New User Special Price Expires in

Let's log you in.

Sign in with Facebook


Don't have a StudySoup account? Create one here!


Create a StudySoup account

Be part of our community, it's free to join!

Sign up with Facebook


Create your account
By creating an account you agree to StudySoup's terms and conditions and privacy policy

Already have a StudySoup account? Login here


by: Mrs. Peter Toy


Mrs. Peter Toy

GPA 3.96


Almost Ready


These notes were just uploaded, and will be ready to view shortly.

Purchase these notes here, or revisit this page.

Either way, we'll remind you when they're ready :)

Preview These Notes for FREE

Get a free preview of these Notes, just enter your email below.

Unlock Preview
Unlock Preview

Preview these materials now for free

Why put in your email? Get access to more of this material and other relevant free materials for your school

View Preview

About this Document

Class Notes
25 ?




Popular in Course

Popular in Physics 2

This 141 page Class Notes was uploaded by Mrs. Peter Toy on Friday October 30, 2015. The Class Notes belongs to PHYS 1120 at University of Colorado at Boulder taught by Staff in Fall. Since its upload, it has received 31 views. For similar materials see /class/232110/phys-1120-university-of-colorado-at-boulder in Physics 2 at University of Colorado at Boulder.




Report this Material


What is Karma?


Karma is the currency of StudySoup.

