyour own formulas no fully worked out problems Page to be stapled to exam One problem preannounced EXAM SCORES EXAM Grade Distr Average 74 Class Participation m w n r Par dpation Elec rr39ic Po ren rial Energy Electrostatic Force is conservative so we can assign an electric potential energy f In general AU W J a i f For electric field AU J qOE 39 aquot Compare with the gravitational potential 7 force due to gravity and contour lines on a map n ror39 lines Whi re y Electric Potential The electric potential energy per unit charge is more useful we call this the electric potential V AVA U lEd q The unit for V is the Volt Joule per Coulomb Yes this is the same Volt as you see on a battery We can now change unit for E from NC to Vm which is more customary Electron Volts Note that qAV has units of energy The electron volt eV is a common unit for energy in particle physics It is the amount of kinetic energy that an electron obtains when it is accelerated by a potential difference of one volt 16v 61V 160X10719C1 160gtlt1019J Masses can be expressed in terms of eVoz which gives 0511x10616x10 1 El t M o511MeV 91103931k 80 ron ass 42 3i0x108 I2 X g Proton Mass 9383 MeV 167x1031 Equipotential Surfaces If you drawthe lines surfaces Where the potential has the same value you get something that is very much like the contour lines on your hiking map Properties Equipotential surfaces are always perpendicular to the eld lines Equipotential surfaces are spaced according to the field strength Equipotential surfaces at different voltages can never cross each other Calculating the Potential If you know the electric field you can calculate the potential from the definition of the potential AVA U jEd q Calculating The Potential poim charge For a point charge we get mg 60 1 1 w Vf 0Vr jEd 39 j7dw4 7 Q 1 ir 47r80r Principle of Superposition Just like E where we can sum the total eld due to a number of sources vector suml we can get the total potential due to a number of sources by adding the different potentials This is easier than summing E since it is a number not a vector Po ren rial due To a Dipole For a dipole the potential can now be calculated as VKK 1 q Kr r 47280 1 7 472290 1 2 Forrsufficiently large 7 r I I dcos 1 qpcos 1 90056 2 V 2 472190 r 4750 r Force due To Magnetic Fields We know about magnetic elds because of the force they produce Similar to the de nition of Electric elds we can de ne the magnetic eld by the force on an elementary paricle gt gt FB X B which you can compare to FE Although these formulas are quite different I d like to note that the underlying phenomenon that causes the electric and magnetic fields is identical One can talk of the electromagnetic field Units for B 1 Tesla 104 Gauss 1 NAm Earth s field on surface is about 1 Gauss 10 4 T For39ce due to Magnetic Fields II F B qv x B Because this formula consists of a cross product between the velocity and the B eld 1 If the particle does not move wrt the eld there is no force 2 The force is always perpendicular to the eld and perpendicular to the velocity No work 3 Only the component of the velocity perpendicular to the eld causes a force 4 Sign of the charge determines the sign of the force vector Magnetic Field Lines Since the magnetic eld is a vector eld we can draw it like lots of little vectors or we can draw it like eld lines Drawing Magnetic Field Lines 1 Magnetic eld lines always form closed loops 2 Magnetic eld lines run from the Magnetic North to the Magnetic South of an object Unfortunately due to history the geographic North of earth is a magneticsouth 3 The eld lines never cross 4 The force due to the eld does not go along the eld lines Crossed E and B fields A particle crossing a region with an electric eld perpendicular to its motion will de ect A particle crossing a region with a magnetic eld perpendicular will also de ect What is the condition so that these 2 de ections cancel E Bquqvx m Independent of q or m Note that the elds must be perpendicular to each other in such a way that E 75x1 Now that we know a way to measure v we could use this in combination with the de ection or acceleration due to an E eld alone to nd the ration m q The Hall Effect The electrons inside a semi conductor also experience the effect of an external magnetic eld so they will also be de ected This sets up a measurable potential difference between the edges of the conductor If the drift velocity is Vd the equilibrium condition is e E evd B Why is this not obvious We can relate the drift velocity with the current density to get a a 1 39 J nevd Svd Esz 1 ESE 133V 133 13 2 d n nel Vle I A solenoid like inductor with 100 windings is found to have L004H You now double the number of windings to 200 The new inductance you would expect is approximately A 002H B 004H C 008H D 016H E Cannot be calculated You need to use LN So it looks like there is only one power of N However there are 2 ofthem so it is N2 the second N is part of the Bflux For a solenoid LuON2Al There will be an N2 for all other geometries as well V0 Hm 73 000900 11 An large ideal inductor R0 is added in series to a circuit with a battery and a lightbulb and a switch When you now close the switch A The light bulb will stay off B The light bulb will burn as normal C The li ht bulb will increase in bri htness until it burns normal D The light bulb will start normal but then decrease in brightness to zero E The extra Emf from the coil will burn out the light bulb s R V0 C I E An large ideal capacitor is added in series to a circuit with a battery and a lightbulb and a switch When you now close the switch A The light bulb will stay off B The light bulb will burn as normal C The light bulb will increase in brightness until it burns normal D The li ht bulb will start normal but then decrease in bri htness to zero E The extra charge from the capacitor will burn out the light bulb WhaT is The ampIiTude of The sTanding wave given by This equaTion y39xtEYm cos wt sinkx j AYm N Ym C27r mm E O WhaT is The ampliTude IT The poinT x 2 ALJEYmsin7z B my C Y m 1 MY cos wt E 0 When 12471 and 9622 ANSE since sin370 Traveling Waves Part I Transver39Se Waves on Strings Chapfer39 17 Sec 15 WAVES Types of Waves 1 Mechanical Waves travel through a material wave properties are governed by Newton s Laws EG Sound Ocean Waves Guitar String Electromagnetic Waves Waves are oscillations of the E and B elds which are governed by Maxwell s Equations Matter Waves Particles also behave as waves but it is not so clear what it is that is waving Governed by Quantum Mechanics Gravity Waves Yes gravity also has waves They are governed by General Relativity Einstein s formulation of gravity Colloquium on this March 24 2003 4pm DeMeritt 209B by Rainer Weiss MIT Two Classi cations of Waves Transverse Waves The movement of the material is perpendicular to the direction in which the wave travels Ocean waves Guitar String Longitudinal Waves The movement of the material is along the direction in which the wave travels Sound Chapter 18 Note that while the MATERIAL of a wave moves back and forth the wave itself moves onward Equations for a Transverse Wave Step by Step Take a string which is moving with a wave on it One small element of the string at the location x0 will be moving up and down We know that type of movement from Chapter 16 simple harmonic motion Thus the equation can be written as ytx0ym sinwt 0 11 The element right next to this one also moves but it is a little delayed that is the phase is a little later for the same t If this element is at xxl the effect of the delay can be put in by setting 1 0 kxl The equation of motion for this element is thus ytx x1 ym sinWkvq o If we keep going for all x we get the general wave solution for a transverse wave yxtym sinat kx 0 12 NOTE Your book has sinkx wt This is really the same as eq 12 except that in the book they left out the phase factor They should not have The phase factor allows the wave to have a value different from 0 when both x and t are 0 Note also that you could have used cos as well as sin the only difference would be the value of The transverse 39 39 wave formula yt x 2y sin kx al nj The different parts of this formula are ytx Displacement Ie the wave s movement is in the Y direction ym The Amplitude sin 06 at IA Oscillating term All harmonic movements must have this 106 mt The Phase This is confusing since 1I is called the phase constant thkw a Angular frequency Wave number and phase constant Properties of a traveling transverse wave m 7 Wave Length y X This is the wave plotted versus x The height is the Amplitude the distance between the peaks or the valleys or the upward zero crossing is the Wavelength y t This looks very much the same However since it is plotted against t the distance between the peaks is the Period of the wave How they relate 27 Angular wave number a 27 Angular Frequency 1 a f Frequency Speed of a wave Look at the equation for a wave once more yxt ym sinkx wt 0 This wave has a peak when yxt is maximum which happens when the