Introductory Physics II
Introductory Physics II PHY 232
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This 189 page Class Notes was uploaded by Quinn Larkin on Saturday September 19, 2015. The Class Notes belongs to PHY 232 at Michigan State University taught by Staff in Fall. Since its upload, it has received 14 views. For similar materials see /class/207634/phy-232-michigan-state-university in Physics 2 at Michigan State University.
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Date Created: 09/19/15
I Warn3139le r W nno Eran 39 39 Hm M 9 I WNW at luIIq39 1 quotI M rcss39art Q IDIBCIIE Ellquot H39Il F51 ul39 EMF 59 ll 4 viieaadnm bm 7 nlcmri cu39rril quotcw PHY232 Spring 2008 Jon Pumplin httpwwwpamsuedupumpinPHY232 Original ppt courtesy of Remco Zegers So far we have looked at systems with only one resistor for ohmic resistor V I r 7 V l Fl I I I o l R R R I 0 Now look at systems with multiple resistors which are placed in series parallel or in series and parallel quiz gt At V10V someone measures a current of 1A through the below circuit When she raises the voltage to 25V the current becomes 2 A Is the resistor Ohmic a YES b NO building blocks battery or other potential source Provides emf electromotive force to the circuit 39 switch allows current to flow is closed u u ampere meter measures current quot3 volt meter measures voltage resistor Ex39 capacHance lightbulb I usually show a realistic picture or resistor instead light bulb made of tungsten 0c48x10393 1K temperature of filament 2800 K so RR01ocTT013R0 consequences 6 A hot lightbulb has a much higher resistance 7 A light bulb usually fails just when switched on because the resistance is small and the current high and thus the power delivered high Pl2R In the demos shown in this lecture all lightbulbs have the same resistance if at the same temperature but depending on the current through them the temperature will be different and thus their resistances assumptions l gt 1 The internal resistance of a battery or other voltage source is zero This is not really true notice that a battery becomes warm after being used for a while gt if this were not the case asystem like this gt should be replaced with internal resistance V interna assumptions ll gt An ampere meter ammetercurrent meter has a negligible internal resistance so that the voltage drop over the meter VA FlA is negligible as well gt Usually we do not even draw the ampere meter even though we try to find the current through a certain line gt Remember that an ampere meter must be placed in series with the device we want to measure the current through ques on Lt If in the above circuit the resistance of the Ampere meter is not zero it will not measure the right current that would t d true the total current will change and thus also the current b not true current cannot get stuck in the line and thus the measurement will not be affected assumptions Ill gt A volt meter has an infinite internal resistance so that no current will flow through it gt Usually we do not even draw the volt meter even though we try to find the potential over a certain branch in the circuit gt Remember that a volt meter must be placed in parallel with the device we want to measure the voltage over assumptions IV gt We can neglect the resistance of wires that connect the various devices in our circuit This is true as long as the resistance of the device is much larger than that of the wires basic building blocks two resistors in series high POISEUllle presswa FlowAPr4l low DI39BSSU FE l 21 399 12 volts e 12 volts voltage 221 The water flow m3s through the two narrow pipes must be equal else water gets stuck so the pressure drop is larger over the narrowest of the two The total pressure drop is equal to the sum of the two pressure drops over both narrow pipes The current I through the two resistors must be equal else electrons would get stuck so the voltage drop is larger over the highest of the two The total voltage drop is equal to the sum of the two voltage drops over the resistors resistors in series ll The voltage over R and R2 Vil IR2TR1lRg if we want to replace RUR2 with one equ39yale Fl 2 l m d hv nm inin 1 and 2 T8 9 For n resistors placed in series in a circuit Req R1R2Rn Note RaqgtRi l12n the equivalent R is always larger than each of the separate resistors gt second building block resistors in parallel The pressure drop over the two narrow pipes must be equal before and after the pipes the pressure is the same but the water prefers to flow through the wider canal ie the flow m3s is higher through the wider canal 39 39 39 i w v 1 quota i A emo 2 gm in parallel reSIstors In parallel II For the current throuah he circuit 1 1 1 71 7 i 2 R1 R3 if we want to replace RvR2 with one equivalent R 1 2 E 1 1 1 and by combining 1 and 2 IT For n resistors placed in parallel in a circuit 1Req 1R11R21Rn Note RmltRi with l12n R I is always smaller than each of the separate resistors ques on gt what is the equivalent resistance of all resistors as placed in the below circuit If V12V what is the current I R13 Ohm R23 Ohm R33 Ohm V12V Fl2 amp Fl3 are in parallel 1R231R21R3131323 R2332 Ohm Fl1 is in series with Fl23 R123R1R2333292 Ohm lVR129224983 A gt A tree is decorated with a string of many equal lights placed in parallel If one burns out no current flow through it what happens to the others a They all stop shining b the others get a bit dimmer c the others get a bit brighter d the brightness of the others remains the same gt A tree is decorated with a string of many equal lights placed in parallel If one burns out no current flow through it what happens to the others gt a They all stop shining gt b the others get a bit dimmer gt c the others get a bit brighter gt d the brightness of the others remains the same Before the one light fails 1Fteq1Ft11Ft21Ftn if there are 3 lights of 1 Ohm Req1l3 IVFteel IiVFti if 3 lights I3V liV1 After one fails 1Fteq1Ft11Ft21Ftn1 if there are 2 lights left Req1l2 lVFteel liVRi if 2 lights I2V liV1 The total resistance increases so the current drops The two effects cancel each other A different Christmas tree gt a person designs a new string of lights which are placed in 39 0 39 t ppens to the others gt b the others get a bit dimmer gt c the others get a bit brighter gt d the brightness of the others remains the same Assume If one fails the wire inside it is broken and Current cannot flow through it any more Kirchhoff s rules gt To solve complex circuits we can use the following rules gt Kirchhof 1 The sum of the currents flowing into a junction must be the sa e the the sum of the current flowing out of the junction 1 5 gt Kirchhof 2 The sum of voltage gains over a loop le due to emfs must be equal to the sum or n it er the loop IR1lFl2 ques on gt Given V12V what is the current through and voltage over each resistor 1 Slide 12 183 A Kirchhof 2 2 VI1R1IZR20 123l1320 R13 Ohm Kirchhof 2 R23 Ohm 3 VI1R1I3R30 123l13l30 R33 Ohm Kirchhof 2 V12V 4 OI3R3I2R20 3l3320 Kirchhoff 5 l12l30 I12l3 10 amp 2 128320 so 432 and I24l3 A demo 1 amp 3 128330 so 433 and 343 A lightbulb Use VR for F1 V18338 v for R2 V24334 V for R3 V34334 V circuit IMPORTANT gt When starting a problem we have to assume something about the direction of the currents through each line It doesn t matter what you choose as long as you are consistent throughout the problem example 0 Kirchhoff I I139I239I3O 1I2I30 Kirchhoff 2 VI1R1IZR2O Kirchhoff 2 VI1R12R2O ques on If U Q I R 3 3 5 5 gt which Of the following cannot be correct gt a VI1R1I3R3IZR20 gt b l1l3l40 gt c I3R3IGR6I4R40 gt d I1R1I3R3IGR6I4R40 NOT A LOOP gt e 3620 ques on What is Kirchhoff I for O b l12l3l40 II o l1I2l3l40 1 39J 24 What is Kirchhoff II for the left small loop with Fl4 and R 29157 I4R4I1R10 bl