You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/30/15
CT 0231 Two uniformly charge spheres are attached to frictionless pucks on an air table The charge on sphere 2 is three times the charge on sphere 1 Which force diagram correctly shows the magnitude and direction of the electrostatic forces on the two b A211 When you rub a Teflon rod with felt the Teflon rod becomes negatively charged and the felt becomes positively charged As a consequence of rubbing the rod with the felt A the rod and felt both gain mass B the rod and felt both lose mass C the rod gains mass and the felt loses mass D the rod loses mass and the felt gains mass E none of the above CT 232 Which charge feels the greatest force Q O o W AI Cl r BI 3q 5q two of the charges tie for the largest force all of the charges feel the same force CT 2511 Consider the charge configuration shown below What i3 the direc oh of the not force on the q charge CT 234 An electric dipole Q and Q separated by a distance 8 is placed along the xaxis as shown A positive test charge q is placed to the right of a dipole The test charge feels a force that is A zero B to the right C to the left D up E down 0 0 0 QSQ arrrrkkRR L 1444Ah RR 41 KN A R R K MVJ a 321 x a Pk frlxr LVLY WW WW K yr Jqu wlvuu gznexltdwW 1 Damnmznm m 03033 ESQ ltm503 lvm much money Will a lypcaperson not all y from this class earn m theirenlre lifetime W 2007 dollarg Don tlust guess try to eslmale the answer CAPA 13 tonight New online participation survey is still up But not for much longer Reading u x u I New pretest available for next Tuesday Vmu mek HIE Last Inductors and AC circuits Today Maxwell39s equations P Punk 2 Next More ofMaxwell M P IV as usual Why does almost everyone use AC Voltage 1 Ease of generating Recall a rotating coil in a B eld creates an AC induced First large scale test at Niagara Falls 2 AC is easy to change from one peak Voltage level to another Transformers Light bulbs and appliances with motors vacuum cleaners blenders use AC Voltage to operate But devices with electronic circuits TV s computers phones need DC Voltage constant to function The power supply in TV s computers etc convert AC Voltage from the wall into DC Voltage typically 215 Volts that the circuitry needs A 600 Watt hairdryer is attached to 120 VAC circuit What is the peak current through the hairdryer to within 5 A0A B5A C7A D10A EOther rms peak15 P IV The instantaneous power consumed by my Deluxe Toaster Oven looks like this as a function of time Which of the following is correct P ave V rms l rms 1200W 120V 10A1200120 1200 W 170 V 7 A 1200170 600 W 120 V 5 A 600120 600 W 170 V 35 A 600170 None of the above is right 3939 IU IO IUU IJgt A 130 s B 160 s C 1120 8 D Other AC Voltage is used to distribute power to homes rather than DC Voltage Electrical power is transmitted from power plants to cities with big aluminum power cables Much energy is wasted because of resistance in the cables the cables heat up i2R losses To reduce this waste power is transmitted at very high voltage 100 W to 1 MV Pfrom plant to city iV constant Set by the needs of the city and the capacity of the plant DC versus AC Lots of good books on the science and the people involved Magnetic Energy Density Recall that for a capacitor C there is stored potential energy in the electric el 1 U 7CV2 2 The energy is stored in the electric eld and the density is E 1 A 5 lEl2 U Volume For an inductor L with current i there is stored energy in the magnetic eld 1 U 7Li2 2 The energy density in the magnetic eld is U ME 7 Volume Jil iz 2 Clicker Question The same current i ows through solenoid 1 and solenoid 2 Solenoid 2 is twice as lon and h twice as many turns as solenoid 1 and has twice the diameter Hint for a solenoid B u n i What is the ratio ofthe magnetic energy contained in solenoid 2to that in solenoid 1that is what is UK F x A2 B 4 C 16 E None ofthese It takes work to get current owing through an inductor You must work against the back EMF which opposes any change in the current That work potential energy stored U 12 Li2 And is thus stored in the inductor s magnetic eld EMF around a stationary closed loop is de ned as efEdi May And thus Faraday s Law can be written as A changing B eld creates a new kind of E eld E If B is decreasing 9 E eld is COW If B is increasing 9 E eld is Clockwise The nonCoulomb E eld is very different from the Coulomb E eld created by single electric charges A 5fEdf 0 Loop If there are only stationary charges no currents then there re no B elds In this case there is no Magnetic Flux in this situation only 5 fEdf 0 specialcasc only Loop A solenoid has an increasing current causing an increasing B eld inside An endon view ofthe solenoid is shown What is the direction ofthe force on an electron shown below the solenoid A D B Bin C EU a electron A long solenoid radius R contains a uniform B eld that increases steadily Bt c t Consider the dashed Amperian loopquot What is E IdL around the loop Whatis ffBdA through the the dashed curve Let s put this together EgcalldB ct Faraday says EdlrErEffBdA Recall the le side is 21rrE Recall ffBdAnB 0 how does E depend on r A E0 B E 1r C E r D E rquot2 E Other Electromagnetism in Full f5 d7 quot I B dt A A dd f3 39 d Helena 080 j 50 fBdA70 CT 2513 Which is the correct direction of the electric field created by charge Q at the location shown below and left of the charge CT 261 Two Charges each Q are equal distances from the origin What is the direction of the electric field at the origin E none of these CT 262 A point in empty space is near 3 charges The distances from the pt to each charge are identical The direction of the electric field at that point is A Some angle lt 45 below the xdirection B 45 below the xdirection C along the y direction D Some other angle E It is zero there CT 2311a A circular ring of radius R uniformly charged with total charge Q is in the xy plane centered on the origin The electric field at position zh on the z axis due to a small piece of the ring with charge dQ is shown What is the magnitude of the field dE A quotQ I I B 019 If C de 7 IR 122 de IR 112 E other CT 2311b What is dEz the zcomponent of dE A dEsin e B dE cos 6 C dE tan 9 D none of these CT 2311C E other CT 2311C Which graph correctly represents the Electric field E2 on the zaxis E non of these are correct HJD l What is the magnitude of the net force on a dipole in a uniform electric field Electric field E A 0 B Q E C2 Q E D Q E sin 6 E Q E d sin 6 24 1 There are no charges in the regions shown but there are charges present outside the boxed regions shown Which of the following are possible electric field line configurations A a only B b only C c only D a and b E other combination CT 268 Which sketch best shows the E field around 2 identical protons A B CT 272 Two open surfaces are in an E field Surface A is a flat circular disk of radius R Surface B is a hollowcup hemisphere of the same radius R Which surface has a greater flux through it A A BB C Both surfaces have the same flux D Not enough info CT 273b Which has the greater flux now A A BB C Both surfaces have the same flux D Not enough info CT 274 What is the net electric flux through the closed cylindrical surface shown B positive C negative CT 278b An imaginary Gaussian surface in the shape of a right cylinder is shown The flux through the surface is radius r B 2EA D 2Achr2 E CT 2721b A sphere of radius R has a total charge Q spread uniformly throughout its volume What is the total charge enclosed by the small centered sphere of radius r A Q igr r2 r3 B QF C QF E None of these CT 2721C If the electric field at distance r from the cee eriof the sphere is E what is the flux QDEodA out of the small sphere AiErcR2 BE47R2 C E 47 r2 Di 39 FZE 3 Ei RzE 3 CT 2721d Within the sphere how does the E field magnitude E depend on distance r from the origin AEocr BEocr2 CEocr3 D Econstant within the sphere E None of these EO within the sphere HL Flux1 HL Flux1 HL Flux1 CT 2719 A negative point charge with charge Q sits in the interior of a thick spherical conducting metal shell which carries 0 net charge What is the total charge on the Inner surface of the shell Outer surface Gaussian surface CT 2719b A negative point charge with charge Q sits in the interior of a thick spherical conducting metal shell which carries 0 net charge What is the total charge on the outer surface of the shell Outer Gaussian surface CT 2719eX CrktCT l Two light bulbs A and B are in series so they carry the same current Light bulb A is brighter than B Which bulb has higher resistance A A B B C Same resistance vAv VTDVAQ B Answer Bulb A has higher resistance Since the resistors are in series they have the same current 1 According to P I2 R if I constant then higher R gives higher P CrktCT 2 Two light bulbs are in series attached to a battery as shown The bulbs are marked 40W and 60W Which bulb is brighter Hints More power brighter When light bulbs are in series they have the same current Light bulbs are intended to operate at 120V A both have same brightness V B 40W is brighter T C 60W is brighter This is a tricky one The answer is that the 40W bulb is brighter The bulbs are labeled 120V2 S o R IF V120V THEN higher R means lower P The bulb with the smaller Plabeled has the larger R The 40W bulb has higher R than the 60W bulb These two light bulbs are in series and light bulbs are not intended to be used this way In series the current is the same so the larger R produces the larger P I2 R assuming that they would be used with a voltage of 120V So Plabded CrktCT 3 A IQ resistor is placed in parallel with a 10000 Q resistor as shown 100009 The total equivalent resistance of these two resistor in parallel is closest to A a little less than 19 B a little more than 19 C 5000 9 D a little less than 100009 E a little more than 100009 Answer a little less than 19 You could use the formula Rtot RT 172 R 099909 Or just think 19 is a very low resistance compared to m 1 100009 so almost all the current will ow through the 19 resistor the circuit will behave almost as if the 100009 resistor is not present and the equivalent resistance is close to 19 The question is is the equivalent resistance a little less or a little greater than 19 Adding the 100009 resistor in parallel provides another current path of the ow of charge More ow means lower resistance CrktCT 4 The four light bulbs shown are identical Which circuit puts out more total light Hint more power more light 1 XT AAA V39V vvv vvv V 12V v12v A B C They both put out the same amount of light Answer A The total equivalent resistance which the battery in the green circuit sees is W2 two resistors each of resistance R in parallel The total equivalent resistance which the battery in the yellow circuit sees is 2R two resistors in series The total power coming from the battery is 2 P L Smaller Run with xed V results in a larger power P tot Ch26l The solid line has length A and makes an angle 9 with the negative yaXis What is the length of the dotted line Y A A cosG B A sinG C A tanG D sinGA E cosGA Ch262 A charge Q is on the yaXis at y d What is the magnitude of the Efield at position X on the yaXis kQ A dX2 y Q kQ X2 dZ kQ d C X2 dz2 X kQ R D E None of these Ch263 TheintegralEZE means Ex ly QEX Often from symmetry one can see that one or more of the three component integrals vanishes An infinite line of charge with linear charge density 9 is along the XaXis and extends to i 00 At the point A shown What can you can about the X and y components of E y A Ch264 A circular ring of radius R uniformly 2 charged with total charge Q is in the xy plane centered on the origin The electric field dB at position 2 h on the zaXis due to a small piece dE of the ring with charge dQ is shown What is the magnitude of the field dE kQ k dQ A F B hz k dQ k dQ C R2h2 D R2h2 E None of these What is dEZ the zcomponent of dE A dE sine B dE cosE C dE tanE D None of these What is cosE A L B L C E 39R2h2 quot184th 39 R 7 h D cos E None ofthese Ch265 A circular ring uniformly charged