sin term is 1 so the phase part of the equation must be 7 Thus the peaks occur at the places where 7239 kx wt 2 13 6x We want to know the speed at which the wave travels Since v 5 if we differentiate the equation with respect to twe get 3x a kE Q0 0gtV We can write this in terms of other variables w vT f 15 Transverse Velocity and Acceleration A different velocity is the speed at which the element of the string or the wave carrying medium move up and down This is the transverse velocity and the change of this velocity is given by the transverse acceleration We obtain it by simply taking the derivative with respect to t of the wave function yxt u sinkx at t 16 aymcoskx at 0u62yxr 2 ay E T w yms1nkx wt 17 Wave Speed on a stretched string The preceding way of nding the speed of the wave did not really tell us anything It only related the speed v in terms of a and k which then begs the question what are a and k W The speed of the wave is really a property of the medium in which the wave travels A sound wave in air goes with the speed of sound a light wave with the speed of light So what about a string There are really only 2 properties of the string that could determine it s speed the mass of the string the length of the string and the tension on the string We can do a dimensional analysis to get the speed of the wave However please note this is just a trick We get the correct formula but we did not M learn why it is the correct formula The speed is v so v has dimensions of lengthtime Note that dimensions are different from units The mass of the string is better expressed as the linear density u 7 with dimensions of mass length The tension 239 in the string is a Force so from F ma we know this has dimensions of masslengthtime2 We want to get lengthtime and there is only one way to do this T has dimensions of u lengthztimez so v 18 There could be a dimensionless constant but in this case that is actually equal to l Speed of a wave in a string the real way The actually correct way to calculate the speed of a wave on a string is really hard The book does another tricky way which is very clever I am not so satis ed with it though Below is a more correct derivation which is quite advanced I am showing you this to get a sense of how to do it The derivation will not be on any test but the result will We look at an element in the wave of size dx If we now zoom in on this element and exaggerate the y displacement of it we get this picture mass ucbc dx The red segment is the part we are interested in We now go back to Newton s Equation F ma and equate each side 02yxt Fy may udx atz 19 But what is F It is the sum of the two tensions The magnitude of the tensions are TL equal but their directions are slightly off We add them as vectors The gure shows how the vector add What you can here because we exaggerated the y displacement is that for a large y the net force F is not quite in the y direction Now we make an approximation that TLX TLy H l39 This holds quite well since the displacement is y is small The vectors sum in x then cancels in y we get Fy TLy TRy we now rearrange making use ofthe fact that the slope of the yxt curve is equal to the derivative with respect to x F Thhrw J 110 6yxt y r 139 6x 6x Li R Setting our 2 equations 19 and l 10 equal to each other at ax L 6x R 2 111 m MM 5yxt i at2 I 6x L 6x R dx As we let dx gt 0 the right hand side term is equal de nition of the second derivative with respect to x We then get the Wave Equation for a wave on a string 52yxaf362yxar 62 I 6x2 Like all differential equations we guess the answer actually we know it already We plug equation 12 into 112 for the derivatives see 17 and we get 0 112 a2 ym sinkx at k2 ym sinkx at 0 113 This will hold true since the terms in are the same when 2 2 T 2 Q T Q T k gtgt 2 114 a k2 V k Which is what we set out to prove Magnetic Field due to a Wire From experiment it was discovered that uol for a lon strai ht wire the B eld is g g 27239R Forjust a little segment of wire the eld is d ampid xffz idssin6 47 r2 47 r2 So now you can just integrate and eld the eld of any current Integrate a long straight wir39e Integrating an in nite straight Wire using the BiotSavart law P so BZJdB idsstZl jJstds Iquot 47139 r2 472150 r2 R Fromgeometry rR2s2 R R s1n6 d V IRZJrs2 B de3WRL l 47r7wR2Sz 47239 R2 RZHZ 27239R Were we used the dx 1 00 00 x 1 And the limit standard Integral x2 a2 a2 x2 a2 002 JrRz 002 For39ce Between 2 Parallel Cur39r39en rs One wire with a current will exert a force on another wire with a current and vice versa We find the force between 2 wires a distance d apart by first calculating the field of one wire and then the force of that field on the other wire i Wire 21 causes a field of Ba h 27rd The force on wire b is then F in X E IuOLl lb a The direction of the force is towards the other wire if the currents are in the same direction and away from the other wire if the currents are in opposite directions Gauss39 Law Gauss Law the net number of field lines through a closed surface must be proportional to the amount of charge inside 1 EdA 80 Breakdown Qenc is the charge inside the Gaussian surface Only the charge inside counts Qenc could be 7L or 0 or p depending on how the charge is distributed Any surface can be used in any of the cases Only those surfaces that use the symmetry of the charge distribution spherical cylindrical planar are useful practical to obtain E E and A are vectors The most useful surface is that surface Where E is either parallel or perpendicular to A everywhere Planar Symmetry Thin nonconducting sheet with a surface charge density 0 Thin conducting sheet with a surface charge density 0 U E 80 Remember the field inside a conductor is always zero Cylindrical Symmetry A long Wire or a long cylinder with a charge per unit length of A E 4 2721901quot Does it matter whether the cylinder is made of metal or of plastic A Yes B No C Don t know Spherical Symmetry Point charge q 80 E 121 gOEgS dA gOE4 r2 gm 3 gene 4 80 r2 Spherical shell with surface charge density of o 2 E201 rZR r E20 rltR A uniform charge distribution Charge Distributions pr0 r2U rm mm b qenc at r is What in i qmvwnpvvdv R 2 4 V 130 per2 cos6d6d dr E R3p0 r gt R0 7127 E eldatr g EdZiEHrzcosad0d 4m2Eqe 0 0 0 55 mo Er I p 380r2 rgtR0 0 Charge Distributions II A charge distribution varying with r p r 1 r lt 139U qenc at r is What L0 j 6d6d dr 2mm rSRO i qmrmpr dV 2 V R If rzcos6d6d dr2 R2A rgtR0 0 739 7127 E eldatr g EdZiEHrzcosad0d 4m2Eqe 0 0 0 A VS 2 R0 Er 2 AR rgtR0 Electric Potential The electric potential V calculated from the E eld AVzA Uz ijd 39 q The unit for V is the Volt Joule per Coulomb Yes this is the same Volt as you see on a battery We can now change unit for E from NC to Vm which is more customary A useful unit for energy is the electron volt the amount of energy it takes to move one elementary charge an electron through a potential difference 0f1V0t 1 eV 41V160x10 19c1160x10 19J Mass can then be expressed as eVcZ or more usual is MeVcZ Electron 0511 MeVc2 proton 938MeVc2 Calculating the Potential point charge For a point charge we get 00 00439 00 Q 1 Q 1 V V Ed 39 d f o 1r l s 4550 r 2 r 472290 r R Principle of Superposition Just like E where we can sum the total eld due to a number of sources vector sum we can get the total potential due to a number of sources by adding the different potentials scalar sum Vz Ii 1 N This is easier than summing E since it is a number not a vector Po ren riol due To a Dipole For a dipole the potential can now be calculated as VVKZL 2L rn 47590 r I 47590 m 2 Forrsufficiently large 7 I I I dcos 1 qd cos 6 1 9 cos 6 V 2 2 472190 r 4750 r Continuous Charge Distributions Analogous to the Efield due to a continuous charge distribution lecture 19 we can find the electric potential due to such a distribution as an integral dV 1 Vzlde lj 4723980 r 4723980 r Name Symbol Integration element Unit Charge q dq q Q C Linear Charge Density A dq Ads 7FQL Cm Surface Charge Density 0 dqadA oQA Cm2 Volume Charge p dqpdV pQV Cm3 Density Finding E from V We can invert the formula which calculates V from E to obtain a formula for calculating E from V however the dot product makes this tricky If take the case where E is along s we find 4 8V V jEd 3 de Eds E 6s In general we can write E 6 V For cartesian coordinates ie xyz this is EX 2 3le aVjEz 5V 3x y 3y 62 Example parallel plates Electric Potential Energy for39 System of Charges A system composed of point charges would require work to put together The amount of work is equal to the electric potential energy self energy of this system Two charges 1 01qu 472290 d l q1 gtU V 472290 d qz 12 V12 Since the potentials sum this can be easily extended to more than 2 particles U q2V12 q3V123 q2V12 q3 V13 V23 Po ren rial of 0 Conductor Since the electric eld inside a conductor is zero the potential everywhere in the conductor must be the same An excess charge on a conductor will distribute itself on the surface of the conductor so that this is true Amme