I4R4I1R10 c I4R4I1R1V0 What is Kirchhoff II for the right small loop with Fl2 and R3 b I3R3l2R20 b I3R3l2R20 I3R3l2R2V0 39i r t i a V l 39 What is Kirchhoff II for the loop with VFl4 and R3 a VI4R4I3R30 h VI4R4I3R30 c VI4R4I3R30 ques on gt what is the power dissipated by R3 PVIV2R2R R1 1 Ohm R22 Ohm R33 Ohm R44 Ohm V5V We need to know V3 andor l3 Find equivalent R of whole circuit 1R231R21R3121356 R236l5 Ohm R1234R1R23R41654315 Ohm 1l4VR12345315 2531 A Kirchhoff 1 12325l31 Kirchhoff 2 I3R3IZR20 so 3l3220 232 3 Combine 32 l3l325l31 so 52 l325l31 l31031 A P2R so P1031231009613031 Js more than one emf R1R2R33 Ohm v1v212 v what is the current through and voltage over each R gt apply kirchhoff s rules 1 l1l2l30 kirchhoff I 2 left loop V1I1R1I2R20 so 123l1320 3 right loop V2I3R3I2R20 so 123l3320 4 outside loop V1I1R1I3R3V20 so 3l13l30 so l1l3 combine 1 and 4 223 and put into 3 129l30 so l343 A and l143 A and l283 A ques on At which point ABCD is the potential highest and at which point lowest All resistors are equal circuit 1 l122R6R a highest B lowest A VA126RR6V I hih Blw D circuit2VB12V d highest C Iowest A Cil CUit 3 123R4R e highest A lowest B Vc12394RR8V VD124RR4RR4V circuit breakers Circuit breakers are designed to cut off power if the current becomes too high In a house a circuit breaker is rates at 15A and is connected to a line that holds a coffee maker 1200 W and a toaster 1800 W If the voltage is 120 V will the breaker cutoff power PV 18001200120 x 300012025 A 25Agt15 A the breaker will cut off power Question Consider the circuit Which of the following isare not true If Fl2Ft32Fl1 the potential drops over Fl1 and Fl2 are the same for any value of Fl1Fl2 and Fl3 the potential drop over Fl1 must be equal to the potential drop over Fl2 The current through Fl1 is equal to the current through Fl2 plus the current through Fl3 I12l3 answer consider the circuit Which of the following F isare not true It Ft2Ft32Ft1 the potential drops over Fl1 and Ft2 are the same for any value of Ft1Fl2 and Fl3 the potential drop over Fl1 must be equal to the potential drop over Ft2 The current through Fl1 is equal to the current through Fl2 plus the current through R3 391392393 1 it R2R32R1 then 1R231R21R31R1 so Ft23Ft1 and l1l23 and potential of Ft1 equals the potential over R23 and thus Ft2 and R3 THIS IS TRUE 2 no this is only TRUE in the case of 1 3 true conservation of current RC circuits Consider the below circuit gt When the battery is connected a current passes through the resistor and the capacitor begins to charge up gt As the capacitor gets more charge and hence more voltage the voltage across the resistor decreases so the current decreases gt Eventually the capacitor becomes essentially fully charged so the current becomes essentially zero gt The maximum charge is given by QCV RC circuit lorthe charge on the capacitor 1 7 67 an Q I 7 mo ior the voltage over the capacitor VG PM i 9 1 i r ior the voltage over the resistor quotR V i 4 V N lorthe current I E14 e 2718 Voltage switched on RC time I E RC voltage a 39 swnched off q Qe r RC LI 5 I 39I I U III Ci 3 2 LI Lil II LI t1 aa ESEEEE EEE gt The value is the time constant It is the time it takes to increase the stored charge on the capacitor to 63 o of its maximum value 1e063 question given V10 V R100 Ohm c1ox1o6 F v in The emf source is switched on at t0 a After how much time is the capacitor C charged to 75 of its full capacity b what is the maximum current through the system a if 1 e 075 then charged for 75 so eV 025 tRCln025 In natural logarithm tRC x n025100x105x139 139x101 seconds b maximum current at t0 it is as if the capacitor C is not present so IVR01 A W V v 391 7 e U Ti17 e W 1 EH PH Y232 Spring 2007 Jon Pumplin thwwwpamsuedupumplinPHY232 Ppt courtesy of Remco Zegers previously Electric currents generate magnetic fields If a current is flowing in a straight wire you can determine the direction of the field with the curly righthand rule Electric current I 1 II B Mugrwlic FinId and calculate the field strength with the equation Bu021td For a loop which is a solenoid with one turn Bu0lN2R at the center of the loop Fora long solenoid is long Bu0l NL anywhere way inside Electromagnetic Induction M The reverse is true also a changing magnetic field can generate an electric field E This effect is called induction In the presence of a changing magnetic field an electromotive force emf voltage is produced demo coil and galvanometer Moving the magnet closer to s N the loop or farther away produces a current If the magnet and loop are held stationary there is no current Necessary definition magnetic flux M A magnetic field with strength B passes through 0 loop with area A M The angle between the Bfield lines and the normal to the loop is 0 M Then the magnetic flux DB is defined as DB BACOSQ Units Tm2 or Weber IonCdpd uses Wb example magnetic flux M A rectongulorshoped loop is put perpendicular to o magnetic field with a strength of 12 T The sides of the loop are 2 cm and 3 cm respectively What is the magnetic flux M DB BACOSQ Bl2 T A002x0036x10394 m2 00 IXI DBT2 x 6x10394x l 72x10394 Tm E Is it possible to put this loop such that the magnetic flux becomes 0 M 0 yes M b no Faraday s law M By changing the magnetic flux Ach in a timeperiod At a potential difference V electromagnetic force a is produced 113 m Warning the minus sign is never used in calculations It is an indicator for Lenz s law which we will see in a bit L r it changing the magnetic flux M changing the magnetic flux can be done in 3 ways M change the magnetic field E change the area M changing the angle MB MEAL men v At m A ACIJB ABAcos N N example x x M a rectangular loop Alm2 is moved 1 info a Bfield Bl T perpendicular x 1 x lo the loop in a lime period of l s quot 39 39 I X X X X How large Is the Induced volfage The field is changing VAABAflxll l V 0 While in the field nol moving the area is reduced lo 025m in 2 s Whaf is the induced volfage The area is changing by 075mi VBAAAflx07520375 V oThis new coil in the same field is rofafed by 45 in 2 s Whaf is the induced volfage The angle is changing cosO l lo cos45 l 22 VBA AcoseAllx025x02920037 V Fordddy s low for multiple loops E If instead of or single loop there are multiple loops N the the induced voltage is multiplied by that number 4333 339 z N L ampt demo loops If an induced voltage is put over or resistor with value R or the loops have or resistance or current lVR will flow resistor R first magnitude now the direction M So for we haven t worried about the direction of the current or rather which are the high and low voltage sides going through a loop when the flux chdnges resistor R Lenz s Law M The direction of the voltage would always make a current to oppose the change in magnetic flux demo magic loops When a magnet approaches the loop with north po39 t39ng towards the loop a current Is Induced As a result a Bfi dis made by the loop Hammad2m so that the field opposes the incoming field made by the magnet Use righthand rule to make a field that is pointing up the current must go counter clockwise The loop is trying to push the mcunel away Lenz s law II M In lhe reverse silualion where lhe magnel is pulled away from lhe loop lhe coil will make a Bfield lhal al39lracls lhe magnel clockwise ll opposes lhe removal of lhe Bfield Bmagnel Blnduced B ma nel Blnduced 4 9 magnel approaching the magnel moving away from coil lhe coil Be careful E The induced magnetic field is not always poiniing opposite lo the field produced by the external mdgnei X X X X X X X X X X X X If the loop is stationary in a field whose strength is reducing it wants to counteract that reduction by producing a field poiniing info the page as well curreni clockwise demo magnet through cooled pipe as N when the magnet