With Z positive charge Q is in the xy plane centered on the origin as shown On the ZaXis E E 2 Which graph accurately represents X y the electric field EZ on the ZaXis E2 E2 Z lt Z A B Ez Z Z C D E None of these is an accurate representation of EZ Ch266 What the magnitude of the vector i j 2 A l B 2 C 0 D Some other number E No answer because 2 37 is not a vector Ch267 There are no charges near inside the regions shown Which of the following are possible electric field line configurations A a only B b only C c only D None are possible b E Some other answer all a and b only etc Ch268 Consider the four electric eld line patterns shown Assume that that are E charges in the regions shown Which if any of the patterns represent possible electrostatic elds A V C A All are possible B 11 only C II and 111 only ll D None are possible E None of the above Ch269 From the figure what can you say about the magnitude of the charge on the bar lQbaI compared to the 9 magnitude of the charge Q of the positive point charge A lQbarl gtQ B3 lQbarl Q C Qbar lt Q From the figure what can you say about the net charge on the bar A Qbar 0 B Qbar gt 0 that is the bar has a net positive charge C Qbar lt 0 the bar has a net negative charge D Not enough information in the figure to answer the question Ch2610 Two infinite planes are uniformly charged with the same charge per area 0 If one plane only were present the field due to the one plane would be E The field in region B has magnitude A zero B E C 2E D depends on exact position The field in region A has magnitude A zero B E C 2E D depends on exact position 2611 A dipole is placed in an external eld as shown In which situations is the net force on the dipole zero 3 A 1 B 2 C 1 and 2 D 3 and 4 E 2 and 4 Ch2612 velocity Vt 4ms 4ms timet O 5 1O 15 For 1D motion if velocity V is constant then the displacement in time t is d V t If V is not constant then the displacement is t d Zt dt The graph above shows Vt t 1 What is the displacement after 10 seconds A 3m B 8m C 10m D 18m E None of these Ch26l3 m Areal mass density 6 of a tile is mass m per area A 5 Z A If a sheet has area A and uniform density 6 then its mass is m GA If a sheet has a nonuniform density 6 chS where Xy is the postion on the sheet then the total mass of the sheet is m Zda Zdzr mygxdy where da d2r represents an infinitesimal area element y dagts Consider a square sheet in the Xy plane with edge length L L and density 6 b kX where b and k are constants V Which expression below correctly represent the integral for III the mass L L L Green L j b 10 dx Yellow L2 j b 10 dx 0 0 L L Pink 1de Blue jbkxdx 0 0 Purple None of these don t know 29813071 SIP physl 120 Electric Potentials We ve been talking about electric forces and the related quantity E Fq the E field or quotforce per unit chargequot In 1110 after talking about forces we moved on to work and energy Ouick Review of work and energV The work done by a constant force F moving F something through a displacement quotdquot is e FHd chos0 d39gt More formally if F varies as you follow some path W f df Eg if you an quotexternal forcequot lift a book at constant speed up a distance d Newton 11 says Fnet 2 ma F eXtT ie FeXt Fg 0 because remember if speed is constant gt a0 0r Fext mg Fg 1 mg You do work Wext FeXtd mgd The sign is because 6 is 0 degrees your force is UP and so is the displacement vector The gravity field does Wfield Fgd mgd The minus sign is because 6 is 180 degrees the force of gravity points DOWN while the displacement vector is UP The NET work done by all forces is WeXtWfield 0 that s just the work energy principle which says Wnet AKE 0 here You did work Where did it go NOT into KE it got quotstored upquot it turned into potential energy PE In other words Fext did work which went into increased gravitational potential energy For gravity we defined this potential energy to be PE 2 mgy so A PE 2 mgyfinal yinitial mgd Wext The change in PE is all we ever cared about in real problems Summary If you do work on an object in a quotconservativequot force field A PE 2 Wquotby youquot Wquotby the fieldquot fFfield 39dF Knight generally uses the symbol quotUquot for quotPotential Energyquot by the way 29813072 SIP physl 120 Now let s drop the book and see what happens I There is no more quotexternal forcequot touching the book like quotmequot 1 in the previous example only gravity acts Neglect friction Fgl mg Energy conservation says i 2 f PEiKEi PEfKEf 1e mgd0 0 2mvf This formula gives a quick and easy way to find vf The concept of energy and energy conservation is very useful Another way of rewriting that equation is PEf PEl KEf KEl 0 ie APE AKE 0 or AEW 0 If only conservative forces act A U ie A PE is independent ofthe path taken End ofqaick review ofwork and energy There is an electric quotanaloguequot of the above examples 4 4 4 4 44 4 4 Consider 2 charged parallel metal plates called a quotcapacitorquot a fixed distance d apart Between the plates E is uniform constant and points from the towards the plate EV v v v v 4f 479 E 1 I I 1 jg Imagineacharge q initially located near the bottom plate The force on that charge is FEqE down do you see why Let s totally neglect gravity here Now LIFT quotqquot from the bottom to the top at constant speed You do work Wext ff d39 Fwd qu The Electric field does Wfield f17 E d FE d cos180 qu 1 Do you understand those signs Think about them ext Just like the previous case you did work but where did it go As before it didn t turn into KE it turned into potential energy We say the charge s electrical potential energy has increased A U Ufinal Uinitial qEyfinal yinitial qE d 2 Wext where y is the distance above the negative plate We lifted the charge from a region of LOW PE near the quot quot plate to a region of HIGH PE near the quotquot plate Note quotupquot and quotdownquot are irrelevant here you could turn the picture on its side or even upside down It s not gravity in this story it s 100 electIical energy 29813073 SIP phys1120 Just like we defined EzFq dividing out q gives forceunit charge let s now define something we call quotelectrical potentialquot orjust quotpotentialquot 2 V PEq 0 Calling this quantity quotpotentialquot is really a VERY bad name because this quotpotentialquot is DIFFERENT from quotpotential energyquot 0 Potential has units of energycharge JoulesCoulomb JC We call 1 JC 1 Volt lV People use the symbol quotVquot for the unit volt as well as for the quantity itself Another bad choice but we have to live with it quotVoltagequot and quotpotentialquot are basically synonyms A change in potential is called a quotpotential differencequot U B AVAB vB vA AB fEdf 1 A and from this the change in potential energy A U APE qAV 0 Potential is a number a scalar You assign this number to places points 0 quotPotential differencequot is defined even if there are no charges moving around 0 The Sign of potential differences is meaningful Examgle A car battery maintains 12 V between the terminals If the headlights contain a 36 W bulb how much charge is the battery moving through the bulb 77 each second And how many electrons is that Answer 36 W 36 Watt 36 Js Each second 36 Joules of energy are dissipated in a bulb This energy all comes from the loss of potential energy as charges flow from one terminal through the bulb to the other terminal If a charge quotqquot drops 12V the energy lost is APE qAV or q12V Each second 36 J are lost ie 36 J q12 V or q 36 J12 V 36 J12 JC 3 C That s a lot of electric charge being moved by a car battery The number of electrons going through the bulb each second is 3Cl6E l9 Celectron 2El9 electrons A heck of a lot I was a little sneaky about signs the charge of an electron is negative just think about it Here s a related question for you given that it s negative electrons that flow out of a battery which way do they go from the quotquot terminal through the bulb to the quot quot or the other way The answer is from r to Electrons are repelled from quotrquot and attracted towards quotquot 29813 074 SIP physl 120 For a parallel plate capacitor we just found two pages ago APE qu which means AV APEq quq Ed Here s another sketch of a capacitor b V is high here d DeltaV VbVa Ed gt 0 l a V 1s low here With gravity you can choose to call quotzeroquot potential energy wherever you want You might choose sea level or the tabletop or the ground It s the same story with electricity you can pick any spot you want and call the electrical potential energy 0 there We usually call this point quotthe groundquot Let s call point quotaquot in the diagram quotthe groundquot or quot0 potentialquot Now put a charge quotqquot at the point quotbquot in that figure That point has a potential given by Vat point b Ed The charge has a potential energy at point b of PE 2 qVat b qu It has quotquot potential energy there which makes sense It s like a pebble up in the air it can do work just let it go The upper plate repels a quotquot charge the lower plate attracts it if you let it go it will run quotdownhillquot in energy from high potential to low If you move an electron charge e from point quotbquot to quotaquot AV Vf ViVa Vb0 Ed 2 Ed This is the same answer if you move a proton it just refers to the difference of voltage at points in space But the change in potential energy of the electron is eAV eEd Moving an electron from to plates gt POSITIVE change in energy you have to DO work on it Notes 0 E points from high V to low V always quotquot charges want to head towards low V if you ll let them quot quot charges want to head towards HIGH V if you ll let them l 0 We talk about the potential at a point or the potential energy of a charge at a point but the numerical value depends on where we chose to call 0 We should talk about potential differences between points and then it does NOT matter where we chose to call 0 29813075 SIP physl 120 What if you re near a point charge 2 rather than a capacitor Q r quotaquot quotbquot What s the potential difference between 0 gt Doints a and b AV ab Vb Va b rb kQ 1 r1 1 1 AVabVb Va fEdr f7dr kQ i in 7 a I r r r rb r a a This is very formal Let s at least think about the signs and make sense of it Since rbgtra this expression is negative which means Vb lt Va This makes sense to me point b is FARTHER from the plus charge so the potential should be LOWER there Remember just like in a capacitor potential is higher near positive charges What if point b is off at infinity Then we d have Vinfinity Va kQra We should really pick a place where V20 and the most natural choice in this case is probably Vinfinity0 Ie far away from this charge there s no potential no potential energy for a test charge In which case my formula is telling us 0 Va kQra or Va kQra or in other words That s the formula for the potential a distance r from a charge Q Remember that we chose V20 to be off at infinity to get that formula This says again potential is HIGH near Q and goes to zero as you run away Also note that this formula has lr not lr 2 Can you make sense of this Think of units WorkForcedistance or dividing out q Voltage 2 E fielddistance If you put a test charge q a distance quotrquot from Q then the potential energy of the test charge will be U qV qur Does this make sense Well if Q and q are both quotquot then PEszqr gt0 Sure two positive charges want to quotfly apartquot they ll DO work if you ll let them The system has positive potential energy Like a rock up in the air And the bigger r is the LESS potential energy we have The charges are far apart they barely feel each other they don t have a lot of stored energy If Q and q are opposite signs then PElt0 This is also correct you would have to do work ON opposite charges to quotpry them apartquot the system has a negative potential energy We might say the system is bound 29813076 SIP physl 120 Graphically we can think about plotting Voltage in the vicinity of a point charge Q The formula says Vr kQr If Q is positive this graph looks like Va The potential gets infinitely large right next to a