rers amp Vol rme rers An ammeter measures the current Amps in a wire or a device To do this it has to be inserted in the wire so that all the current ows through it in series with the device The ideal amm eter has itself a very small resistance approaching zero so that the voltage drop is negligibly small A voltmeter measures the voltage Volts across some other device eg a resistor or a battery To do this it has to be connected across in parallel to that device The ideal voltmeter has itself a very large resistance approaching infinite so that the current owing through it is negligibly small RC Circuits When the switch is open there 4 is no charge on C so V50 At t0 when the switch is rst closed this condition still holds since there was no time yet for any charge to ow So V the voltage drop across R is V0 0 7d and 10VER Ifwe wait for a long time then C charge will have owed to C until the potential across C equals that across the battery VCV0 The capacitor is now fully charged and there is no current RC Cir39cui rs II What happens between 10 and Ivery large Charge will flow at a rate of ildqdl from the battery through the resistor to the capacitor The voltage across the capacitor will be VqC We can write the circuit equation going around the loop as VoiiRi q0I07qu7103 C C a q5 dt RC R This is a differential equation which has a solution in the general form if q A Be AC A and B are constants which we determine from the initial conditions At 10 q0 from which we learn that 0AB so BA and at 100 VcqCVU so we must have qVUCA0 thus AVUC The solution for a charging capacitor is thus qZVOCleltzQCgt RCCir cui rs III A charging capacitor thus has a time dependent charge 7 on it that follows the e uation RC q q t VOC l e j The voltage across the capacitor is just VI qC which gives V I 2 V0 1 el aclj To get the time dependent current we differentiate ql wrt I which gives get C The term RC has units of seconds and is called the time constant 1 After one time constant t 1 the capacitor will have charged to le39163 maximum value and the current will have dropped from IfVDR to le37 of that value RCCir cui rs IV A discharging capacitor has fairly similar equations The solutions are 1 t Q N V0Celt Ac R V t V0e C C lt Voe ec 69 R Physics 408 Chapter 22 Electric Charges John Dawson February 28 2003 Electric Charges In materials an excess or lack of electrons Conductors electrons carry the charge In semiconductors electrons and holes Elementary particles carry charge Integral charges e e p nn07r Nonintegral charges quarks u d s t b Only two types of charges and What is it Force Law Action at a distance F k q1 q2 r2 Charge is in coulombs C mC uC k14nso9x109 Nm2C2 Superposition principle F F1 F2 Looks like gravity but is not the same Charges can repel as well as attract whereas all masses attract each other This is impossible for electric charges More about the force law Vector form of the law F12 is the force of charge 2 on charge 1 Example 1 Ch22p5 Forces on a charge Example 2 Ch22p8p14p15 Equilibrium of Electric forces Example 3 Ch22p27 lonic bonding in CsCl cesium chloride Lecture 7 Thermo Part I Chapter 19 14 Introduction The whole topic of temperature can on the face of it seem very obvious You feel warm or cold you look up the temperature outside on a thermometer These are everyday experiences The physics of temperature is a bit more complicated though It deals first of all with what this phenomenon of temperature is and how we measure it We will then relate temperature to other physical quantities heat expansion specific heat work and thermodynamics Next chapter deals with the kinetic theory of gasses and finally we will get to entropy While temperature seems simple and everyday the concept of entropy is usually considered rather difficult to grasp On the whole the topic of thermodynamics is far from trivial Zeroth Law of Thermodynamics If bodies A and B are each in thermal equilibrium with a third body C then they are in thermal equilibrium with each other In more common language Temperature is a fundamental quantity of objects When 2 objects have the same temperature they are in thermal equilibrium Measuring temperature So we know something about temperature and if we can measure this quantity then we can predict if two objects will be in thermal equilibrium We can measure temperature by any one of the physical property of an object some material that changes with a change in temperature The most obvious one is expansion but change of color radiation spectrum works too especially for hot objects and change in resistivity is used in digital thermometers The rst thermometers where made of mercury in a glass tube The rst scales were set by Daniel Fahrenheit and Anders Celsius in the 18th century The Fahrenheit scale which is really only still used in the US was based on a zero point at the temperature of a mixture of water ice and salt This was the coldest temperature that Fahrenheit was able to make in his laboratory The other point he chose was his own body temperature which seemed very stable to his thermometer which he set to 96 on his scale Guess is that he chose 968 12 because it is easier to divide than 100 On his scale the melting point ofice is at 32 and boiling at 212 Note that later the scale was recalibrated using the 32 and 212 point and unless Fahrenheit was a very cold blooded person the body temperature was thus moved to 98 Celsius started out with melting ice for 0 and boiling water for 100 This scale has some benefits over the Fahrenheit scale since these are well defined points and easy numbers Kelvin Much later Sir William Thompson Lord Kelvin came to some very fundamental discoveries about temperature First he realized that there is an absolute zero a temperature l2 272003 below which you can not go This is quite profound With most other measures you can always keep going more and more negative but apparently not with temperature This insight is closely related to the workings of a constantvolume gas thermometer about which you should read in your book He took this point for his zero and the triple point of water he set at 27316 This number was chosen so that the spacing of the degrees on his scale are the same as those for a Celsius thermometer Converting different scales To convert from one scale to another the following rules apply From KelVin scale to Celsius To T 273 15 From KelVin or Celsius to Fahrenheit TF TC32 T 24115 22 272003 Calculating the Capacitance Recipe II The alternate method which is not in your book Use Gauss Law to calculate the relation between E and q Use EdV to relateEtoV C21 ds V Parallel Plate Capacitor 93Ed qenc 3E q 1 80 A80 so L 13 A d d E 3E 90 dz d V1480 3C q Calculating the Capacitance Kb 1 Cylindrical geometry Gauss Law 27rrLE gm SE 9 0 2723980L r r C dV dV 3 1n 2723980L E ds dr 9 AVVa Vb MOL 9 lna lnbmln Calculating The Capacitance IIIb l Gauss Law 4 39F2E 1 E 12 80 4723980 r Ez dlsz1C dr 47rsor AVVa Vb 1 Ll 1 I m 47590 a I 47590 ab 47rg V 2C147rs0 ab 7 0 V b a ab Capacitors in Parallel and Series When you combine capacitors in a circuit in a parallel or series configuration the result can be simplified as a single equivalent capacitor of different capacitance C1 02 C1 12V 7 7 Q C2 12V Parallel Capacitors Visualize using a parallel plate capacitor Two identical capacitors in parallel the C lici resulting capacitor would have the same 12V distance between the plates but twice the area so the equivalent capacitor is C 90 at 80 A1 142 80141 tot d d d More accurate for any two nonidentical capacitors is that the total charge stored on the capacitors sums while the voltage across them stays the same qcz d CMV qm q1 q2 q3C1VC2VC3V 2 N Cm C1 C2 C3 2Q 11 Capac i ror39s in Series Visualize using a parallel plate capacitor C1 Two identical capacitors in series the resulting capacitor would have the same area 1 I but twice the distance between the plates so 12V CZ the equivalent capacitor is 50A 50A 1 1 00 C d ddd d 11 CC O 1 Z ttle AACl zlz 1 1 1 Ctat Cl CZ More accurate for any two capacitors is that the total voltage across them sums while the charge on them is the same 1 1 Vm V1V2V3 1 1 1 C i q q 1 C2 C3 V 202 Energy S ror39ed Energy stored in a capacitor q 1 CV2 Volts Energy Density for E eld Voltsl 8014V2 U 1 CV2 d 80 V2 1 2 u 2 go I Ad 2161 2 d 2 Volumel D Ie lec r r39I cs If the space between the plates of a capacitor is lled with something other than vacuum or air then the capacitance will increase by a constant factor which is called the dielectric constant Ck kCair A primer of Circui rs Circuits are build up of a number of different element each of which perform a speci c function In the type of circuits that we will be dealing with these functions are independent of each other that is each element responds only to the speci c input and output it receives not how this input is generated Wire carries current ie allows charges to move Connection The dot signi es that two wires are connected to each other allow current to ow either way