passes through the tube a current is g induced such that the Bfield vmagnet produced by the current loop i opposes the Bfield of the a magnet opposing fields repulsive force this force opposes the gravitational force and slow down the magnet cooling resistance lower current higher Bfield higher opposing force stronger can be used to generate electric energy andstore it eg in a capacitor demo torch light question x x x x A rquotIB r1 I I 39 x gtlx 1 X 39l I I I I I X X X X A rectangular loop moves in and then out of a constant magnet field pointing perpendicular into the screen to the loop Upon entering the field A a current will go through the loop a clockwise b counter clockwise When entering the field the loop feels a magnetic force to the a left b right question X X X X A l39 39 39 I B l39 39 39 1 I I 39 x gtI x x 39I I I I X X X X A rectangular loop moves in and then out of a constant magnet field pointing perpendicular into the screen to the loop Upon entering the field A a gh the loop a clockwise b counter clockwise The loop will try to make a Bfield that oppose the one present so out of the screen Use second righthand rule counterclockwise When enterin the field the loop feels a magnetic force to the Ia left I b right Method 1 Use first right hand rule with current and Bfield that is present left Method 2 The force should oppose whatever is happening in this case it should oppose the motion of the loop so point to the left to slow it down Eddy currentdemo M Magnetic damping occurs when a flat strip 0 conducting material pivots inout of a magnetic field E current loops run to counteract the Bfield K At the bottom of the plate a force is directed the opposes the direction of motion strong opposing force weak opposing force no opposing force x x x x x x x x x x Bfield into the page applications of eddy currents M brakes apply magnets to or broke disk The induced current will produce or force counteracting the motion M metal detectors The induced current in metals produces a field that is detected A moving bar Bfield into the page x x x x x x x X X X E Two metal rods green placed parallel al a distance d are connected via a resistor R A blue melal bar is placed over the rods as shown in the figure and is then pulled lo the right with a velocity v E a whai is the induced vollage b in whai direclion does the current flow And how large is ii 0 E c whai is the induced force magnilude and direction on the bar What can we say about the force that is used lo pull the blue bar 0 IZI answer Btield into the page x x x X X X X X X X X X Ad A BA 0 M a Induced voltage V 4 B w At At B constant cosei AAAtv x d so A BAtBvdinduced voltage 0 B Direction and magnitude of current The induced field must come out of the page ie oppose existing field Use 2nd right hand rule counterclockwise V RBvd R answer II x x x x l x x x x x x E Induced force Direction Method I The force must oppose the movement of the bar so to the left Method ll Use first right hand rule for the bar force points left Mognitude Findlucedl BlL see chapter 19 B x l x d This force must be just as strong as the one pulling the rod since the velocity is constant Doing work E Since induclion can cause a force on an obiecl lo counler a change in lhe field lhis force can be used lo do work E Example jumping rings demo 0 Q snlrd ring currenl can low c urrenl cannol llow The induced currenl in lhe rirg produces a eld opposile lrom lhe one produced by lhe coil lhe opposirg poles repel and lhe rirg shools in lhe air applicalion magnelic propulsion or example a lrain generating current a The reverse is also true we can do work and generate currents By rotating a loop in a field by hand wind water steam the flux is constantly changing because of the changing anale and A atvoltage is produced emt with 0 angular velocity 0321rf 21rT f rotational frequency T period of oscillation Ac00 NBAwsinlwi A 439 4v 1 ABA At 1B4 At demo hand generator Time varying voltage V ANBAAwosG ANBAA toswt At NBAmsinlwi At side View of loop M Maximum vollage V N B A co M This happens when lhe change in flux is largesi which is when lhe loop is jusl parallel lo lhe field Question M A current is generated by a handgenerator If the person turning the generator increased the speed of turning A the electrical energy produced by the system remains the same B the electrical energy produced by the generator increases C the electrical energy produced by the generator decreased question A current is generated by a handgenerator If the person turning the generator increased the speed of turning a the electrical energy produced by the system remains the same b the electrical energy produced by the generator increases c the electrical energy produced by the generator decreased The change of flux per time unit increases and thus the output voltage Or one can simply use conservation of energy More energy put into the system more must come out Self inductance Iquotquot v M Before the switch is closed l0 and the magnetic field inside the coil is zero as well Hence there is no magnetic flux present in the coil M After the switch is closed I is not zero so a magnetic field is created in the coil and thus a flux M Therefore the flux changed from 0 to some value and a voltage is induced in the coil that opposed the increase of current Self induclance l rquotquot39 K The se induced currenf is proportional lo lhe change in flux K The flux IB is proportional lo 8 eg Bcer e pol Nlenglh for a solenoid K B is proporlional lo lhe currenf lhrough lhe c M So lhe self induced emf volfage is proporlional lo change in currenf A L induclance proporlionalily conslanl 4L Unils VAsVsA E Vse f mduced At usually called Henrys H Induction of o Solenoid E flux ofaooil I I 10an E Change of flux with time At AltIgt AI E induced voltage V ANE l onNA An RL circuit V A solenoid and cr resisior are placed in series Ai i0 the switch is closed One can now set up Kirchhoff s 2nd law for this system AI VAIRALE 0 If you solve this for I you will get V L I 4 8 13 The energy siored in ihe inducior W V2 L 2 RL Circuit II I 2Ft is released I I energy i M When the switch is closed the current only rises slo I because the inductance tries to o ose the flow 4 A pp 1 1ee39n R E Finally it reaches its maximum value lVR When the switch is opened the current only slowly drops because the inductance opposes the reduction V L M T is the time constant 5 I e t R E question R L Iquotquot v M What is the voltage over an inductor in an RL circuit long after the switched has been closed IXI 10 b VR c LR d infinity question R L Iquotquot v M What is the voltage over an inductor in an RL circuit long after the switched has been closed E a 0 b VR c LR at infinity Answer Zero The current is not changing anymore so the change per unit time is zero and hence the voltage ta f C 1 Stat z39ndrrr2ead At example Given R10 Ohm and L2x102 H and V20 V a what is the time constant b what is the maximum current through the system c how long does it take to get to 75 of that current if the switch is closed at i0 a T LR Use given L and R time constant is 2x10393 b maximum current after waiting for some time VR2 A tltLRgt c I14 4 075 2 2xi e IZIIZIIZIIZI 025equotlt39R so i39tLR and ti39 x 2x10393278x10393 PHY232 Remco Zegers zegersnsclmsuedu Room W109 cyclotron building thwwwnsclmsueduzegersphy232hfml MICHIGAN STATE U N I v E R s I T v Resources for PHY232 wwwnsclmsueduzegersphy232html 39SYIIClbUS lectures pre and post lecture notes old exams ohelproom hours oexom amp homework schedule grdding homework is done in LONCAPA HITT clickers are used for extra credit oextro credit quizzes stdrt week 3 oneed to enroll in LONCAPA owill use clicker frequently Suggestions for studying for each chapter make an overview 12 pages based