charge That makes sense imagine trying to move a test charge right up next to Q The r closer you get Coulomb s law says the Force 2 qur Z is getting bigger and bigger keeping you away You have to quotpushquot harder and harder adding more and more energy to the system The voltage is very high near a positive charge it takes lots of work to bring a test charge there and you ll GET lots of energy when you release it No wonder the signs say quotDanger High Voltage Far away as we ve decided by convention the potential goes to zero There s no energy associated with being very far from a point charge If Q is negative the graph is Va r This one requires more thought but we ve seen a graph like this before it was the gravitational potential energy near a massive object Like planets orbiting the sun If you are in orbit you have positive kinetic energy but negative potential energy and in fact negative TOTAL energy You are bound it would TAKE some energy to separate the orbiting object off to infinity Same with opposite charges you would have to ADD energy to the system to separate the charges ie to get one of them quotoff to infinityquot at which point the system has NO energy So you must have started with negative energy 29813037 SIP physl 120 What if there are a bunch of oint char es what s the otential The potential quotvoltagequot at any point is just the sum of the voltages arising from each of the individual charges this is quotsuperpositionquot again MUCH easier than superposition of E fields because we rejust adding numbers not vectors C11 1 If you have charges ql q2 q3 etc and ask about the 0 r1 voltage potential at a point P it s just VP V1V2V3 k qlrl k q2r2 C1 r1 is the distance from ql to P r2 is the distance from q2 to P etc q Oqg If you have charges distributed smeared you d just write V fampdq r In practice it s really quite easy to nd the voltage at a point because of this Example Three charges 1 2 and 3 with charges Q 2Q and 2Q respectively are 2 ZQ arranged as shown What is the potential V at the origin Answer VVfrom 1Vfrom 2Vfrom 3 r kQr k 2Qr k2Qr 3 2Q kQr The answer is positive if you put a positive charge there it would be happy to run away off to infinity if you d let it The math is reasonably simple No vectors or components to worry about finding E at the origin would be a lot more trouble 29813078 SIP physl 120 Example A test charge quotqquot is moved from point quotaquot to quotbquot in the figure There are two other charges present Q and Q fixed in position at the corners of a square as shown How much work does this lake Answer We APE PEb PEa Remember at any point the potential energy is just PE qV PE q Vat b q k Q kZQ H b K r r r PE WW aqw 1 r r r Subtracting we find We APE PEb PE 0 It doesn t take any net external work at all Depending on how you move you might do some work part of the way and work part of the way but in the end you do zero total work going from this particular quotaquot to quotbquot 1 Trying to figure out the work by thinking of forcedistance along the path would be HARD because force changes all the time Using voltages makes this much easier to figure out 29313079 SIP physl 120 It s often useful to find all the points in a diagram that have the same voltage Eg consider a capacitor again Everywhere along the top surface the potential is the same VzEd V is high here Everywhere along the bottom the d V is in between herb potential is the same V20 1 V is low here And every point on that dashed line A has the same potential something between 0 and Ed We call such a line an quotequipotentialquot egual potential line If you move a test charge along an equipotential line eg the horizontal dashed line in the figure the potential is the same everywhere so no work is required It s like walking along a flat surface where there s no change in the gravitational potential Or eg traversing sideways on a ski slope There can t ever be a component of E parallel to an equipotential line if there was a nonzero Eparallel you d do work moving along it since W Fparalleld q Eparalleld This means that E eld lines are always perpendicular to equipotential lines The equipotential lines are shown as I 39l39 quotIf 39I39 39l39 39l39 39l39 39lquot quotdashedquot in this figure 39 J quot quot quot quot higher V They re like contour lines on a topo map midoue V which show quotgravitational equipotentialquot hwerv lines Constant height on a topo means v constant PEgrav Anywhere along a dashed line the potential is constant 0 Inside any static conductor we know E20 That means no work is required to move charges around anywhere inside or along the surface So metals are equipotentials throughout their volume 0 Real life is 3 D those lines are really surfaces 29amp30710SJ39Pphy51120 More examples of equipotential lines Remember voltage near a charge Q is v xr lower voltage given by V kQr equipotential line I 1 x I I 1 The farther away you get the lower the l 6 1quot I voltage higher voltage x CQUiPOtemial line If quotrquot is fixed ie a circle V is constant quotquot39 It takes no work to follow an equipotential Again in 3 D these would be equipotential surfaces rather than lines In this case they d be spheres surrounding the charge Q Another example here s a dipole with a couple of equipotential lines shown Look back in Ch 23 we sketched the electric field lines before Now we re adding in the equipotential surfaces We usually choose to define that center equipotential line as V20 Remember it s up to you to pick where V0 is and that line extends out to infinity That s a pretty common choice Far away from everything the potential is considered zero To find the numerical value of the potential on one of those dashed surfaces you d do a quick calculationjust like 2 pages ago find the distance rl to Q and r2 to Q and then V kQrl k Qr2 Anywhere on the center line you re equidistant from both charges so rlr2 and the two terms cancel V 0 It s all consistent 29amp30711SJ39Pphy51120 ENERGY and units Example Release a proton charge e from point a in the figure Suppose the voltage difference between the plates is 5000 V a typical voltage in eg a normal TV set Howfast is the proton going atpoint b Answer I d use conservation of energy here PEl KEl PEf KEf ie eva 0 eVb mv Solving for vb we get vb 22va vb m J21610 19C5000J C16710 27 kg 106 msCan you check that the units worked out ok Note that Va Vb 5000 J C ie Va is a HIGHER potential However the change in potential of the proton as it moves is AV Vb Va 5000 J C Think about the signs objects spontaneously move to LOWER potential energy if they can That means objects like to go to lowerpotential ie lower voltage What was the final KB of the proton in this example It s easy enough to find using conservation of E KEfKEiPEi PEf 0eVa eVb e5000V 16103919C5000J C 839103915J Many people prefer to change units here like converting 254E 2 m to 1 inch metric to non metric We can define a new unit of energy The quoteVquot is also called an quotelectron Voltquot but it is NOT a volt which is a unit of potential or JC recall It s just a name ee vee The eV is defined so as to be the energy loss of a particle of charge quotequot like a proton dropping across 1 Volt of potential Since energy change is qDelta V this is an energy change of e1 V 16E 19 C1JC16E 19 Jjust like we said In the little problem at the top of the page we could ve done it without a calculator if we d used eV s instead of J Namely KE KEiPEi PE 0eVa eVb e5000V 5000e V f f You can check that this agrees by doing a simple unit conversion 16 10 19 J 5000eV 5000eV 81045 wh1ch 1s what we got above leV 29amp30712SJ39Pphy51120 Review Potential voltage is electric potential energy per charge Vr k qr Due to a single charge q a distance r away Vr EEqi Sum of k qiri Due to a bunch of charges l i Vr f dq if you have a continuous distribution of charges Example You have a rod length L with total charge Q uniformly distributed over it What s the value of electric potential a distance a to the side dqzkdx We solved this problem back in Ch 23 for E there x390 x x39L XIaL It s similar for voltage k H k V L id A d X a fr q x aL x x Here AzQL is a constant I can do the integral it s a log getting VaL 79 lna L 10 s kTQlna lna L If you re not familiar with this kind of integrating work it out yourself We can rewrite this as VaL kQL lnaLa kQL lnlLa Let s stop and think if this is reasonable The units are right kQL is volts If a is huge agtgtL then La is tiny and we can use the calculus trick that says lnl Did you know that So VaL kQLLa kQa which is right if you re really far away the rod just looks like a point charge with charge Q and you re a away 293130713 SIP physllZO We ve seen that if you know E you can find voltage differences because b AVab fEdr a Physically knowing E everywhere tells you the force on a test charge and so if you imagine pushing a test charge around you can figure out the work which tells you the potential energy Knowing Er tells you Vr But we can go the other way too If you know the voltage as a function of position then you can figure out what the E field is Imagine that you move just a tiny distance ds In that case we don t have to do any integral wejust have N E dg If you choose to move a tiny distance in the direction ofE then Eds E ds which says dV 2 E ds as long as you go the right way moving parallel to E or But you really have to think carefully about this E is the rate of change of voltage as long as you consider that change only along the direction of E perpendicular to the equipotential contours of V Note the units E NC and dVds 2 Wm JCm NmCm works You can use NC or Vm for electric field it s the same thing Example Suppose Vr kQr We know this case It s the potential of a point charge at the origin But suppose you forgot that Staring at that function I see that Vr will be constant if r is constant ie on spheres centered around the origin Those are the equipotential surfaces So the direction of E must be perpendicular to those surfaces that is radial Now we have E dVds as long as quotdsquot points radially So E dVdr kQr Z Which is correct that s the E field from a point charge On a contour map any place where V changes rapidly ie dV is big for a small ds then you get a big E field That s like a top map places where the height gravitational potential changes rapidly is a cliff an object released there will have a big acceleration 29813 0714SJ39Pphy51120 Potential near conductors We already know that E20 throughout any conductor if you re in electrostatic equilibrium We talked about that in the Gauss law chapter If E was NOT 0 charges would freely move around tending to cancel out E This happens so quickly that in any static situation you know E20 inside the conductor Since AVab fEdF this also means V is the same everywhere inside a a conductor Because the integral gives you zero whichever way you go in there no change in V Also in the Gauss law chapter we saw that any excess charges will always sit on the surface of a conductor and the E field will always point normal perpendicular to the surface Otherwise the tangential component of E would move charges around on the surface cancelling out that component very quickly But since equipotential surfaces are always perpendicular to E that means there will be an equipotential surface just outside the conductor of the same shape as the surface Try to picture that Eg let s put a metal sphere inside a capacitor Before the ball was there the E field lines point uniformly from left to right But the presence of the conductor will distort the field lines Charges will accumulate on the sphere on the left attracted to the plate and on the 1i ght attracted to the plate These charges set up a quotcounter fieldquot inside the sphere where E20 throughout The sphere is an equipotential V is constant throughout it The E field lines have to be perpendicular to the sphere everywhere at its edge The dashes show some equipotential surfaces in this little figure I Remember our discussion from Gauss law