Switch which is open does not allow current to ow Switch which is closed allows current to ow Battery with potential indicated Abattery sets up a potential difference between the two wires and maintains that potential Constant current source provides a current in a wire and maintains that current Example Cir39cui r Storing Electric Energy Several methods for storing electric potential energy 1 Battery Energy is stored and extracted with the help of chemical reactions N Capacitor 7 Energy is stored in the electric eld itself E Magnetic 7 Energy is stored in a magnetic eld we ll learn about this later Capacitance We call the capacity of a capacitor to store electric energy it s capacitance q gt C 2 Unit F 0N Farad C depends only on the physical properties of the capacitor geometry and dielectric material L 7 O Circuit symbol for capacitor Calculating the Capacitance Recipe Use Gauss Law to calculate the relation between E and q Use V lE d or EZ V torelateEtoV C21 3 V Parallel Plate Capacitor 95 M 2 qm we 90 V1480 a V 3 q T V jEd EdE 7 d Ag 0 C1 V d Calculating the Capacitance II Cylindrical geometry 2mLEMgtEe 1 6 0 a Z gorL b 2 c Z ioL mil q dr q 11mg 1 1n Inl l b Z gorL Z EOL b Z EOL a Spherical geometry 47590 r2 V4126 q 1quot 4 L1 4 ba 3 b47rsor2 47590 rb 47590 b at 472290 ab ext 60 A spherical non conducting shell carrying a uniform charge distribution 60 is placed at the origin in an external uniform electric eld in the x direction I What is the correct expression for the eld at A inside the spherical shell x c 4 R2 a A 2 AE 0 BlEEexrl C E UO 2 f D EEmiUOR2 9 01 01 E You cannot use Gauss Law because of the external eld so this is too complicated to calculate II What is the correct expression for the eld at B outside the spherical shell The electric eld is always the sum ofall the electric elds caused by the various con gurations of charges In this case in region A the contribution due spherical shell is zero from Gauss Law so the total electric field isjust the eld produced by the external eld In the case of region B we need to sum the contribution of the shell to the contribution due to the external eld The contribution due to the shell comes from Gauss Law again The correct answer is thus D III Would the answers for I amp II be different if the shell was conducting instead of non conducting A I amp II would be different B Only I would be different C Only II would be different D Both would stay the same E Not enough information to answer Both would change For region A the answer changes because there is never any electric field inside a conductor 80 inside a metal sphere the field is zero even if there is an electric field outside This is so because the charges on the metal sphere redistribute themselves to cancel out any external field For region B the answer changes because as just stated above the distribution ofthe charges is no longer uniform once the shell is placed in the external eld 1 Which resistors are in parallel A R1ampR2 B R1ampR3 C R1ampR3 R2ampR4 D R2ampR4 E None 11 Which resistors are in series A R1ampR2 B R1ampR3 C R1ampR3 R2ampR4 D R2ampR4 E None 111 If R40 Ohm can you immediately write down the answer for the current through R2 A No you need to do the loops B Yes it is VoR1R2 C Yes it is V1R2 Vow R112 sz D Yes it is R1 R5 0 R112 E Yes it is R1 R3 a IV If R4oo Ohm can you immediately write down the answer for the current through R2 Same choices as above 1 Which resistors are parallel A R1ampR2 B R3ampR5 R4ampR6 C R1ampR3 R2ampR4 D R1ampR3ampR5 R2ampR4ampR6 E None 11 Which resistors are in series same answer choices as above 111 The currents indicated in graph 101112 relate as follows A1021112 101112 10 1112 10 11 12 None ofthese For39ce due to Magnetic Fields II The Hall Effect The electrons inside a semi conductor also experience the effect of an external magnetic eld so they will also be de ected This sets up a measurable potential difference between the edges of the conductor If the drift velocity is Vd the equilibrium condition is e E evd B Why is this not obvious We can relate the drift velocity with the current density to get ineadvdijJ ne neA EZZ BEKZESVZESIlz neA d neA nel We Cool experiments Allows to discover the sign of the charge carrier Allows to measure the drift velocity Allows to measure the number of charge carriers Allows to measure the magnetic field Particles moving in Circles Since the magnetic force on a particle is always perpendicular to the direction of motion the particle will move in a circle 2 v FC ma3FC m 2 r EIggtmV qugt r rzmvi amp 613 613 lfthere is also a parallel component to the velocity we get a helix for path If we know B we can measure the ratio of a particle s momentum to it s charge Particle Detectors CLAS Cyclo rr39on Frequency How long does it take the particle to go around the circle 27rr27r 27rm T v v qB qB l B B fq wq T 7rm m Magnetic Force on a Wire The current in a wire is essentially a large number of charged pa1ticle moving with vd In a magnetic eld they all contribute to a force on the wire For a wire segment with a volume of Vand length 133 q17gtlt 13m ner7dx j nex7d gtiAJAnevd gtVnevd 139L E0tiigtltl So the force on a wire in a magnetic eld is proportional to the current the length of the wire and the strength of the eld I3B 9 D320 m vquotquotquotquot 5 X x xx x m o a quotquot quot 39 L 39 ofa solenoid it such that the current comes out ofthe plane for the top and goes into the plane for the bottom The current in the solenoid is Varied so that the magnetic eld changes as BBut where Bu is a constant There are 3 different sets ofwindings with this solenoid Winding A is right at the windings ofthe solenoid so close all the B eld must go through it WininsD r t quot quot 39 39 eldAnd windings C t completely inside with a radius 2 that ofthe solenoid Each ofthese 39 39 L t s 39 urn How does the current in these windings compare to each other Answer BAgtC Even though the coil B is in an area where there is no magnetic eld it still encloses a region that does have a magnetic eld The ux throu h B is thus not zero actually the ux through B and A are the same the entire ux through the solenoid The gyfor A and B is then g 7 dig rerolmw rer mMBn and the current is this divided by R the resistance not the radius For C the area of the loop is smaller so the enclosed ux is smaller 14 to be exact so the g is also 14quot Lecture 4 Transverse Waves Part III We finish Chapter 17 and start 18 Interference When 2 waves interfere we can use the principle of superposition to add up the resulting waves y xty1xty2xt 11 For the special case of two sine waves that travel in the same direction with the same frequency the same wavelength and the same amplitude but a different phase this result simpli es Using the trig formula sin a sin 2 sin Ga cos Ga for two waves we get y xt ym sinkx atym sinkx at 12 2ym cos smkx wt Note that it does not make a difference whether a kx wt and kx mt or the other way around since cosxcosx In equation 12 the red part is the new amplitude and the blue part the new oscillating term Phasors When the two waves on a string do not have the same amplitude we can still add them In fact we can add any two waves using equation 1 1 One trick for adding two waves with the same frequency and wavelength is a phasor diagram Because of the sine term the displacement of a wave can be seen as the y component of a vector that is rotating with an angular velocitya Two waves can then be added as two vectors which is easier than two sine terms It is far easier to add two vectors and then look at the resulting vector s angle and y component than it is do the math for the sum of two sine term l4 1302003 Standing Waves When two waves travel in the opposite direction of each other we can still perform the same trick as equation 12 except that the term arrange themselves slightly differently y xt ym sinkx atym sinkxat 13 Zym cosar s1nkx Note that really just the at term hopped from the sin to the cos term The result is a very different wave though Note that since the kx at term is gone the resulting wave not longer appears to travel The blue part equation 13 gives the shape of the wave with respect to x This shape means that the amplitude of the oscillations with frequency at of each element of the string changes with position Another way to look at the same thing is that the blue part defines a shape with respect to x and the green part causes the amplitude of this shape to change with time Such a wave is called a standing wave Properties of a standing wave The locations along x where the nodes occur the point where the amplitude is zero are found when the sin kx term is equal to zero thus with n0l2 L A A kx n7r sz n k2 n7 14 Note that the book sets the phase term equal to zero which simplifies the equation We can always find a clever location for the origin ie at a node so that we do not need the phase factor Similarly we can find the maxima which results in n7r n k 2 47 Trick question What happens with the energy in