on the lecture notes and textbook Do not only include equations but also main conceptual points Try not to tall behind after making the overview start doing the loncapa using just the overview it you can t do a problem using the overview refer back to the lecturesbook and update your overview use the guide for doing the homework problems provided in the lectures to cut down on the time to start a problem it you get stuck or it takes long to get started with a problem go to the helproom or use the comments from others in loncapa but make sure to try yourselt tirst The helproom is not only for homework go with broader questions as well suggestions by the time an exam comes you have good and tested overviews of each chapter which you can use to make your equation sheet test for weaknesses with the old examsreview problemsbook problems Review the loncapa it you have trouble with unitsconversionsmath etc add to your equation sheet memorizing loncapa problems will likely not get you very tar It you feel that your efforts are not paying off spending a lot of time but get poor scores contact me Don t wait until late in the semester electric charges gt There are 2 types of charges glam gt positive basic carrier proton gt negative basic carrier electron gt In atoms the nucleus consists of gt protons positive H gt neutrons neutral 7 gt And is surrounded by a cloud of electrons neigative gt It the atom is not ionized it is neutral gt By removing electrons it becomes ionized and positively charged since there are more protons than electrons gt Note that the mass of the electron is much smaller than that of the proton and neutron gt me91093826x103931 kg mp167262171x103927 kg electron 1111 C question gt A neutral atom has a more neutrons than protons b more protons than electrons c the same number of neutrons and protons d the same number of protons and electrons e the same number of neutrons electrons and protons Charged objects gt Like charges repel each other 00 00 gt Unlike charges attract each other O 0 Conservation of charge gt In a closed system charge is conserved This means that charge is not created but rather transferred from one object to another gt Charge is quantized there are only discrete amounts of charge The electron carries one unit of negative charge e and the proton carries one unit of positive charge e 1e160219x103919 C Coulomb conduclors gt In conductors Le consisting of conducting malerials electric charge can move freely The resislivily against the flow of charge is very small gt example melals eg Copper one of the electrons can move freely Copper Wira apper Atom 39 elae lran aluctmn flow electron shall mpr a39mm nlac39trnn nucleus insulators amp semiconductors gt In insulators charge can not move freely The resistivity against the flow of charge is very high gt Semiconductors are materials whose properties are in between that of conductors and insulators gt What makes a material a conductorinsulator gt Depends on the shell structure of the atoms involved We will discuss this later demo charge through air demo transterring charge charging by conduction gt An object can be charged by conduction Charged neutral charged neutral charged charged charge has moved by conduction charge is induced but object is still neutral charging by induction ChGrged neutral charge is induced object is still neutral but polarized xexcess charge can escape L x r I I I I I I 39 l 39 I 39 39 39 O GOF I I I I I I I connected charged neutral obedient ruler charged 39 ChGrged etc The earth is an infinite sinksource of electrons demos bending water question a large negatively charged block is placed on an insulated table A neutral metal ball A is rolled towards it and stops before it hits the block Then a second neutral metal ball B is rolled towards ball A After the collision ball A stops closer to the block but without touching and ball B stops further away from the block The block is then removed What is the final charge on balls A and B Ball A is positive ball B is negative Ball A is negative ball B is positive Both ball remain neutral Both balls are positive V V V Coulomb s law 1785 r q1 F21 F12 qz Coulomb sldw F k q1zqzl e directed along the l ne joining the two gt objects q1 qz is d rdciive if the charges have the opposite Sign O 0 double the distance force drops to H4 Is repulswe IHhe charge If the same Sign q OO p lt gt double the charge force increases by factor of 4 ke Coulomb considn18 9875x1 0 7 Nm2C2 s 14 nke885X1 03912 CZNmz 10 be used later superposition principle gt When more than 1 charge acts on the charge of interest each exerts an electric torce Each can be computed separately and then added as vectors q1q3 q2IIQ3l r13 r23 E3Zke 2 F23ke 2 W 7 13 7 23 13 23 q1 q3 q2 Add F3 2 E3 23 gt in this case F13 and F23 are along the same line and can be added as number but be careful with the sign gt example q11nC Cl22nC CI32nC what is the net force on q3 r13r231 m superposition principle ll Remember forces are vectors so treat them accordingly quot313 r 7 r 23 r13 23 13 gt gt gt FF 23 Cl2 F3 Fi3F23 13 3 ql Add In this case you need to take into account the horizontal and vertical directions separately and then combine them to get the resultant force le like parabolic motion questions true false AO 0C 0 B a it A and C are positive B is pushed away from A and C b it A and B are positive A and B will move further apart c it A is neutral and C is positive B will move along the line BC cl it AB and C have the same charge they will separate turther Ioncapo 1 2 do LC problems 1341012 useful for LC9 a srmple electroscope demo gt Two equal masses are charged positively both 1 uC and hung from massless ropes They separate as shown in the figure What is the mass of each gravitational amp electrical force gt two objects of mass 10 kg are placed 1 meter from each other Each has 1x101o excess electrons What is the net force on each of the objects gt do problem 11 PHY232 Remco Zegers loncapa Electric Forces amp Fields electric fields gt Instead of a force acting on an object A by an object B somehow magically over the distance between them one can consider that object A is situated in a field arising from the presence of object B gt Because object A is in the field created by object B it feels a force gt The electric field produced by a charge Q at the location of a smau test charge go is defined as a F a a The magnitude of E only E E gt F qu depends on the go charge of Q and not the Q sign and size of the test W ke 2 charge electric fields II gt elecfric fields and forces due fo a charge Q on fesf charges of differenf charge and af differenf disfances fesf charge E F direcfionquot direcfionquot A keQr2 kquor2 B keQr2 kquor2 C keQzrv kquozrv means poinfing away from Q electric fields lll gt To determine the electric field at a certain point 3 due to the presence of two other charges 1 and 2 use the superposition principle q E13 ke lqlzl E23 ke I 2239 r13 C13 r23 r13 r23 2339Cl2 E32E13E23 E13 E3 quotCh E3 is independent of the charge q3 The field would also be present it q3 were not there Note that in this case F13 and F23 point in the opposite direction of E13 and E12 question gt 2 charges equal in magnitude are lined up as shown in the figures A third point no charge P is defined as well In which case is the magnitude of the electric field at P largesl The distance between neighboring points is constant CCICICCICCCI P P A B C D elecfric field lines gt To visualize elecfric fields one can draw field lines fhaf poinf in fhe direc on of fhe field af any poinf following fhe following rules gt The ele cfric field vecfor E is fangenf fo fhe elecfrical field lines af each poinf gt The number of lines per unif area fhrough a surface perpendicular fo fhe lines is proporfional