chapter If you dig a little hole in the middle of that sphere nothing changes There is no E field in there If you crawled inside you d be quite safe feeling E20 29amp30715SJ39Pphy51120 What happens if you dump charge on a conductor It will spread out as best it can but the charge will tend to concentrate more at quotsharp spotsquot Why There are lots of ways to argue this all a little handwavy for now Heres s one model a curvey surface as locally like a sphere of radius R We know V kQr for spheres but Q o 4nr 2 so V goes like 0 r If you have an equipotential constant V then where r is small sharp spots 0 is big and big 0 means big E field In a lightning rod you have a conductor with a sharp tip If you dump some charge onto this rod it will distribute itself on the outside surface and wejust argued that you ll get lots of o and big E at the sharp spot If E is big enough you can ionize air causing an electrical discharge called quotcorona discharge This allows the excess charge to quotbleed offquot It actually helps discharge the big potential difference buildups So lightning rods can be effective not by drawing lightning to them but instead by quotbleeding offquot excess charges earliery and thus preventing a strike from building up to catastrophic proportions There are lots of other remarkably cool applications of these ideas including electrostatic precipitators that can clean dust from the air to photocopy machines that use electric fields to charge up ink drops 29amp30716SJ39Pphy51120 Comments on electrical potential energy of a system Remember that Voltage 2 potential energycharge or turning it around PE 2 qV That s a quotone particlequot kind of view you take your q plunk it down somewhere and say it has potential energy presumably because of all the other charges in the universe that built up quotVquot Another way of thinking about this is to look at your entire system all the charges and argue that that whole system has a certain total potential energy That energy represents the amount of work required to move those charge from far away at infinity and bring them to form this system So eg if you bring ONE charge in from infinity and there s nothing else around that takes no work at all Now you have a charge ql sitting somewhere you call the origin Next bring in a SECOND charge q2 from infinity keeping ql fixed Now you have to do some work In fact to bring q2 to a distance r away will take k ql q2r of work So this SYSTEM of two charges has total potential energy k ql q2r If you bring in a third charge q3 so some position that s r13 away from 1 and r23 away from 2 you will have to do more work k ql q3rl3 JUST because of ql and k q2 q3r23 because of q2 Thus the total energy of the system of three charges is Usystem k ql q2r12 k ql q3rl3 kq2 q3r23 That s the total stored energy of the system If it s positive it means you ll get it back when you let them all fly apart to infinity If it s negative you would have to DO work to pull this system apart to infinity And as usual we ve decided to call U20 when everything is widely separated 29amp30717SJ39Pphy51120 CAPACITORS are any physical objects that can store charge I guess they re called capacitors because they have a capacity to store charges They re useful your flash camera has a big capacitor in it to store up a bunch of charges which it quickly releases to flow through the flash bulb Any time you have two conducting surfaces near each other you have a capacitor We call the surfaces quotplatesquot even if they re funny shapes When we say a capacitor has quotcharge Qquot we mean Q on one plate and Q on the other like in the picture here Of course the net charge is really zero but we still say it s charged up These two plates have a voltage difference between them It takes WORK to move a little test charge from one plate to the other Let s think about this What if you double Q on both plates gt you ll double the number of E field lines gt which means you ve doubled E gt since voltage difference between the objects comes from integrating E see last chapter that means doubling Q doubles AV They re proportional Q AV or in other words Of course you can also turn it around and define C QAV C is a constant and called the quotcapacitancequot C depend on the precise shapes and the materials involved but not on Q A big C quotbig capacitancequot means you can store a LOT of charge for a SMALL voltage difference In a sense big C means it s quoteasyquot to store up a bunch of charges A capacitor can store charge and energy for you that s why they re useful Think of a closet with lots of bowling balls up on the shelf at high gravitational potential You can let them fall down whenever you like to do lots of work or damage later Your closet shelf has only a limited capacity to hold bowling balls Similarly metal plates of a given voltage difference have a certain limited capacity to hold charge and that s what quotCquot tells you The units of capacitance are CoulombsVolt CV We call 1 CV l Farad l E The units are getting confusing now 1 V l JC so 1 F21 02 too 0 l E is a really big capacitor It would hold 1 C a lot of charge with only one volt difference between the plates Most normal capacitors are more like micro or nano Farads 29amp30718SJ39Pphy51120 39 39 H 39 H My favor1te example 1s the parallel plate capac1tor T Q Last chapter we showed the potential difference between the plates is AV Ed 1 We often drop the quotA quot for convenience and say VEd 39Q It s a bad notation but standard V really means quotvoltage differencequot here For large parallel plates area A the E field between them is given by a simple formula derived last chapter from Gauss law E 050 2 QA 0 The formula says the more Q you have the bigger E is That makes sense yes Also the bigger the area of the plates the more the charge is spread out and that should weaken the E field so the A area dependence in that formula also seems reasonable The formula doesn t depend on quotdquot how far apart the plates are This is perhaps a bit surprising but it s correct E is UNIFORM it doesn t change with distance from either plate We know AVEd for a parallel plate capacitor f Why Let s review it s because AV fE df That39s a key formula from last chapter For a parallel plate capacitor E is constant this just gives AV E d If you integrate FROM the negative plate TO the positive one ie moving quotagainstquot the direction of E that dot product will be negative because E points opposite d The two negatives cancel to give positive AV 2 Ed Which is what I claimed Combining this with the formula for E just above gives AV i d 30 Finally C QAV that defines C remember So that last equation for AV yields can you check the algebra yourself C This formula is only for ideal infinite parallel plate capacitors As advertised C is a constant depending only on the size and shape Since AVEd and E is uniform the closer the plates the smaller AV is Smaller AV for a fixed charge means CzQAV is BIGGER If you can get plates close together you get a better bigger C capacitor So that quotldquot dependence in the formula makes physical sense If the two plates are really close the s on one plate are tightly attracted to the s on the other If I may anthropomorphize they are happy and you can fit LOTS of them on the plate That s what big C means Similarly if the area of the plates is bigger the formula says you have more quotcapacityquot to hold charge That makes good intuitive sense to me you have 29amp30719SJ39Pphy51120 more area to spread those Charges out onto so you can easily hold more Charge If they re allowed to spread apart from each other they re happier 29amp30720SJ39Pphy51120 If you connect a wire up to a capacitor like on the left the charges on the 0 capacitor plates are free to travel The and attract one another the capacitor will discharge quickly possibly with Q some nice sparks There s energy in 0 there It takes work to charge up a capacitor Consider e g a battery charging up a capacitor A battery quotwantsquot the plates to be at a certain voltage difference V Eg a 12 V car battery always tries to make things connected to it to be 12 V apart To do this it forces charges to go from one plate onto the other plate fighting against an E field all the while How much work does the battery do charging the capacitor up from 0 to V The first charge moves easily it takes almost no work because at first the plates have no charge x sist There s no E field and charges are pretty much chargefrom one plate to the other I free to move around wherever they want On the other hand if the capacitor is fully charged with a voltage AV across it you d expect that moving Q charges across that voltage difference AV would cost energy QAV Because after all V means quotpotential energy per charge But the real answer is the average of 0 and QAV on average while charging the voltage difference is only AV2 See next page for a rather more formal argument to get the same answer Bottom line Energy to charge up Energy stored in capacitor 2 U QAVZ Remember the symbol quotUquot is used for potential energy Don t confuse potential energy with potential Since QCAV we can rewrite this as U CAVA22 or UQA22C Depending on what s held constant these forms are sometimes useful Please don t confuse the symbol C Capacitance with C Coulombs 293130721 SIP physllZO How much work does it take to charge a parallel plate capacitor from 0 to Q This will also be the total energy stored in that charged capacitor Each charge that you move from the bottom plate to the top plate has to move from a low potential to a higher potential That takes a little work call it dW Each time you move one more charge the E field gets bigger and the potential difference between the plates increases So the next charge will need even MORE dW 4Q The total work 251W zde Lzo AV dq Think about that last one I m saying that the work dW to move a little bit of charge dq from one plate to the other with voltage difference AV is dq AV That s from the basic definition voltage difference 2 potential energy difference charge So total work I E d dq fOQAL d dq 30 The first step came from remembering AV 2 Ed and the second step from recalling our formula for E derived 2 pages ago We can do this integral most of the stuff in there is constants 2 2 2 U total energy stored total work i f9 q dq i1 dQ A80 0 A80 2 0 2A 0 We can rewrite this in several ways First remember that we showed Capacitance C so convince yourself U 12 QAZC Here s another way which is very cool Remember that E QAeo d d 2 1 So It39s also true that U 2 7 EA 7 Ad E2 2A 0 Q 2A 0 8 2 8 Think about that last line since A d is the volume of the capacitor we vejust seen that U volume 30 E2 This leads me to think of 30 E2 as the quotenergy per volumequot It implies that ELECTRIC FIELDS STORE ENERGY It s a very important and different way of thinking about things Instead of thinking of the charges as storing the energy think of that E field as storing it This is no proof but it s a very general result Anywhere you have E field you have energy and the energyvolume is 30 E2 Even in empty space 29amp30722SJ39Pphy51120 Q2 These can each be useful depending on what you knoware interested in Eg If you have a battery involved you ll probably have a fixed quotAVquot Recap Energy stored in any capacitor is U g CAV2 QAV between the plates and so U CAV2 might be most useful But they are all very general true for any capacitor geometry Also general U volume 30 E2 E fields store energy BUT only for parallel plates can we argue C NOT a general formula Examgle A 12 V car battery is hooked up to a capacitor with plates of area 03 m 2 a distance 1 mm apart How much charge builds up on the plates Answer The battery will charge the capacitor up until the voltage is 12 V Since QCAV we only need to find C But we know that C 80 Ad So we re all set here I write quotVquot for quotAVquot as a shorthand 2 2 C 0373m 12V 10 m Q CV 30 V 8851042 Nm2 9 C2 3210 V 32 nC Nm 1 used 1 V1 J C check for yourself that