a standing wave Does it still get transported like it does in a traveling wave kx n7l39 gtx 15 Re ections at the end of a string When a wave hits the end of a string which is clamped down fixed this wave can not travel any further The energy can not travel any further either since the point can not do any work W F Ax and Ax is zero The only option for this energy is to re ect back which is what it does The wave will invert itself and travel back where it came from It is inverted so that the fixed point can act as a node When the same thing happens on a loose end a ring sliding on a rod the energy can not go further either There is a Ax but now there is no F The wave is re ected back without inverting 24 1302003 Standing Waves and Resonance An ideal string clamped between to xed points can have a wave on it that continuously re ects back and forth If the string is not ideal the wave will damp out and we would somehow need to add some energy to keep it going For certain frequencies called resonant frequencies all these re ecting overlapping waves will perfectly add up to one large standing wave For off resonant frequencies they will tend to cancel each other out The lowest frequency for which such a standing wave occurs the fundamental mode would have exactly one half wavelength between the two clamped points 2 2 L 3 2 2L The next one up would have 1 wavelength the next one 15 etc So 2Lgt 16 2 n Or using v if we can write v v 2 17 f A quotZL When nl we talk of the fundamental mode or the first harmonic with 712 we get the second harmonic etc Sometimes n is called the harmonic number Since we count the nodes at the end the first harmonic has 2 nodes one loop The second harmonic has 3 nodes and 2 loops etc Standing Waves on a clamped String n2 second harmonic L n3 third harmonic Loops nodes 34 1302003 Sound Waves Sound is a longitudinal wave that travels through a material in the form of small compressions and expansion along the direction of travel of the wave Speed of Sound To get to the speed of sound we rst need to know something about the compressibility of the the material the sound travels through This is given by the amount of pressure that is created when a certain volume is compressed a little bit The quantity is called the bulk modulus and comes from the equation AV A B 18 P V VJE 19 p where p is the density of the medium The speed of sound in air is between 331 at 0 C and 343 ms at 20C the speed of sound is then given by 44 1302003 Two metal spheres 81 with a radius of 3 cm and 82 with a radius of 9 cm are separated by a large distance They are connected by a thin conducting Wire If a charge Q is placed on 81 what is the final equilibrium charge on each of the spheres A QlQ2 Q2Q2 b g 21 24 223 24 c Ql3Q4 Q2Q4 D Q1Q Q20 E I don t know The voltage on both spheres must be equal and charge is conserved so V1 V2 becomes Q1R1Q2R2 and Q1Q2Q If I double the area of the plates of a capacitor and I also double the distance between the plates the total new capacitance will be A 2 times b four times c 12 times d 1A times e same 2 so if both A and d are twice E Q 3Vszz jc z2 80 80A 8014 V the capacitance stays the same In a capacitor energy is stored in A The charge on the capacitor b The electric field c The potential on the capacitor d The metal of the plates e none of the above Alternating Current AC So far we have dealt with batteries for which the Emf and thus the voltage is constant This is often called DC which stands for Direct Current The wall outlet and many other devices use AC which stands for Alternating Current For AC circuits you have a constantly changing voltage and thus a constantly change current The gwfor an AC circuit would then take the form 3 sinat Note in terms of music this is a pure note having only a single frequency fan21 However any complicated signal can be broken down into single frequency components this is called a fourier transform so what you learn by analyzing circuits with single frequencies can be extended to any complicated wave form Ref Ch336 A r eSIS l39Of39 in AC A resistor is the simplest circuit S R element in an AC circuit In the simple 1 I I a l R circuit on the right the current runs immediately as the switch is closed so there is no time component Since there is no time component for a resistor it behaves the same as in a DC circuit so Olt1 AAAAA VVVVV l VR 2 VR sin wt VR 12F51nat IRsmat Rt VR IRR AAAAAA VVVVV Ref Ch338 P779 A capacitor in AC 8 1 R A capacitor in an AC circuit does have time dependent behavior In the circuit on the right at t0 the capacitor has a voltage across it VC0 in the long run VCV0 A V0 C different way to say this is that the capacitor l is a conductor R0 for rapid changes high frequencies and an insulator Roo for slow changes low frequencies Instead of R we use the capacitive reactance X 7 1 C 7 a C VC 2 VC sin wt d g C i sinat90 21C sinat90 lN VC 2 I CX C amplitudes Ref Ch338 P781 Induc ror39 in AC In the circuit on the right at t0 the inductor S has a voltage across is VLV0 in the long run 1 the voltage becomes VL0 A different way to say this is that the inductor is an insulator Roo for rapid changes high frequencies V0 L and a conductor R0 for slow changes low frequencies Instead of R we can use the inductive reactance XL de 76 0000 VL VL sin wt 000000 isinwt 90 ICsinat 90 L VL I LX L amplitudes LC Oscilla rions 0 bl ll 000000 LC Oscilla rions Analyzed L L0 d2 1 VGl 2 I 2 g q0 C dqjdi dzq dz LC 2 L dz dz dz2 L dz qQsinat wz q Q V t V v C s1na C lag dz Icoswt l wQ wVC Radios oooooo L I You have one battery of 5V and 2 identical 025 Watt 5V light bulbs You hook these up in the two ways indicated above Which gives more total light In the right hand side picture this would be the sum of the light produced by the bulbs A The left hand side con guration B The right hand side con guration C They are the same PV2RV2R in the right hand R istwice as large V is the same 11 You try again with the con guration above Which now produces more light A The left hand side con guration B The right hand side con guration C They are the same Now V is the same but I is twice as large in the right hand side III In the circuit on the left what is the current through the lamp A 025 Amp B 1 Amp C 001 Amp D 005 Amp E 05 Amp The bulb is 025 Watts at 5V so PV 025l5 l005Amp thus the resistance ofthe bulb is RVl 5005 100 Ohms You could get that directly from PV2R so 02552R so R25025100 Ohms I A toroidal transformer has two linked coils All the ux from coil 1 will go through coil 2 and Visa versa Coil 1 has N1100 turns Coil 2 has N21000 turns Ifthe change in current in coil 1 is 01Nsec the Emfmeasured in coil 2 is 100 Volts Ifthe same changing current was applied to coil 2 what would you measure for Emf in coil 1 A 100 Volt B 1000 Volt C 10 Volt D 1 Volt E nothing The M of the transformer is the same whether you are hooked up to the 100 turn coil orto the 1000 turn coil The Emf in a coil due to a change in current in the alt1 n2 change in the current is the same the Emf will also be the same other coil is given by M since M is the same and it is given that the L l Out VVVVV pd AAAAAA II This circuit will predominantly pass A All Signals B High Frequencies C Low Freguencies D Only a speci c narrow range of frequencies E All frequencies except for a narrow range F No Signals The inductive reactance is given by XLDL For a resistor the reactance is just R Note that the reactance behaves like resistance in those situations like this one where you can ignore the phase ofthe signal 80 the inductors reactance changes with the frequency ofthe signal For a low frequency it is zero so all the low frequency signals are passed through For a high frequency XL is large so very little if any of these signals can make it through Out 0 WT VVVVV 0 III II This circuit will predominantly pass A All Signals B High Freguencies C Low Frequencies D Only a speci c narrow range of frequencies E All frequencies except for a narrow range F No Signals Same argument as for the previous question except that the capacitive reactance is given by XC1DC and so reactance is small for high frequencies Electric Field for a dipole On one of the axes perpendicular On the same axis as the two to the 2 Charges charges 5 E 1 5 1 16 3 3 3 472290 Z 27z80 Z 47z80 x 6 C Where the dipole vector is defined as p q m pointing from the charge to the charge Continuous Charge Distributions When we have to deal with many charges on a conducting or non conducting surface adding up the field for each charge is not practical instead we integrate over the charge density Name Symbol Formula Units Charge q q Q C Linear Charge Density 7L AQL Cm Surface Charge Density 0 6QA Cm2 Volume Charge p pQV Cm3 Density Integrating charge densities For a linear charge density small segment of a line ds would have r e cha g dq ids Coulomb s law then gives us 1dql dE 2 2 472290 r 472290 r Integrating this 1 dq 1 Ads39 E J 47r230 r2 47230 r2 For a surface charge density dq O39dA For a volume charge density dq