fo fhe field sfreng gt field lines sfarf from a posifive charge gt field lines end af a negafive charge gt field lines never cross gt I electric field lines II gt Following these rules one can draw the field lines for any system of charged objects E J EMU5 EMV GE 1 E EA En A E E E E A A a En Er TA En E eleciric field lines II gt examples nummmumm mmmwmfm 39 13M 7 1 39 I 4 4 7 rNN NV 411 questions charge P is a positive b negative charge Q is a positive b negative charge P is a larger b smaller than charge Q a negative charge at Rwould move a toward P b away from P c toward Q do problems 6789 PHY232 Remco Zegers Ioncapo Electric Forces amp Fields conducfors gt In fhe absence of any exfernal charges an insulated conducfor is in equilibrium which means gt fhe elecfric field is zero everywhere in fhe conducfor gtsince nef field would resulf in mofion of charge gt excess charge resides on fhe surface gtsince elecfric force 1r2 excess charge is repelled gt fhe field jusf oufside fhe conducfor is perpendicular fo fhe surface gt ofherwise charge would move over fhe surface gt charge accumulafes where fhe curvafure of fhe surface is smallesf gt charges move aparf rnore af flaffer surfaces Millikan oildrop experiment I No Efield battery off mgkv k drag constant known so mkvg oConsider first the case where the battery was switched off 00quot droplets will fall and reach a constant velocity v which can be measured At this velocity the gravitational force balances the frictional drag force which equals kdmgx velocity From this the mass of the droplet can be determined Now the Efield is switched on Millikan s oil drop experiment II t 15 Oil drop mg The droplets are negatively charged By tuning E one can suspend them in air If that happens the electrical force balances the gravitational force and qEm Millikan found qmgEn 16E l9 and thus discovered that charge was quantized Nobel prize 1923 Area m Electric flux normal Jquot rA um 15 The number of field lines N fhrough a surface A is proportional fo fhe electric field NEA The flux ltDEA Nm2C If fhe field lines make an angle wifh fhe surface gt ltIgtEAcos9 where 9 is fhe angle befween fhe field lines and fhe normal fo fhe surface For field lines going fhrough a closed surface like a sphere field lines enfering fhe inferior are negafive and fhose leaving fhe inferior are posifive Gauss Law gt Consider a point charge q Imagine a sphere with radius r surrounding the charge The Etield and flux an here on the sphere are r 6 Ezker z CIDzEAzkeiZMr2 47zkeq2q80 r gt It can be proven that this holds for any closed surface Gauss Law D Qinside 80 Efield insideoufside a sphere Faraday s cage gt consider imaginary surface A gt Qinsideao0EA so E0 no field inside charged sphere gt consider imaginary surface B gt Qinside80Q 80EA so EQeoA nef field oufside charged sphere question A point charge q is located at the center of a spherical shell with radius a and charge q uniformly distributed over its surface What is the Etield a anywhere outside the shell and b at a point inside the shell at distance r from the center CD M 2 EA Gauss39 Law 90 question A point charge q is located at the center of a spherical shell with radius a and charge 2q uniformly distributed over its surface What is the Etield a anywhere outside the shell and b how much charge resides on the inner and outer surface of the shell CD M 2 EA Gauss39 Law 90 question A neutral object A is placed at a distance r001 m away from a charge B of 1 uC a What is the electric field at point A b What is the electric force on object A c What is the flux through the sphere around object B that has a radius r001 d A is replaced by a charge object C of 1 C What is the force on C e What is the force on B PHY232 Spring 2008 Jon Pumplin httpwwwpamsuedupumpinphy232 MICHIGAN STATE original ppt courtesy of Remco Zegers WW work gt A force is conservative if the work done on an object when moving from A to B does not depend on the path followed Consequently work was defined as w PEi PEf APE gt This was derived in Phy231 for a gravitational force but as we saw in the previous chapter gravitational and Coulomb forces are very similar FgGm1m2r122 with G667x103911 Nm2kg2 Fekeq1q2r122 with ke899x109 Nm2C2 Hence The Coulomb force is a conservative force work amp potential energy gt WABchose with 6 the angle between F and direction of B movement so d gt WABFd gt WABqu since FqE 3 gt work done BY the field ON the charge W is positive gt APEWABqu negative so E field from A to B over a distance decreased D We can ignore gravity why gt Conservation of energy gt What is the work done by the field gt gt What is the change in PE gt APEAKE 0 2 2 gt If initially at rest what is its speed at AKE12mvf quotVi gt 12mvf2qu B gt v 2qum work amp potential energy II gt Consider the same situation for a charge of q gt Can it move from A to B without an external force being applied assuming the charge is initially A and finally B at rest gt WABqu negative so work must be done by the charge This can only happen if an external force is applied Note if the charge had an initial velocity the energy could come from the kinetic energy le it would slow down If the charge is at rest at A and B external work done qu If the charge has final velocity v then external work done W12mvf2qEd Conclusion gt In the absence of external forces a positive charge placed in an electric field will move along the field lines from to to reduce the potential energy gt In the absence of external forces a negative charge placed in an electric field will move along the field lines from to to reduce the potential energy ques on 39 2m 1m gt a negatively charged 1 pC mass of 1 g is shot diagonally in an electric field created by a negatively charge plate E100 NC It starts at 2 m distance from the plate and stops 1 m from the plate before turning back What was the initial velocity in the direction along the field lines answer gt Note the direction along the surface of the plate does not play a role there is no force in that direction Kinetic energy balance gt Initial kinetic energy 12mv2050001vx2vy2 gt Final at turning point kinetic energy 050001vx2 gt Change in kinetic energy AKE050001vy2 5x10quot vy2 Potential energy balance gt Change in Potential energy APEqu1x10quot5100110394 J gt Conservation of energy APEAKE0 so 5x104vy21040 gt vy044 ms Electrical potential The change in electrical potential energy of a particle of charge Q in a field with strength E over a distance d depends on the charge of the particle EQEd For convenience it is useful to define the difference in electrical potential between two points AV that is independent of the charge that is moving AV The electrical potential difference has units JC which is usually referred to as Volt V It is a scalar Since AV Ed so E AVd the units of E NC before can also be given as Vm They are equivalent but Vm is more often used Electric potential due to a single charge M 9 r 0 1C gt the potential at a distance r away from a charge q is the work done in bringing a charge of 1 C from infinity V0 to the point r Vkeqr gt If the charge that is creating the potential is negative q then Vkeqr gt If the field is created by more than one charge then the superposition principle can be used to calculate the potential at any point example 1 2 0 1m 0 1C 392C r a what is the electric field at a distance r b what is the electric potential at a distance r a EE1E2keQ1r2keQz1r2ke1r2ke21r2 ke 1r221r2 Note the E is a vector b VV1V2keQ1rke021rke1r21r Note the V is a scalar ques on gt a proton is moving in the direction of the electric field During this process the potential energy and its electric potential b increases decreases c decreases increases d increases increases e decreases decreases 41 APEWABqu so the