that nasty combination of units simplifies like I claimed to Coulombs Also note the capacitance involved here 2 Nil 3 m2001m 226 nF Fairly small How much energy is stored in the capacitor now UQAV2 2 3213 9 C 12V 2 02 micro Joules Not so much You must use high technology to get big capacitances you need BIG areas and VERY small distances between the plates 1 mm is not really that close C 2 8851012 d C2 Coul Coul Coul Farad 2 7 Nm N m m J m Volt m m Coul Coul Units 30 When I see a formula like C it makes sense to think of 80 as being quotFaradsmeterquot It also tells me that need Ad 10quot11 meter to get a capacitance on the order of 1 Farad Reminder Where exactly is the energy stored in a capacitor It s stored in the E field Wherever you have electric fields there is stored energy The energy is stored in the quotspacequot between the plates in the form of electric field energy 293130723 SIP physllZO In diagrams we will represent capacitors with a simple symbol like this even if the capacitor in reality isn t physically parallel plates Sometimes the C is left off the picture too Parallel and series capacitors We might combine capacitors eg we could take two capacitors and connect them side by side called quotparallelquot like this C1 C2 Solid lines represent quotideal wiresquot perfect conductors l l Remember in electrostatics conductors are equipotentials throughout so every point along an ideal wire is at the exact SAME voltage In our diagrams we often indicate a voltage drops as shown to the I left here the top plate is at Vtop the bottom plate has a voltage Vbottom and what we mean by quotVquot in the fig is instead really AV 2 Vtop Vbottom This is so important I claim that our two parallel capacitors are really equivalent to ONE capacitor with a quottotalquot or c1 c2 quoteffectivequot or quotequivalentquot capacitance given by Cparallel C1C2 I CC1C2 Informal proof Suppose C has charge Q on it and C2 has charge Q2 on it Let s call the voltage difference between top and bottom of the picture quotAVquot Notice that this voltage difference must be the same for both capacitors So AV 2 Q1C1 Q2C2 or if you prefer Q1 2 C1AV and Q2C2AV Thus Q1Q2 C1C2AV or AV 2 Q1Q2C1C2 QtotC1C2 I now just read off the denominator Ceffective Cparallel C C1C2 v equiv to l For capacitors in series lc1 equiv to Icla1m 1Csenes1C1 1C2 I C Teames Informal Proof39 This time since C1 and C2 are connected I claim Q1 2 Q2 Has to be Think about it the section in the middle starts off uncharged and then as charges shift around at the top and bottom charge can SEPARATE to go up or down but whatever s go up the same of s must go down Now suppose you have a total voltage drop AV from the top to the bottom AV 2 AV1 AV2 the total drop 2 the sum of the two drops added up Since Q1 2 C1AV1 and Q2C2AV2 AV 2 AV1AV2 Q1C1 Q2C2 But calling Q1Q2Q tells us AV 2 QC1 QC2 Q1C1 1C2 Finally we set AV 2 QCse1ies which really DEFINES Cseries andI see 1Cseries 1C11C2 29813 0724SJ39Pphy51120 Optional fun stuff for those who want to go further DIELECTRICS Any insulating material paper plastic etc can be called a dielectric Most real capacitors have dielectric materials between the plates This helps to keep the plates apart After all the plates are 0 oppositely charged and so attract each other strongly If they ever touched the capacitor would discharge or quotshort I u H y in outquot and be useless HlililililiHIHI HHIHIHWHHHWif lHiiililHlHHli So with a dielectric in there you can make quotdquot quite small Q and remember that helps make the capacitance bigger But the main reason for putting a dielectric in there is something different it actually decreases the E field in the capacitor Why Real materials like dielectrics are quotpolarizedquot by a strong E field That means the E field in the capacitor effectively pulls some quot quot charge towards the top of the dielectric nearer Q the quotquot plate and some quotquot towards the bottom nearer the quot quot plate In the region throughout the middle it looks like some of the quotQquot around you has been weakened or shielded or canceled out I The net effect is that E is reduced throughout the dielectric And since VEd the voltage between the plates is reduced Q Since CzQV ifQ is fixed and V is reduced C gets bigger In this way dielectrics make the capacitance bigger Your capacitor can hold MORE charge for a given voltage with a dielectric in there because the surface of the dielectric effectively shields out some of the E field from the middle region 1 A A Recall the formula C so 475k d d That s if there s no dielectric If there IS a dielectric wejust argued C is bigger It turns out for most dielectric materials C is bigger by some constant factor which depends only on the material ie C K 1 475k 94gt A with dielectric K 80 E K will be some constant for any given dielectric material Bigger K means you get a bigger C Paper eg has K23 roughly 29amp30725SJ39Pphy51120 Capacitors are everywhere in circuits radios computers TV s It s handy to have a quotcharge storagequot device The final brief topic of this chapter is a real world application of capacitors CRT s or quotcathode ray tubesquot vertical capacitor hot metal quot quot cathode screen quotquot anode with holes horizontal capacitor A quotcathode rayquot is an old fashioned name for electrons The quotcathodequot is a heated piece of metal set at a very low voltage The quotanodequot is set at a high potential so the quotcathodequot and quotanodequot basically form a capacitor Electrons boil off the hot cathode and then they are accelerated towards the high voltage quotquot charged anode The anode is a grid with lots of holes so many electrons can fly right on by and cruise towards the screen They pass through a pair of capacitors vertically and horizontally oriented which have a voltage that quotsweepsquot As electrons pass through these capacitors they feel the force from the E field which bends the path of the electron Since the voltage is swept the electrons are also swept They fly on by and hit the screen which glows where the electrons hit So you see them sweeping by and this makes the whole screen glow By turning the anode on and off you can make the electrons go through or not thus making bright or dark spots which allows you to make an image This device is used in old computer monitors and TV oscilloscopes EKG traces you still see them all over We ll play with an oscilloscope in lab where you can control the voltages by hand and mess around with manipulating electron beams 321 Which arrangement of two coils has the larger mutual inductance B C They are equal 322 A largediameter twoturn coil coil 2 is wrapped around a long smallerdiameter solenoid coil 1 as shown Which is larger M12 or M21 322b Now the number of turns in coil 2 is doubled What happens to M C132I1 A M is half B M remains constant C M doubles D M quadruples E None of these Electric toothbrush CT 3321 lnductor1 consists of a single loop of wire lnductor2 is identical to 1 except it has two loops on top of each other What is the B field at the center of coil 2 B2 compared to the field in the center of coil 1 A B2 B1 CB24B1 I D B2 312 CT 3322 Inductor 1 consists of a single loop of wire Inductor 2 is identical to 1 except it has two loops on top of each other How do the selfinductances of the two loops compare Recall L CIDl N B A l A L2 2 L1 BL2 gt 2L1 CL2 lt 2L HLIN I What is the inductance of a solenoid with length d area A and turnslength n A no n2A B HO n2Ad C no n2 d D HO nAd E HO nA CT 324 Two long solenoids each of inductance L are connected together to form a single very long solenoid of inductance Ltotal What is Ltotal O V V V39 v V v V v V V V v V v V V V L CT 3328 The switch is closed at t0 What is the current through the resistor at t0 R20 2 A 0A V10V B 05A C 1 A D 10A E other CT 3327 The switch is closed at t0 What is the initial rate of change of current didt in the inductor at t0 Hint what is the AVinit across the inductor R 209 A 0A V10V B 05 NS C 1 NS D 10As E other CT 3329 The switch is closed at t0 What is the current through the resistor after a very long time R 209 A 0A V10V B 05A C 1 A D 10A E other HLN2 What are the units of LR AV BA CAs D S E 18 HLN2 What is the voltage across an inductor as a function of time for a Charging LR Circuit 1 V A T EL R6 D ELz V L1 e E EL V1 6 CT 3330 At time tO the switch is Closed What is the current thru the inductor L just after the switch is closed t 0 R1 100 A0A BO5A C10A CT 3331 At time tO the switch is Closed What is the current thru the inductor L after a long time R1 100 BO5A C10A CT 3332 Now suppose the switch has been closed for a long time and is then opened Immediately after the switch is opened the current thru R2 is R1 100 B Not 0 amps CT 327 The same current i flows through solenoid 1 and solenoid 2 Solenoid 2 is twice as long and has twice as many turns as solenoid 1 and has twice the diameter Hint for a solenoid B no 11 i What is the ratio of the magnetic energy contained in solenoid 2 to that in solenoid 1 that is what is A2 B 4 E1 O C 8 2 I 23 line of 0llllllllllllllll llll HLAC What is the average power dissipated by the resistor zero positive A B C negative D Not enough information CT 3334 A light bulb is attached to a wall plug 120 VAC 60 Hz How many times a second is the instantaneous power output to the bulb equal to zero A Never there is always power B 30 timessec C 60 timessec D 120 timessec E Other CT 3335 A 100 W light bulb is attached to a wall plug 120 VAC 60 Hz What is the peak power output to the bulb A Sqrt21OO w 141 w C D Other CT 3315 A transformer is attached to a battery and a resistor as shown The voltage difference across the resistor R is iron core VN1N2 CVN2N1 Other CT 3317 A stepdown transformer is attached to an AC voltage source and a resistor How does R compare to the current that is drawn from the AC source in input A IR gt input B IR lt input C IR input RII l A point charge q is brought from infinity to a point b near 3 other charges Q Q and 2Q The charge q is brought along 3 different paths in turn path 1 path 2 and path 3 as shown Along which path is the most work done by the external agent carrying the charge q 2QO O Q Path 1 b Path 2 A l B 2 C 3 D Same work on all three paths Only the three charges Q Q and 2Q are present q is absent and the point b is midway between Q and Q The voltage at point b is A positive B negative C zero RHZ A voltage sensor moves from position a to b and b measures a constant unchanging voltage at all points along the path What can you say about the electric field along that line a A The electric field along that line must be zero B The electric field might be zero or it might be nonzero and perpendicular to the line ab D The electric field can vary along the line ab but it must have the same value at points a and b E You can conclude absolutely nothing about the electric field RIl3 Two equal and opposite charges q and q are a distance r apart or a distance 2 r apart Which configuration r or 2r is the higher energy configuration higher energy larger electrostatic potential energy A separation r is higher energy r B separation 2r is higher energy q q C the energy is independent of separation O I Zr I O q q RH4 A rod with total charge Q is bent into a quarter circle of radius R What is the voltage at the center of the quartercircle RHS A positively charged metal sphere or radius R has total charge Q and is in electrostatic equilibrium What is the voltage difference between the surface Q and the center of the sphere k A EQ B zero kQ 2kQ C E D T E depends on where we put the zero of voltage RII6 In three different regions of space labeled I II and III the electric fields are accurately indicated by the electric field lines shown Notice that the separation of the field lines in II is the same as in the right part of I and the left part of III The distance from point A to point B is the same in all three regions In which region is the magnitude of the voltage difference lAVABl the smallest 4 0 0 