pdV Note E is really a vector So you need to take into account the direction of r The charge density integral II Mathematically more accurate is to take into account that x point to the point of observation P and x points to the charge element For a volume charge we then get 5 4780 V pde39 a 2 xxl aza a39 f2 quot x x Fl Also note that we get QldqlpdV Some results From your book you will find some results of such integrals qZ E 4 80 z2 RZ Charged ring on 2 axis 039 Z Charged disk on 2 aXIs E 1 250 V22 R2 J Infinite sheet 28 0 Hoop segment and Bowl Two more for which the integral is relatively easy are the hoop segment and the bowl The hoop segment is worked out in the book also For the bowl 1 2 3990 6 dq o dA gt E J39 1 472290 0 0 R Remember surface of Z Z w a sphere comes from A Rsm szd 1392 39E dzp In 17 0056 47d2 1 we cos9o R2d dicos 9 039 1700829 1 4 0 E 4713980 0 0 R2 80 E0licos90lcos90 72 EL1cos 07A727IRZ17c05490 875901132 0 Sheet of current A large thin sheet of current is produced by a thin charged plastic sheet with surface charge density UQA that moves with a velocity v What is the magnetic field The field will be both above and below the sheet in opposite directions From Jvalme nevd and ne NeVQV we can write the surface current density as rimvQA la The enclosed current is then PAW101 qSRdE j39EdEj39 d iEd j d uoim 2 a 17 c d 17 d IEd IEd BhBhLthO39VDBJDO39V Current Loop as Magnetic Dipole You cannot use Ampere s Law so we need to integrate the current using the BiotSavart Law rR2zzcosa5LdsRd 9 V IR2Z2 Bu JdBH IcosadBImj 2Xf 27F 2 I 0 1 5 Rdg R 2 H Er Zr 0 47239R2Zz BuDz39 R2 uDz39RZ uDz39A un H 3 2 H 2R2Zz 72 3 EZ3 27239 Magnetic Flux The same way in which we de ned the electric ux through an area we can de ne a magnetic ux c1gt3j1 7d21 In the case where B is perpendicular to the area parallel to the area vector and B is uniform we can simply write ch BA Induced Currents An EMF is induced in a conducting loop when the number of eld lines through this loop are changing More precisely An EMF is induced in a loop when the magnetic ux through the loop changes The magnitude of this EMF is equal to the rate at which the ux changes with time dch g dz If you have N turns the result is somewhat like stacking N batteries the EMF increases by a factor of N 61an dt N Lenz39s Law What is the direction of the EMF that is produced in a loop with a changing magnetic eld Lenz s Law The induced current EMF is such that the magnetic eld caused by this current opposes the change in the external magnetic eld Gauss39 Law In rr39o While integrating Coulomb s Law always works for any situation the calculations can be very lengthy and complicated Gauss Law is valid in all those situations also however it is not always useful Gauss Law is very useful in those cases where there is symmetry so that we know something about the electric eld already Area as a vector A The size of the vector is proportional to the area the direction is perpendicular to the plane of the area You can construct an area from two vectors The area vector is then the cross product of the two vectors kA Az xd 04 7 0 The flux of an electric field though a flat area A CD E A 2 EA cos 6 Flux through a very small flat differential area dA dd 2 E dA Flux through an arbitrary area Flux through a closed surface 24356017 Gauss Law Gauss Law the net number of field lines through a closed surface must be proportional to the amount of charge inside 801 5OC Edjl qm Gauss39 Law for poin r charge 30 Edgl gOEC dA 60E47rr2 qenc gt qenC 47rgor2 Boltzmann s Entropy Equation The entropy and the number of microstates of a specific system are connected through the Boltzmann s entropy equation 1896 S k In W 2 d Law of Termodynamics AS 2 0 For a closed system entropy can only increase it can never decrease For an irreversible process the entropy increases For a reversible process the change in entropy is zero Some Definitions The con guration of a system is some measurable state in which the system resides Examplel 6 molecules are distributed 4 on the left 2 on the right ExampleZ A gas on n mole in a volume V has temperature T and pressure P The number of microstates for a configuration is the number of ways in which the configuration can be put together This number of ways is called the multiplicity W of the configuration We find the number of microstates by counting them There are mathematical methods for counting combinatorics that help us The factorial NJ is one such tool N is the number of ways in which N items particles spin states can be arranged It is equal to NNNlN2 2l When counting microstates we have to be careful to not count states twice or more Since usually we deal with molecules that can not be distinguished from each other since they are identical we need to make sure we divide out all the ways in which the molecules can be arranged in their state Some Examples Sixty particles in a box no division between left 60 and right Then the number of microstates is 6 0 Ten particles in a box with 2 distinct sides so a 39 particle is either on the left or on the right We W i 210 have 6 particles on the left 4 on the right 6l4 Ten particles in a box with 3 distinct regions left W 7 l 7 2520 middle right filled with 532 particles resp 7 533m 7 100 92 One hundred particles in a box 10 regions each W 10 23610 region has 10 particles 10 Entropy Thermodynamic View Rudolf Clausius 1865 Entropy is the amount of thermal energy put into a system that can not be extracted as work f dQ Aszsf Sj7 t In a cyclic process you can convert heat in to work Le a steam engine However you can never convert all the heat fully into work some of it is lost At constant temperature we have AS T Entropy as State Function If entropy is a state function then the entropy of a system is the same Whenever it is in the same state Thus a cyclic process must have AS0 For an ideal gas we can write down the change in entropy between 2 states as Vf Tf AS nR1n nCV 1n V T i Engines In a thermodynamic sense engines turn heat into work No engine can convert all the heat at the input into work some heat must be discharged Thus and engine needs a hot reservoir for heat input and a cold reservoir for heat output You cannot extract work when you have no temperature difference Any Ideal Engine For any ideal engine we use a gas that does work in a cyclic process each step of this process is reversible thus for one full cycle for the gas AP0 AV0 ATO AE 0 AS0 mt Carnot Engine Ideal Carnot Engine In a Carnot Engine the gas absorbs heat during an isothermal expansion at TH then adiabatically expands to TL then releases heat during an isothermal contraction at TL and finally contracts further adiabatically to come backto the initial temperature TH Pro erties p AE39nLZOSWZQSWleHQL 1 IQHI AS ASHASL 03 TH TL 03 TH TL ZMZIQHI IQLI1 TL 8C 1 IQHI IQHI IQHI TH Ligthing sTrikes you see The flash and 10 seconds aTer you hear The rumble of The Thunder How far away did The ligthing hiT A 11 Km B 22 Km C 33 Km D 44 Km E 10 Km When The oscillaTing elemenT of a sound wave is moving righTward Through The poinT of zero displacemenT equilibrium poinT The pressure is A AT Equilibrium B Beginning To increase C Beginning To decrease D The pressure does noT change E I don39T undersTand Two uncharged metal balls X and Y stand on glass rods A third ball Z carrying a positive charge Q is brought near the rst two Next a conducting wire is run between X and Y The wire is then removed and ball Z is nally removed When this is all done what is the nal result A Both X and Y are still uncharged B Balls X and Y are now charged positively C Balls X and Y are now charged negatively D Ball X is positive ball Y is negative E Ball X is negative ball Y is positive Answer X Y X Y It is true that X and Y start uncharged but that does not mean they do not have charge on them They have and charges all mixed up to produce a neutral result When the charged ball is brought near the two neutral balls will be polarized slightly because of the electric eld that is set up by ball Z What that means is that the charges in the neutral balls are attracted to the charge of Z the charges are repelled but in fact they don t move The positive charge on the left of X and Y are caused because the charges moved away from there When the wire is connected some electrons will ow from X to Y You can look at it as one object now where the charges are trying to get even closer on the whole to the of Z When the wire is then disconnected the charges stay divided so X will have charge and Y will have charge 11 The situation is no different if the wire is connected rst and then the charged ball Z is brought near The charges are still induced on the sphere as in picture 3 111 If the ball Z is removed again before the wire is removed the nal situation is different The charges on the ball X and Y now have a chance to ow back and so they will very rapidly too The nal result would then be two neutral balls B C 1 One ideal battery is connected to the circuit