potential energy decreases proton is positive AV APEq so the electric potential that the proton feels decreases Note if the proton were exchanged for an electron moving in the same direction the potential energy would increase electron is negative but the electric potential would still decrease since the latter is independent of the particle that is moving in the field eguipotential surfaces 1r Tim 39a 41w x any l 39II un a A m ugh miniRut Juan 139 7 3 V lyh39umlrml39iL Hum va 1 if i quotZ L39mssmtnnsd fgqnipnl n al 39 hrn1n quotPM hm I I Tim zsz rutRim amp vy1ml mrmmm r nahnu mummij V VVVVV A capacitor symbol for capacitor when used in electric circuit 3 w r f if i 9 5 u Equot g is a device to create a constant electric field The potential difference VEd is a device to store charge and in electrical circuits the charge stored Q is proportional to the potential difference V QCV C is the capacitance units CV or Farad F very often C is given in terms of uF 10396F nF 10399F or pF 103912F Other shapes exist but for a parallel plate capacitor CeoAd where 2014 pi k 885x103912 Fm and A the area of the plates electric circuits batteries ASimpleElectricCimnit gt The battery does work eg using chemical energy to move positive charge from the terminal to the terminal Chemical energy is transformed into electrical potential energy gt Once at the terminal the charge can move through an external circuit to do work transforming electrical potential energy into other forms Symbol used in electric circuits Our first circuit 10nF 1V Capacitor r Th u l 39 5 L L L L n 1 I v u y 1 r 1 1 1 r example a 12V battery is connected to a capacitor ol 10 nF How much charge is stored answer QCV10x105 x12V120 nC 39 OTE Q on one plate Q on the other total is 0 but Q is called the chargequot r il the battery is replaced by a 300 V battery and the capacitor is 2000nF how much charge is stored answer QCV2000x109 x 300V0GC u L L unwhich is the same as a1 kg ball moving at a velocity ol134 rns capacitors in parallel C110nF At the points 0 the potential IS fIxed to one value say 12V at A and 0 V at B C210nF I E This means that the capacitances C1 1 V and C2 must have the same Voltage I The total charge stored is QQ1Qz gt We can replace C1 and C2 with one equivalent capacitor Q1C1V amp 0202V is replaced by QCqu since QQ1Q2 C1VC2VCqu so gt CeqC1C2 gt This holds for any combination of parallel placed capacitances CeqC1C2C3 gt The equivalent capacitance is larger than each of the components capacitors in series I J The voltage drop of 12V is over both C110nF CZ1OnF capacitors VV1V2 I l The two plates enclosed Inn 39are 1 V not connected to the battery and must I be neutral on average Therefore the charge stored in C1 and 02 are the same gt we can again replace C1 and C2 with one equivalent capacitor but now we start from VV1V2 so VQC1QC2QCeel and thus 1Ceq1C1 1C2 gt This holds for any combination of in series placed capacitances 1Ceq1C11C21C3 gt The equivalent capacitor is smaller than each of the components ques on gt Given three capacitors of 1 nF an capacitor can be constructed that has minimally a capacitance of d 1nF e 15 nF f 3nF The smallest possible is by putting the three in series 1Ceq1C11C21C31 1 1 3 so Ceq1l3 nF Fun with capacitors what is the equivalent C c AHA CBquot Juli CLquot 39 l 12v STRATEGY replace subgroups of capacitors starting at the smallest level and slowly building up gt step 1 C4 and C5 and C6 are in parallel They can be replaced by once equivalent C4 6C4C5C6 step II C I quot39 C4554 Juli CLquot I I 12V gt C3 and C456 are in series Replace with equivalent C 1C34561C31C456 so C3456C3C456C3C456 gt C1 and C2 are in series Replace with equivalent C 1C121C11C2 so C12C1C2C1C2 step Ill C 3 Cll C1 123456 n fl 1rv gt C12 and C3456 are in parallel replace by equivalent C of C123456C12C3456 problem A C4 I I 3 39 39I G C110nF I f 0220nF C C C310nF 1 4 C410nF What is Vab C520nF 39 l 12v V12V34512V C45C4C510nF20nF30nF C345C3C45C3C453004075nF Q345V345C34512V75nF90nC Q450345 V45Q45IC4590nCI30nF3V check V3Q3IC3Q345IC390nCI10nF9V V3V4512V okay VVVVVVV energy stored in a capacitor 0 V o mm 0 the work done transferring a small amount AQ from to takes an amount of work equal to AWVAQ At the same time V is increased since VQAQC The total work done when moving charge Q starting at V0 equals W12QV12CVV12CV2 Therefore the amount of energy stored in a capacitor equals EC12 c v2 example gt A parallelplate capacitor is constructed with plate area of 040 m2 and a plate separation of 01mm How much energy is stored when it is charged to a potential difference of 12V answer First calculate CsoAd885x103912 x 040 00001 354x108 F Energy stored E12CV205x354x10quot3x122255x10396 J Now let s assume a 2000uF capacitor being charged with a 300V battery E12CV290J This is similar to a ball of 1 kg being fired at 134 ms capacitors ll A material 1 vacuunt 100000 039 air 100059 ghss 55 paper 33 vvater 80 gt the charge density of one of the plates is defined as oQA gt The equation CeoAd assumes the area between the plates is in vacuum free space gt If the space is replaced by an insulating material the constant 80 must be replaced by Keowhere K kappa is the dielectric constant for that material relative to vacuum gt Therefore C1lt20Ad Inserting a Dielectric molecules such as lhose in glass can be polarized when placed in an Elield lhey orienl lhemselves along lhe eld lines lhe negallve plales allracl lhe posi 39ve side of lhe molecules near lo posilive plale nel negalive charge is collecled near lhe negalive plale nel posilive charge is collecled al polenlial difference V belween ll a ballery was connecled more charge can be added increasing lhe capacilance l39rom Clo x limes C is palled the Wcot fQ i g e aterial problem gt An amount of 10 J is stored in a parallel plate capacitor with C10nF Then the plates are disconnected from the battery and a plate of material is inserted between the plates A voltage drop of 1000 V is recorded What is the dielectric constant of the material answer step 1 Ec12CV2 so 1005x10x109V2 V44721 V step 2 after disconnecting and inserting the plate the voltage over the capacitor is equal to Voriginall 1 So 44721 1 00044721 1c K1023 problem gt An ideal parallel plate capacitor is connected to a battery and becomes fully charged The capacitor is then disconnected and the separation between the plates is increased in such a way that no m off The energy stored in the capacitor has c decreased d not changed e become zero answer Ec12CV2 with C1lteoAd If d increases C becomes smaller The charge remains the same and QCV So if C becomes smaller V becomes larger by the same factor Rewrite Ec12CV2120V Since G is constant and V goes up Ec must increase PHY232 Spring 2008 Jon Pumplin httpwwwpamsuedupumpinPHY232 Ppt courtesy of Remco Zegers Electric current So far we have studied Static Electricity Now consider the situation where charge can move and hence produce an electric current A 9 l 1 9 J 9 Current amount of charge AQ that flows through an area A divided by the time interval At I At Electric current II gt A matter of convention The direction of current is the direction in which positive charges flow even though the flow is often of electrons negative High v 9 Low v High V O gt Low V gtRemember positive charge moves from high potential to low potential electric current Ill what really happens gt When electrons move through a wire they undergo many collisions and a typical path looks like Typical Path of anElectrun gt Because of the collisions the velocity is on average constant gt The drift velocity of the electrons is actually very slow less than 1 meter per hour So why can we have hinh currents demo model of resistance Because there are so many electrons electric current IV gt let s