4 0 0 4 0 0 x 4 4x x 4x I II 111 A region I B region II C region III D the voltage difference in regions I and III tie for smallest E none of these RII7 A capacitor is attached to a battery which maintains a constant voltage V across the capacitor plates While the battery is attached the plates are moved further apart The energy stored in the capacitor A increased B decreased C remained constant RHS A cylindrical carbon resistor placed between two copper terminals and a steady current I ows through the system The carbon resistor is tapered so that it has a fat end and a thin end w Igt Inside the carbon resistor the current density j is A constant B not constant quotConstantquot meaning constant in space throughout the resistor the conductivity 6 is A constant B not constant the electric field E is A constant B not constant RIl9 Consider the following circuit containing three ordinary household light bulbs a 100W light bulb a 60W bulb and a 40W bulb mmm Which one statement is true A The 100W bulb will be brightest B The 60W bulb will be brightest C The 40W bulb will be brightest D All the light bulbs glow with the same brightness RH10 Consider the circuit below How much current ows through the battery A 1A 1 19 B 2A 1V 4 C 05A R1 1 D 025A 29 2 2 E none of thesedon t know 29 In the circuit above the resistor R2 is increased to SQ What happens to the current through R1 A increases B decreases C remains constant RIIll In the circuit below which resistors have the same current owing through them Do not assume that the resistors are identical A R2 R5 only 1 2 B R3R4 only 1V R1 35 c R2 R5 and R3R4 only R3 D R1 R2 R5 only E none of these R5 Which resistors have the same voltage drop 39 R2 R5 Ol lly R3R4 only R2 R5 and R3R4 only R1R2 R5 Ol lly none of these ECO RHlZ 0 A capacitor C is charged up to a voltage V The switch is closed at time t0 connecting the capacitor to two identical resistors as shown Immediately after the switch is closed how does the current through the second resistor on far right compare with the current through the capacitor AIRIc B1Rl2lc C IR21c E IR 0 Immediately after the switch is closed how does the voltage across the second resistor on far right compare with the voltage across the capacitor AVRVC BzVR12Vc szR2vc DzVR0 Where does the word quotlaseIquot come from A The name ofthe physicist who invented it The name ofthe dog of the physicist who invented it C ight Ampli cation by Stimulated Emission of Radiationquot D It39s a mix of quotlightquot and quotp quot E It wasjust made up because it sounded cool m m g CAPA14 Tonight Set 15 in bins short Reading lfyou haven t already gt 33 9 347 8 There is a new and nal online participa ion available don t forget about it Last Maxwell39s equations light nd EM radiation Today More of Maxwell and EM radiation Next Waves LC circuits and resonance Prof Gurarie A Maxwell s equations 31 E dA QM km 51 1 d8 0 e dd E39dl B Theyareoneofthe 11 major triumphs of modern science The axioms of electromagnetism s dd B d1 P39 mmm HEB th Visible Light Example Yellow light Frequency 5E14 Hz x 500550 nm yellc1 Traveling wave m lt gt DOSItIOI l Wavelength MTC fAc Visible Light x 400 nm violet 9 700 nm red Example f A C x 550 nm yellow f CD 3X108550X10399 5X1014 HZ Television transmissions are in the radio range x 15 meters fcgt 108 Hz 100 MHz On Television different channels have different frequencies and thus different wavelengths When you set the channel knob you amplify just that frequency range A radio wave of wavelength 2 meters passes by a person with a radio receiver The person watches the electric and magnetic fields go up and down as the wave travels past After 1 second the number of waves that moved past the person is A 1 wave B 3X108 waves C 15x108 waves D 6X108 waves E Not enough information A radio wave of wavelength 2 meters passes by a person with a radio receiver The person watches the electric and magnetic fields go up and down as the wave travels past After 1 second the number of waves that moved past the person is A 1 wave B 3X108 waves C 15x108 waves D 6X108 waves E Not enough information A radio wave of wavelength 2 meters passes by a person with a radio receiver The person watches the E and B elds go up and down as the wave travels b Later a new radio wave passes the person They observe that the E and B elds go up and down 10 times faster than the original wa What is the best conclusion The second wave has 110 the 39equency More than one or none ofthese is correct Polarization of Light EM waves have a direction ofthe Electric eld vector E As viewed 39om an observer E Ordinary Light from the Sun or a lightbulb is unpolarized meaning it is a mixture of waves with Electric eld vectors in random directions within the plane perpendicular to the wave direction Unpolarized Polarized Polarizer Polaroid lter lter that passes light with the E eld along the pass axisquot ofthe lter only Filter Polarized 39 Light Passes Through Polarizer Polaroid lter lter that passes light with the E eld along the pass axisquot ofthe lter only Filter Polarized Light Does Not Pass Through All the EM energy is used up as work in moving charged particles in the lter If the E eld is not all parallel to the pass axis only the component of E along the pass axis gets through Only Ecose is transmitted intensity is related to ENERGY And thus to eld strength squared Filter Axis SW Sn 00529 Clicker Question An unpolarized beam oflight passes through a Polaroid lter The intensity ofthe original beam is In What is the intensity of the light coming through the lter A 11 4ID c B 12ID c 14ID D lEl E None ofthese Clicker Question An unpolarized beam oflight passes through 2 Polaroid lters oriented at 45El with respect to each other intensity of the original bea is In What is the intensity of the light coming through both lters L A 11 4ID An unpolarized beam of light passes through 3 Polaroid lters each oriented at 45El with respect to each other The intensity of the original beam is ID Does ANY light pass through the last lter L A Yes some makes it thru B None makes it thru A b 30 m W 0 HI C m Q I xx G gt l quot quotquot cl llll Unpolarized light of equal intensity a d gt b c reaches four pairs of polarizing lters B b 2 c gt a d Rank in order largest to smallest the C dgt gt b gt intensities transmitted through the a C second polarizer of each pair D d gt a gt b C E Something else Copyrighl O 2004 Pearson EdllL illlUll Inc publishing 1 Addison chlcy Why do polarizing lter sunglasses work E F f lfsun light is unpolarized then itjust component However scattering can polarize light Then the sunglasses work very well at reducing this light glare Signal picture information is carried by a range of frequencies Af bandwidth centered on a carrier frequency f Information is encoded either by Amplitude Modulation AM ENVELOPE WARVlNG mmmnn mama WAVE cmsuw rnmumcy Frequency Modulation FM I consum Aururuns meme Fnzouzucv Ex r EM sinkx wt 30c t B sinkx 6002 peak Who are these people and what do they have in common FOb It is the Friday before Spring Break Yet I am here dutifully attending lecture Shouldn t I receive some clicker points A Yes give me the points please B No I am not here so no points for me thankyou Fl A bar magnet is inside an imaginary closed dA surface S What can you say about the magnetic u I I flux through the surface NB di is I S A zero B positive C negative I D impossible to tell from the information given F2 A loop of wire is moving rapidly through a uniform magnetic field as shown True or False there is a nonzero emf induced in the loop 8 69696969 696969696969 69 88 69 A loop of wire is spinning rapidly about a stationary aXis in uniform magnetic field as shown True or False there is a nonzero emf induced in the loop A TT BFF CTF DFT F3 A loop of wire is sitting in a uniform constant magnet field as shown Suddenly the loop is bent into a smaller area loop During the bending of the loop the induced current in the loop is A zero B clockwise C counterclockwise F4 A bar magnet is positioned below a horizontal loop of wire with its North pole pointing toward the loop Then the magnet is pulled down away from the loop As Viewed from above is the induced current in the loop clockwise or counterclockwise A eyeb all t N A clockwise B counter clockwise U F5 A square loop is rotating in a fixed external magnetic field into the page At the instant shown the loop is out of the plane of the page with left side of the loop above the page and coming out of the page the right side in going in The induced current is axis of rotation C44 Bin C Neither F6 A loop of wire is near a long straight wire which is carrying a large current I which is decreasing The loop and the straight wire are in the same plane and are positioned as shown The current induced in the loop is A counterclockwise B clockwise C zero I to the right but decreasing loop F7 Two loop of wires labeled A and B are placed near each other as shown A large currentl in loop A is suddenly turned on This causes an induced current in loop B which causes A A net replusive force the two loops repel B A net attractive force the two loops attract C whether the force is attractive or repulsive depends on whether the current in first loop is CW or CCW D No net force F8 A wire loop moving right enters a region where there is a constant uniform magnetic field pointing into the page B As the loop enters the Bfield the current induced in the loop is A CW B CCW As the loop enters the Bfield the direction of the net force on the loop is A right gt B lt C up T D Ldown E there is no net force F9 An electric motor consists of a coil free to turn on an aXis and located in an magnetic field B created by an arrangement of permanent magnets With the current and field directions as shown which way will the coil rotate A CW axis B CCW C the coil won t rotate when at this particular orientation FlO A solenoid has an increasing current causing an increasing Bfield in its interior An endon View of the solenoid is shown below An electron is directly below the solenoid While the currentl in the solenoid is increasing what is the direction of the force on the electron 69 A 8 Bin D B electron C Hint What is the direction of the Efield at the position of the electron Think Lenz s Law Fll A long solenoid of radius R contains a uniform magnetic field B which is increasing at a steady rate Bt A t where A is a constant For distances r lt R from the center of the solenoid how does the magnitude of a d a the Electric field depend on r Hint d1 daJ AEzero B Eoc lr CEocr DEocr2 EEocr3 Ch 23 Electrostatics Coulombs Law Fk Q1 Q2rA2 f39 1 k 43180 w1ll of course be g1ven on the exam Electric charge is conserved Charge of electron is negative e e will of course be given on the exam Electric Field EzFq force per unit charge Units NC points away from towards charges E fields superpose gt know how to add vectors find components Terminology Conductor insulator dielectric dipole Ch 24 Gauss39 law Interpret field line diagrams Higher density of field lines means stronger E field counting lines to find charges Definition and computation of flux 1 fEdA Gauss Law ngdA QCl lClOSCd80 dA points outward Whenhow does symmetry help Whenhow does superposition help E field is zero inside conductors in steady state All excess electric charge sits on surfaces of conductors At surface of conductor E is normal and EO80 Do you know WHY all of the above are true Ch 25 Electric potential Work is Fdcos0 Unit is J Nm V Uq Voltage Potential energycharge V JC B AVABVB VA EdL Isolated systems conserve energy PEiKEiPEfKEf Where KE 12 m VAZ If speed is constant Wext AU q A V Point charge V k Qr Use superposition if more than one charge adds up like a number V is neg if Q is neg You choose Where you want to call PEzO or V20 Here it was at infinity Constant E means AVAB E AL 1 eV 16E 19 J unit of energy This Will of course be given on the exam Equipotential graphs not on this exam Point P in empty space is equidistant from 3 charges as shown