shown above All 3 light bulbs are identical What happens with the brightness of bulbs A and B when the switch is closed A A stays the same B dims B A gets brighter B dims C Both A and B get brighter D Both A and B get dimmer E A and B stay the same 11 How does the total output of light from all 3 bulbs change after I close the switch Assume that these are ideal light bulbs which have a constant resistance R A The total light stays the same B The total light increases C The total light decreases D I have no clue Ming Ming A B 1 Which of the light bulbs is properly connected to the battery A A mB C Bamp C D B amp C amp D E All of them 11 Rank these circuits in order of brightness of the bulb A A B gt C mBgtAC C A lt B C D B gt A gt C E A B C V0 39 0 II II This lab setup allows an experimenter to test the charging of capacitor C The resistor can be changed to any value from 0 to in nity and the meters will indicate the current and voltage 1 After the switch Sl has been closed for a long time the experimenter changes the value of the resistor from 10 kOhni to 100 kOhni The voltnieter now reads A 10 times higher B 10 times lower C The same 11 Now the switch 1 is opened and the switch 2 is closed The Volt rneter will now read A V is and slowly decreasing B V is and slowly decreasing C V is and slowly increasing D V is and slowly decreasing E V is zero III The Aninieter will now read A A is and slowly decreasing B A is and slowly decreasing C A is and slowly increasing D A is and slowly decreasing E A is zero Engines In a thermodynamic sense engines tum heat into work No engine can convert all the heat at the input into work some heat must be discharged Thus and engine needs a hot reservoir for heat input and a cold reservoir for heat output You cannot extract work when you have no temperature difference Any Ideal Engine For any ideal engine we use a gas that does work in a cyclic process each step of this process is reversible thus for one full cycle for the gas AP0 AV0 ATO AEim0 AS0 Carnot Engine Ideal Carnot Engine In a Carnot Engine the gas absorbs heat during an isothermal expansion at TH theh adiabatically expands to TL then releases heat during an isothermal contraction at TL and finally contracts further adiabatically to come backto the initial temperature TH Properties AEth 0 3W Q 3 W QHQL ASAS AS 0 Q03 H L T T T T H m T 8C 2 1 1L IQHI IQHI IQHI TH H L Refrigerators A refrigerator uses work to transfer heat from the low temperature reservoir to the high temperature reservoir It is much like an engine running in reverse Carnot Refrigerator Same cycle as the carnot engine but running in reverse The efficiency of a refrigerator would be the amount of heat extracted from the cold reservoir for the amount of work you have to put in QL IQLI T L C W iQHiQLi TH TL Efficiency of Engines No engine can be more efficient than the Carnot engine so conversely no refrigerator can be more efficient than the carnot refrigerator Proof of this theorem Suppose an engine X can do better than the Carnot engine In that case putting this engine between the hot and cold reservoir and then coupling its output to a Carnot refrigerator would effectively pump net heat from the cold reservoir to the hot reservoir violating the 2quotd law Charge Electric charge is either positive or negative The absence of electric charge or the perfect balancing of and is called electrically neutral Opposite charges attract same charges repel Charge can ow in the form of charged particles Both particles and particles can ow The ow of charge is called an electric current Since electrons are very small and light and because ofthe structure of molecules they flow more easily so usually but not always the flow of charge is due to electrons Materials that allow the ow of charge are called conductors materials that do not allow the ow of charge are called insulators Coulomb39s Law szqlqzz 1 611612 r2 471290 r2 The constant 80 is very well de ned in terms of the speed of light which is known exactly since it de nes the meter So this is a constant we know exactly 1 l N 712 C2 50 Z Z88541878xio A Z 00 4nx10 7 NA39Z 4299792458 ms39l m 9 This is quite large As a result two charges of k 899X10 l Coulomb one meter apart would result in a 7280 tremendous force Usually charges are much much smaller gt gt Fm FFFg The A39romic Nucleus Krzchl pm 1 unwimzmg The UHHK chm From gluon g unwimting C 7 atoom Krzc zmn y unwimling Fr39om A39roms a Future Sizes are 8 orders of magnitude different 39om the atom to the size ofa nuc eon Inductors Sr Inductance Similar to capacitors which give a certain amount of electric field for a given amount of charge we can define a quantity for coils that expresses their property to produce a certain amount of magnetic flux for a certain amount of current The inductance is defined by 2 L NQDB H Tm 39 A l The term NltIgtB is called the flux linkage For a solenoid the inductance per unit length is L Nqu nluoinA 2A z 1139 11 W Self Induction Since a coil produces a magnetic field it creates it s own flux If the current through that coil is changing it will thus change it s own flux creating an 323 This process is called selfinduction The produced is given by dt dt dt The direction of this is such that it opposes the change in current Lenz s Law RL Cir39cui rs An inductor in a circuit will produce an Emf whenever the current through the inductor is changing This induced Emf will be such as to oppose the change in current Right after the switch is set to a a S the inductor completely opposes the I I change in current As a result the i 1 current is zero The circuit behaves as f I if the inductor is not there A long time after closing the switch the current no longer changes so the inductor behaves like a wire If the switch is then thrown from a to the inductor keeps the current going for a brief while until the current decays to zero RL Cir39cui rs II So what happens between t0 and too Use the loop rule g iRi g 0 dl39 R g i L L L dz This is the same differential equation as we saw for capacitors The solution is thus ig l ei i 1 R L R RL Circuits III If the current is going and we then set the switch to position b how does the current decay i 0 gt izfei i dt L R Ener39gy Stored in a Magnetic Field The energy stored in the eld of an inductor with current i 1 2 For an electric field and a UB 3L1 capacitor these equations were Energy stored in a capacitor 2 The energy dens1ty is given by U Li E 2 C Energy density of the electric field 2 1 L1 1 B 1 E2 uB LIE 350 Induced Currents BA CD Foraconstant magnetic eld B q this simplifies to The EMF induced in the loop is quB then given by Faraday s Law ah LenZ s Law The direction of the EMF is such as to oppose the change in the magnetic field Example a solenoid with some windings Inductors and Energy Transfer Energy is conserved So when we set up a current in a resistive wire loop by changing the flux through that wire loop the force we use must do an equal amount of work as is dissipated in the resistance If we pull a loop out of region with a magnetic field dog dt BL BLV dt V2 1 P Rl392 BZLZV2 R R Work done by the force Fin amp LBiBZL2v R R dW dx Wj d FxP F air air Fv P iBZLZV2 R Eddy Currents Magnetic Brakes When the object that is pulled out of the magnetic field is not a loop but a continuous piece of conductor there will still be currents set up in it These currents are called eddy currents since they look like the eddies of a river This is how magnetic brakes work Induced Elec rr39ic Fields The trouble with the induced Ef of a changing magnetic field is that it is so unlike a battery The X is ok but potential difference does not make sense around the loop What Faraday really wanted to say was dch Ed The path of integration on the left is around the area used for calculating the flux on the right hand side of this equation This is more clear if we write out the formula for PB 9 a d r a c Edp EjBdA Inductors St Inductance Similar to capacitors which give a certain amount of electric field for a given amount of charge we can define a quantity for coils that expresses their property to produce a certain amount of magnetic flux for a certain amount of current The inductance is defined by NQDB H Tm2 1 A L For a solenoid the inductance per unit length is B Ilu0n2jA Nd nluoinA 1 li ll A Leyden Jar consists of a piece of foil wrapped around the outside of a glass jar and another piece of foil lining the inside of a glass jar The energy stored in a charged Leyden Jar is actually stored A On the metal foil inside the jar B On the metal foil outside the jar C On both the inside and outside foil D Inside the jar itself E In the glass between the foils The electric field stores the energy and the field resides in the glass In addition the polarization of the molecules which reduces the field also stores energy Consider a simple capacitor made of a pair of conducting plates in close proximity Supposed the plates are charged and and then discharged producing a spark Next the plates