assume the average electron speed is v 39 gt consider one electron at point x gt after time t it will have moved gt a distance D vt I gt in fact all the electrons over the distance D will quot have moved gt the volume of the cylinder V A D A v t gt if n number of electrons per unit volume the I r number of electrons moved is nVnAvt gt the charge AQ that has been moved nAvtq gt current I AQt n A v q I question r A currenl ol1 A is running lhrough a Copper wire wilh cross seclion 1mm2 Each Copper alom produces 1 free eleclron c How many free charge carriers per unil volume are lhere Given lhal lhe molar mass of Cu is 635 g and lhe densily of copper is 892 gcmz b Whal is lhe drill velocily gta The volumetaken by 1 mol of Cu atoms mm 7 I k 9 V 7 712 cm 712 gtlt 10quot mquot1 7 u p 892 yvm the number of electrons is also 1 mol in this volume NA602x1 021 so gt I 7 602 x 1013 u 7 W 540 x 10 w question I 39nq gt b Use I nqvA so 1 gt with n846x1028 m393 A1mm2 1x10396m2 q16x103919 C and 1A1CS gt so v246x10395 ms gt Le this is 0089 m in one hour wait a second gt Wasn t charge supposed to be collected on the surface of a conductor That only happens when the conductor has a Net Charge more electrons than protons or fewer electrons than protons The conducting wires we are talking about are neutral batteries gt A battery can produce a potential difference between the anode negative and cathode positive When connected le using a wire or via a device current can flow gt The charge is created through chemical reactions Once the chemical fuel is used the battery is empty gt commonly used are zinccarbon batteries A simple circuit A basic eleclric circuil consisls of a connecled via a wire and some device As long as lhe circuil is open no currenl will flow and hence lhe quot 39 device nol wo Power sources can be DC Direcl Currenl or AC Allernaling Currenl We will deal wilh DC circuils rsl ques on gt Which of the following lights will not shine after the switches e closed a 2 lights 2 and 3 will not shine since there is b 23 no potential difference over the contacts c 234 d 1234 how to measure current gt The current anywhere between A and B must be constant else electrons would quot accumulate at a certain point in the line gt A device to measure current in the light should therefore be placed in line in series with the light Either side gt The device is called an Ampere meter am meter how to measure voltage gt To measure the voltage to the light realize that we need to measure the potential difference between A and B gt A device to measure voltage to the light should therefore be placed in parallel with the light gt The device is called a Volt meter Resistance l Typical P l If an Electron gt When electrons move through a material they undergo many collisions which hinders the motion like friction gt Without such collisions the electrons would accelerate since there is a force acting on them gt The resistive force counterbalances the electric force so the drift velocity is constant gt When the resistive force is high the current will go down if the voltage difference that drives the motion remains the same Resistance ll Compare with water flow through a pipe If the pipe becomes narrow flow is reduced If the length over which the pipe is narrow becomes longer flow is further reduced 1 so resistance R R N 4 Resistance Ill 39v gt voltage is the equivalent of pressure and current the equivalent of flow gt If pressure voltage difference increases the flow current will increase gt If the resistance increases the flow current will go down if the pressure difference remains the same Ohm s law and resistivity gt Ohm s law V V f 7 l l IR I R R I gt For a specific material the resistance R can be calculated using l RIPE gt where R resistance in VAQ Ohm p the resistivity material dependent in 9m the length of the object and A the cross section of the object Ohm s law Ohm s law implies that I is proportional to V which is true for many materials but not for all mm 7 Ohmic resistance Nonohmic resistance ques on gt A voltage of 100V is put over a thick wire of unknown material The current is measured is 45x103 A The cross section of the wire is 1cm2 and the length is 10m What material is the cable made of R Vl 0022 plA so p0022A A1cm200001m2 10 m p22X10397 Ohmm Lead Material Resistivity Ohmm Silver 1 59x10398 Gold 244x108 Lead 22x1043 Silicon 640 Quartz 75x1016 a resistor bank is an adjustable resistor ad39ust length of wire demo ques on gt A person measures the resistance over a 10 m long cable through a measurement of V and l He finds at V10 V that 1 A A second cable made of the same material and length but with a radius that is 2 times larger than the original cable is then studied At a voltage V10V what current is measured 2 a 1A R PE A m If radiusx2 then Ax4 and Rx025 d 8A VR so lx4 and thus 4A superconductors For some material the resistivity P drops to nearzero below a certain temperature the critical temperature For such a material current would continue to flow even if the potential I is zero l T Element Tc K Mercury 415 Tin 369 Lead 726 Niobium 92 Aluminum 114 Cuprate Perovskite 138 resistors in a circuit resistors are commonly used in circuits their resistance is usually much higher than the resistance of the connecting wires and the wires are usually ignored deviceslights etc are also resistjii The symbol used for a resistor i ques on gt a resistor of 10 Ohm is put in a circuit 10V is put over the resistor The resistor is replaced by one of 100 Ohm By what factor does the current through the resistor change gt a 01 gt b 1 unchanged gt c 10 if if irgsf I 5 If Rx10 then 10 so a 01 7 tungsten amem bulb 51316 1 I I at E 397 Hail Inyulu39rmn elemmsm ism zumlum electrical energy and power gt consider the circuit The potential energy lost by a charge AQ falling through a potential difference V is V Fa APE ll ixtg 393 gt The energy lost per time unit the g 7 power dissipated is APE A a V2 P V 91r IIER in At R P Watts Js For the energy consumed Eth often kWh kilowatt hour is used 1 kWh energy consumed in 1 hour at a rate of 1000 W 1 kWh1000W x 3600 s 36x106 J ques on gt A voltage of 10 V is put over a wire with cross section A and length l The wire is then replaced with one of the same material that has cross section 2A and length 4 At the same time the voltage is increased by a factor of 2 By what factor does the dissipated power change gt a the same gt b doubles factor of 2 gt c quadruples factor of 4 gt d halves factor of 05 l Ax2 and lx4 so Rx2 R U T Vx2 note that lVRconstant PV2R so Px2 PHY232 Spring 2008 Jon Pumplin p wwwpamsuedupumpinphy232 MICHIGAN STATE original pp courtesy olHemco egers U N V E R S T Y Electric Charges gt Two types of charge in atom gt positive carrier proton gt negative carrier electron gt Nucleus consists of gt Protons positive gtneutrons neutral a gt Nucleus is surrounded by cloud of electrons negative gt If the atom is not ionized it is neutral gt By removing electrons it becomes ionized and positively charged since there are more protons than electrons gt Mass of the electron is much smaller than that of the proton or neutron gt me9109x103931 kg mp16726x103927 kg gt Question A neutral atom has more neutrons than protons more protons than electrons the same number of neutrons and protons the same number of protons and electrons the same number of neutrons electrons and protons Electric Forces gt Unlike charges attract each other That s what keeps the electrons attached to their atom O0 gt Like charges repel each other A different strong force keeps the protons attached to their nucleus OO 00 Conservation of charge gt In a closed system charge is conserved This means that charge is not created but can be transferred from one object to another fa gt