Note the 4 quotanswerquot arrows are all 45 T W degrees from the axes HX GREEN L BLUE What is the direction of the electric r field at point P 2Q rgtIltr PURPLE Some completely other angle not sure YELLOW PINK B If a negative test charge q is now placed at P which direction is the force on that charge A point charge Q is at is the origin Point P is located at the point X23 y4 What is the X component of Electric field at point P BLUE kQ 25 PINK kQ 9 YELLOW kQ253 GREEN kQ2535 PURPLE None of these What is the voltage at point P BLUE kQ3 PINK kQS YELLOW kQ535 GREEN kQ25 PURPLE None of these The diagram shows a positive point charge Q21 mC near a metal bar The electric field in the Vicinity of the point charge and the bar are shown by the field lines in the figure A From the figure What can you say about the net charge on the bar BLUE Qbar 1 mC YELLOW Qbar 1 mC PINK Qbar 025 mC GREEN Qbar 25 mC PURPLE None of these Three closed surfaces enclose a point charge Which surface has the smallest outward ux through it PINK Small cube YELLOW smaller sphere GREEN larger sphere PURPLE Need more info BLUE All three have the same ux If the pink cube was moved to the side so that it no longer enclosed the charge which surface now has the smallest outward flux F l A bar magnet is inside an imaginary closed surface S What can you say about the magnetic ux through the surface I I NB di is s A zero B positive C negative I l D impossible to tell from the information given I x Answer 0 The positive ux caused by the B s fields lines coming from the Npole and exiting the surface is exactly cancelled by the negative ux due to the B field lines entering the surface heading into the Spole The N I I equation NB XdA 0 is always true under all circumstances It is one of Maxwell s 5 equations F 2 A loop of wire is moving rapidly through a uniform magnetic field as shown True or False there is a nonzero emf induced in the loop 969 6969 6969 8 298 A loop of wire is spinning rapidly about a stationary aXis in uniform magnetic field as shown True or False there is a nonzero emf induced in the loop A TT BFF CTF DFT Answer FT In the rst case there is no changing ux so no induced emf In the second case the ux is changing because ux F BA oosqand the angle 9 is changing F 3 A loop of wire is sitting in a uniform constant magnet eld as shown Suddenly the loop is bent into a smaller area loop During the bending of the loop the induced current in the loop is A zero B clockwise C counterclockwise 6969 69 88gt 910 Bin Answer Clockwise The ux is decreasing as the loop area decreases To ght the decrease we want the induced B to add to the original B By the right hand rule version 11 a clockwise induced current will make an induced B into the page adding to the original B F 4 A bar magnet is positioned below a horizontal loop of wire with its North pole pointing toward the loop Then the magnet is pulled down away from the loop As Viewed from above is the induced current in the loop clockwise counterclockwise or zero A eyeb all N A clockwise B counter clockwise U C zero Answer Counterclockwise The B eld from a bar magnet points out of the North pole As seen from above the eld through the loop is out toward the observer As the magnet is pulled away the ux is decreasing To ght the decrease the induced B eld should add to the original B eld F S A square loop is rotating in a xed external magnetic eld into the page At the instant shown the loop is out of the plane of the page with left side of the loop above the page and coming out of the page the right side in going in The induced current is axis of rotation Q4 Bin C Zero Answer B At the moment shown the ux through the loop is decreasing since the amount of B eld quotthreadingquot the loop is decreasing To ght the decrease the induced eld should add to the original eld F 6 A loop of wire is near a long straight wire which is ca1rying a large current I which is decreasing The loop and the straight wire are in the same plane and are positioned as shown The current induced in the loop is A counterclockwise B clockwise C zero I to the right but decreasing loop Answer clockwise At the moment shown the ux through the loop is decreasing since the amount of B eld quotthreadingquot the loop is decreasing To ght the decrease the induced eld should add to the original eld F S Two loop of wires labeled A and B are placed near each other as shown A large current I in loop A is suddenly turned on This causes an induced current in loop B which causes A A net replusive force the two loops repel B A net attractive force the two loops attract C whether the force is attractive or repulsive depends on whether the current in rst loop is CW or CCW D No net force Answer A repulsive force the two loops repel B increasing B induced Assume a CCW current I in loop A When the current I is turned on there is an increasing B eld pointing out of the page in the middle of both loops A and B This increasing B eld out induces a m current in loop B so as to create an induced B eld into the page ghting the increase in ux Now you have to remember that antiparallel currents repel parallel currents attract The two currents are antiparallel so the loops repel If the current in loop A was the other way CW then everything would be reversed and the two currents would still be antiparallel F 9 A wire loop moving right enters a region where there is a constant uniform magnetic eld pointing into the page B As the loop enters the B eld the current induced in the loop is A CW B CCW As the loop enters the B eld the direction of the net force on the loop is A right gt B lt C up T D Ldown E there is no net force Answers Question 1 CCW As the loop enters the eld the ux increases To ght the increasing ux the induced B eld out of the page should be opposite the original B into the page A CCW current will create an induced B out of the page This current is an example of an quoteddy currentquot Question 2 Method I the easy way The magnetic forces on eddy currents are always in the direction which opposes the motion Otherwise you could make a perpetual motion machine Method 11 harder way Work out all the directions As the loop enters the eld the ux thru the loop is increasing By Lenz39s law the induced current in the loop will be CCW so that the induced B eld out of the page ghts the change in ux There will be a magnetic force on the right side of the loop l I I X B The direction of that force is to the left 6060 6969 B field Concept Tests B l Anegative particle and a positive particle are moving with certain velocities in a constant uniform magnetic field B as shown The direction of the Bfield is to the right The particle is moving directly left the 7 particle is moving directly up L B lt 3 The force on the positive particle due to the B field is in into page out out of page A in B out C zero D right E left 7 V7 V7 7 V The force on the negative particle due to the Bfield is A in B out C zero D right E left Answers The particle is moving antiparallel to the B field The angle 9 is 180 and the force is FBqu sine 0 The 7 particle is moving at right angles to the field By the righthand rule the direction quotv cross Bquot is into the page but the particle has a negative charge q so the force is out ofthe page B 2 Apositive particle is released from rest in a region of space where there is constant uniform electric field E and a constant uniform magnetic field B The electric field points up and the magnetic field points out of the page in the diagram below Which path will the positive particle follow All paths shown are in plane of the page oog out D it will remain stationary Answer The particle will feel a force FE qE due to the E eld along the direction of the E eld As it starts moving along the E eld direction it will acquire a velocity and it will start to feel a force FBqVB due to the B eld The direction of the force is to the right by the righthandrule B 3 A charged particle with an initial speed v0 is moving in a plane perpendicular to a uniform magnetic eld B into the page There is a tenuous gas throughout the region which causes viscous drag and slows the particle over time The path of the particle is A a spiral inward B a spiral outward 293 C something else If the time for the particle to complete the rst revolution once around is 1 second the time for the rst 5 revolutions is Agt5s Blt5s C5s Answers It will spiral in As the particle39s speed v decreases its orbital radius decreases according to R mvqB The period is 5s The period T and the frequency f lT are independent of the radius R or the speed v The quotcyclotron frequencyquot is f qB27Em As the speed v decreases the radius R decreases in proportional It moves more slowly but around a smaller circle so the time to go around once remains constant B 4 A square loop of wire carrying current I is in a uniform magnetic eld B The loop is perpendicular to B B out of the page What is the direction of the net force on the wire A out ofthe page 9 B into the page B C T D gt E None ofthese The same loop is now in a nonuniform eld B Bi where B By A y where A is a constant The direction of the net force is B s onger B weaker E net force is zero Answers When the Bf1eld is uniform the net force is zero The force on each side is F I L B where L is the edge length The forces on each edge each point away from the square and so they cancel In the nonuniform Bf1eld the net force is upward since the upward force on the top wire is larger than the downward force on the bottom wire B S Aparticle with unknown charge but nonzero q and moving left with speed V enters a regions where there is a uniform electric eld down and a uniform magnetic field out of the page The particle is observed to go in a straight line The charge of the particle must be A positive B negative C impossible to determine B G G Suppose the particle is a proton Ifthe speed of the proton is increased it will A still undergo no de ection C0 9 G B de ect out of the plane of the page A T V C stay in the plane of the page and de ect upward D stay in the plane of the page and de ect downward G 0 C0 6 E V V V V Answers Impossible to tell the sign of the charge It is possible that the charge could be either positive or negative If the charge is positive the force from the Efield is down the force from the B eld is up and the forces cancel But if charge is negative both forces switch direction and the forces still cancel In either case the fact that the particles is moving with constant velocity implies that Fnet 0 Since the net force is zero the magnetic force magnitude lqlvB must cancel the electric force magnitude lqlE So we have vB E the lql s cancel so only particles with speed v EB pass straight thru regardless of the sign or magnitude of the charge This device is called a velocity selector only particle with this speed go straight thru any other speed will result in a curving trajectory FB qu FE qE B v v Q FB qu E FE As the speed of the proton increases its trajectory stays in the plane of the page and de ects upward If the speed V is increased the upward magnetic force FE qVB is increased but the downward electricfield force FE qE remains the same


Buy Material

Are you sure you want to buy this material for

25 Karma

Buy Material

BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.


You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

Why people love StudySoup

Steve Martinelli UC Los Angeles

"There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

Jennifer McGill UCSF Med School

"Selling my MCAT study guides and notes has been a great source of side revenue while I'm in school. Some months I'm making over $500! Plus, it makes me happy knowing that I'm helping future med students with their MCAT."

Bentley McCaw University of Florida

"I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"


"Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

Become an Elite Notetaker and start selling your notes online!

Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.