are charged again exactly as they preViously were only this time after they are charged they are pulled further apart This time the spark that is produced will be A Bigger than the first spark B The same as the first spark C Smaller than the first spark D There won t be any spark Pulling the plates apart requires a force working against the electrostatic attractive force This force over the distance that the plates are pulled apart means that you did work which means the energy went somewhere It went into the Efield You can also look at it as the charge stays the same the capacitance is reduced so the voltage goes up Stored energy is U 12CV2 2Q2C so as C is reduced U goes up The capacitor of a Vandergraaf s generator consists of just a large metal sphere If the size of the sphere is doubled the capacitance A Remains the same B Halves C Doubles D Quadruples E Don t know For a sphere V L 3 C 2 47r 0R 4723980R V To make an extra large spark when discharging a capacitor one could use several capacitors Assume that you have 4 capacitors with a rating of l uF at lOOOV and a power supply that can charge them to lOOOV The biggest spark is obtained when they are all A parallel because C is now 4X B in series because V is now 4X C 2 X 2 2 sets in series that in parallel D Makes no difference E Don t know Each capacitor at a certain voltage stores a certain amount of energy Four capacitors store four times as much energy it does not matter how they are connected In parallel the total capacitance Ctot4C and VV0 so Utot20V02 in series CtotC4 but now V4V0 so Utot12C44V0220V02 A magnetic eld points up from the oor in the laboratory A proton and an electron are shot into this eld parallel to the oor Looking along the direction of travel the particles will curve A Proton to the right Electron to the left B Proton and Electron to the left C Proton to the left Electron to the right D Proton and electron to the right E None ofthese Remember that the proton has charge e and the electron has charge e Using the right hand rule you find Fqva but q changes the direction ofthe force vector for a negative particle A magnetic field points up from the oor in the laboratory A proton and an positron positively charged antielectron are shot into this field parallel to the oor with the same velocity The ratio of the radii of the curvesproton to positron will be approximately A 1 B 11000 C 1000 D 12000 E 2000 The mass of the proton is approximately 1000 MeVCZ the mass of the electron is approximately 05 MeVcz the radius of curvature goes like rmvqB Since dB is the same the relevant part to look at is mv The particles have the same velocity so r scales like m rprotonrelectronmprotonmelectron2000 A magnetic field points up from the oor in the laboratory A proton and an positron positively charged antielectron are shot into this field parallel to the oor with the same momentum The ratio of the radii of the curvesproton to positron will be approximately A1 B 11000 C 1000 D 12000 E 2000 This time we are told the momentum is the same Rememberthat pmv so if mv is the same and qB are the same the radii are the same A strong magnetic field points in the same direction as a beam of electrons The electrons will be A De ected up B De ected Down C De ected Left D De ected right E Unaffected The B field and particle velocity are in the same direction so sine is zero A strong magnetic field points in the same direction as a beam of electrons Some electrons scatter in a target which changes their direction to 45 degrees with respect to the beam These electrons will now travel A In a curve away from the beam B In a curve towards the beam C In a lefthanded helix around the beam D In a righthanded helix around the beam E They remain unaffected This is a tricky question If you look in the direction ofthe beam particles are moving away from you then the eld is also pointing away from you A particle that scatters down at 45 degrees will have a force exerted on it Now va is pointing towards the right but these are electrons so qva points to the left We thus get a clockwise moving helix which we call a left handed helix Lec rure 12 Kine ric Theory Chap rer39 20 Announcements 0 This weeks LAB You will be asked to do an evaluation off your understanding of ElectroMagnetism This is for research purposes You will get 10 bonus points towards your Lecture participation Transla rional KineTic Energy 3RT 3kT RMS veIOCIty vrms M m Average Kinetic K 1 2 1 3 energy avg 2 ms 2 2 m A 2 J57de NV Mean free path Distribution of Speeds Not all the molecules in a gas will have the same velocity Even if they started with the same velocity collisions will soon spread the velocities so that there is a distribution of speeds 1852 James Clerk Maxwell Maxwell s speed distribution law va4 Mv2 36 2 2RTJ ve Z RT The fraction of molecules with a Properties of a distribution function velocity between v1 and v2 is equal to the probability of finding a particular molecule with a velocity between v1 IPvdv1 and v2 v 0 fracwz IP02 dv OO Pv10393 sm Dis rribu rion of Speeds II Speed Distribution for 02 T 00K 4 35 3 25 2 15 1 05 0 0 200 400 600 800 1000 1200 v ms Average RMS Mos r Probable v We can now make several interesting conclusions The probability of nding a molecule with v0 is O The probability of nding a slow molecule is not zero it gets larger for lower temperatures The probability of nding a molecule with very large V say v2000ms is not zero It gets larger for higher temperatures P0ooTPvdvl 3 8RT 8kT vavg vPvdv 2 TM 2 Tm 3 3RT 3kT 2P d vIms v v v M m dPv 2RT 2kT VP gt 0 gt VP 2 dv M m Internal Energy of a 605 A gas can store energy in 1 Translational Kinetic Energy 2 Rotational Kinetic Energy 3 Molecular Excitations vibration modes A monatomic gas is the simplest case Quantum mechanics tells us that it can not store energy in rotation and having only one molecule limits possibilities for excitation So the energy is stored in kinetic energy For N molecules we get T E NKan gva m Nmi gNkT gnRT m 1nt Molecular Specific Heat at constant volume The molar specific heat C is defined by Q nCAT Capita39 quot We ase It does not matter as long as you keep track For constant volume we use CV and using the 1st law of thermodynamics and WO we can write AEmt Q W Note that the result AEth is general as long as you use the correct CV 3 3 For a monatomic gas we then have 3 3 CV 3 R Molecular Specific Heat at constant pressure At constant pressure the heat that is added to the gas Will not only increase the internal energy of the gas but also do work The amount of work done is just W pAV nRAT Combining this with the 1st Law of Thermodynamics and the previous result AEint Q W 2 Q nCVAT nRAT Definition of molar specific heat at constant pressure Q quotCPAT nCpATanVATnRAT gt Cp CVR Monatomic Gas C R Equipor ri rion Theorem The available energy to be stored in a gas will distribute itself equally over the available degrees of freedom 0 Each degree of freedom will have on average 1sz of energy associated with it Monatomic gas 3 degrees of freedom Eint NkT gt CV R Diatomic gas 5 degrees of freedom Eint 3 translation 2 rotation NkT gtCV R Diatomic gas 6 degrees of freedom Eint 3NkT gt CV 2 3R 3 translation 3 rotation Adiobo ric Expansion of Gas We can now find an equation for adiabatic expansion Using dEQW and setting Q0 WpdV and using our previous results plus some math wed in pV constant 3 TV 1 constant Admmx iQ 0 Pressure Immenm 70 K 10 Spins either up or down are IIin a boxquot What is the total number of possible microstates all configurations together a 10 b 20 c 10 d2 10 e Not enough info This is a bit of a tricky question You can check on your calculator that the sum of the microstates for all 11 possible con gurations starting will all up 9t 1t 8t 2t 7t 3l 2f 8t 1f9l ml is 1024 l l l l l l Thus 1039 1039 1039 ll 1039 1039 1039 1024210 you can check this on 100 91 8l2 2l8 19 010 your calculator Because we are counting microstates the con guration fffffffffl is distinct from fffffffllt etc So from that you can tell just like binary numbers that there are 2quot10 combinations 10 Spins either up or down are IIin a boxquot What is the number of microstates for the configuration 4 up 6 down a 46 b 1046 c 1046 d 416 e Don39t Know This is a little more straight forward It is like the problem in the book a box with a left and a right side and like the one we did in lecture The key here is that ffffllllll and lfffflllll are not distinguishable So we need to count the number of combinations microstates ofl and T that make 4 up and 6 down The answer is c There are 10 ways of putting the spins in if each spin was numbered so you know the difference between 12345678910 and 21 345678910 However since they are NOT numbered and we can not tell the different microstates apart we need to divide out the number of different ways in which we can assemble 4 up which is 4 and the number of ways in which we can assemble 6 down which is 6
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