Charge is quantized there are only discrete amounts of charge The electron carries one unit of negative charge e and the proton carries one unit of positive charge e 1e1602x1019 C Coulomb conductors gt In conductors ie conducting materials electric charge can move freely The resistance to the flow of charge is very small Example metals like Copper one of the electrons from each atom can move freely EDWIN Wira Eupplr Atom 39 39 7 l V7 I stun IKE Insulation Iquot 139quot 7 1 INquot Ha I I q H r quotI l l 7 1 EH l I i l f h t i ll at V l i 7 I W I m r imam IIF l or ll 51 A 391 1 i nucleus 1 1 Fg quot1 U electron I 3 alactmn flaw electron nucleus electron shall mppm atuml Conductors Insulators amp Semiconductors gt In conductors charge can move freely The resistance to flow of charge is very low gt In insulators charge cannot move freely The resistance to flow of charge is very high gt Semiconductors are materials whose properties are in between that of conductors and insulators used in transistors What makes a material a conductor or insulator or semiconductor It depends on the shell structure of the atoms involved We will discuss this later in the course charging by conduction gt An object can be charged by conduction Charged neutral charge is induced but object is still neutral charged neutral contact charged charged charge has moved by conduction charging by induction charge is induced object is still neutral but olarized charged neutralx p excess charge can escape Iquot I l gt quot l to earth charged neutral charged 39 Charged The earth is an infinite sinksource of electrons ques on a large negatively charged block is placed on an insulated table A neutral metal ball A is rolled towards it and stops before it hits the block Then a second neutral metal ball B is rolled towards ball A After the collision ball A stops closer to the block but without touching and ball B stops further away from the block The block is then removed What is the final charge on balls A and B Ball A is positive ball B is negative Ball A is negative ball B is positive Both ball remain neutral Both balls are positive answer although A is neutral near the block one side is positive the other negative The same will happen for ball B When A and B collide some electrons will jump from A to B conduction Correct answer A Coulomb s law r lqlllqzl 391 F F12 it e r2 F2 7 r1 F12 gt directed along the line joining the CI El two objects gt is attractive if the char es have 6 a 9 0 6 the opposite sign double the distance force drops to ii4 gt IS repulswe If the charge If the same sign gt ke Coulomb constant9x109 6 Nm2C2 5 do uble the change gt 801 nke885x1012 force increases by Factor of4 to be used later Superposition Principle When more than one charge acts on the charge of interest each exerts an electric force Each can be computed separately and then added as vectors lqllql lqllql 1 3 2 3 k F k r r 13 e r 23 er 132 232 q1 F13 q3 F23 q2 Add F32F1323 in this case F13 and F23 are along the same line and can be added as numbers but be careful with the sign Choose a coordinate system and stick to it Superposition Principle ll Remember forces are vectors so treat them accordingly 399 9 lq IIQI F k 1 3 F k 13 e Iquot 23 e 2 r 13 23 F 2F F 3 13 23 q1 Add In this case you need to take into account the horizontal and vertical directions separately and then combine them to get the resultant force a b C 039 questions true false A0 5 QB if A and C are positive B is pushed away from A and C if A is positive and B is positive A and B will move further apart if A is neutral and C is positive B will move along the line BC if AB and C have the same charge they will separate further Answers to questions A0 5 QB a if A and C are positive B is pushed away from A and C b if A is positive and B is positive A and B will move further apart c if A is neutral and C is positive B will move along the line BC d if AB and C have the same charge they will separate further answers b false if B is negative it will move towards A and C c false if C is negative and the absolute charge much larger than A and B A and B could come closer d false B might be neutral and not move at all e true the will all feel an outward pointing force A simple Electroscope gt Two equal masses are charged positively both 1 uC and hung from massless ropes They separate as shown in the figure What is the mass of each 1m tanoc0011FeFg Fekeq1qzlr122 coulomb force Fgmg981m gra3itational force with q1q2q and r122001002 m k899x109 Nm2C2 so mFe001gkeq2001gr122 m229 kg The electric force is very strong compared to the gravitational force of the masses on each other Compare FgGm1m2r122 with G667x1011 Nm2kg2 Fekeq1q2lr122 with ke899x109 Nm2C2 Electric Fields gt Instead of a force acting on an object A by an object B magically over the distance between them one can consider that object A is situated in a field arising from the presence of object B gt Because object A is in the field created by object B it feels a force gt The electric field produced by a charge 0 at the location of 39 defined as The magnitude of E only depends on the charge of Q and not the sign and size of the test charge electric fields II gt To determine the electric field at a certain point 3 due to the presence of two other charges 1 and 2 use the superposition principle E lqll q2 13 e r 2 23 gr 2 13 23 23 q2 E3ZE13E23 E3 is independent of the test charge q0 quesuon 2 equal charges are lined up as shown in the figures A third point P with no charge is also defined The distance between neighboring points is constant In which case is the magnitude of the electric field at P largest PPP A B C Answer to question gt 2 equal charges are lined up as shown in the figures A third point no charge P is defined as well In which case is the magnitude of the electric field at P largest The distance between neighboring points is constant A B C D Cl CTC 02 0 P P p gt gt 4 gt C is correct electric field lines gt To visualize electric fields one can draw field lines that point in the direction of the field at any point following the following rules gt The electric field vector E is tangent to the electrical field lines at each point gt The number of lines per unit area through a surface perpendicular to the lines is proportional to the field strength gt field lines start from a positive charge or infinity gt field lines end at a negative charge or infinity electric field lines Following these rules one can draw the field lines for any system of charged objects t A t t wv Gm we electric field lines quotI examples mmmmmmmmmmmmm K x w r w questions charge P is a positive or b negative charge Q is a positive or b negative charge P is a larger or b smaller than charge Q a negative charge at R would move a toward P b awayfrom P c toward Q d none of the above answers charge P is a positive charge Q is a positive charge P is a larger than charge Q a negative charge at R would move d toward a point between P and Q conductors ElectroSTATlCS an insulated conductor is in equilibrium 9 gt The electric field is zero everywhere in the conductor gtsince net field would result in motion gt Any excess charge resides on the surface gt since electric force 1r2 excess charge is repelled gt The field just outside the conductor is perpendicular to the surface gt otherwise charge would move over the surface gt Charge accumulates where the curvature of the surface is smallest gt charges are farther apart at flatter surfaces ques on A point charge q is located at the center of a spherical shell with radius a and charge q uniformly distributed over its surface What is the Efield g anywhere outside the shell and h at a point inside the shell at distance r from the center
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