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Popular in Physics 2
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Date Created: 10/03/15
1272011 What are the units of potential difference Electric Potential and Electric A Amperes Potentlal Energy cont 3 potentiometem C Farads D Volts E Henrys The change in potential energy depends on the sign of a charge as well as on its magnitude What are the units of the electric eld A VC lt B NC C Vm D Jm2 E Qm u1 upwmc hc Llncclmnm I m l mmrg H llhu ncgum39u mm H 7 14 Hmu xlmuy 1mm m mgwc mm Exercise A proton is fired with a speed of 200000ms from the midpoint of the capacitor toward the positive plate Since the force on a negative charge is opposite to the field direction Positive charges accelerate in the direction of ashow that this is insuf cient speed to reach decreasing electric potential the Positive Plate bWhat is the proton s speed as it collides with v Sun V39 Negative charges accelerate in the direction of the neative late increasing electric potential In both cases the charge moves to a region of lower potential energy 1 i 7 3 ll Llll H Ill ri 1272011 111e Potential Energy of Point Charges Consider two point charges 41 and 42 separated by a distance r h H x 41 um m 1 my gum mum rmm m mm 205 magma polenllal ola polnlchaue mmammmwmmu 1 The charges have equal magnitude Rank in order from most positive to m negative the potentials Exercise charge q 411x109C is placed at the orig and a second charge 2q is pl ed atx 1m Find the electric potential midway between the two charges I d the value of x where the electric potential vanishes somewhere between in a charges In each case the chargesare the same distance f rom the origin Rank in order of increasing potential 1 111e Electric Potential ofa Charged Sphere In practice you are more likely to work with a charged sphere of radius Rand total charge Q than with a pointcharge Outside a uniformly charged sphere the eleclric potential is identical to thatof a pointcharge Qat the center Thatis l V 79 7 39 hequot nl39churuc r 2 R 4175 I p L 1272011 Or in a more useful form the potential oulside a sphere that is charged to potential lEI is R 7 VH sphere charged 0 potential Vquot This eq tells us that lhe potential of a sphere is V0 on t e surface and decreases inversely wi the distance Exercise A proton is released from rest at the surface of 1cm diameter sphere that has been charged to 1000V aWhat is the charge of the sphere bWhat is the proton s speed at 1cm from the sphere Belm39c After 1 n Ihum o v r 4H lcm Midterm exam1 review 1There are two and only two kinds of charge and 2Two like charges repel each other 3Two opposite charges attract each other The atom is held together by the amac ve elecm39c force between th positive nucleus and the negative electrons Electrons and protons have charges of opposite sign but exac yequal magnitude e 16x103919C TABLE 25 Protons and eleclrons Particle Mass kg Charge Prolon Io 7 X 0 1 u Elcclmn 0 x m r Since all stable matter consist of gtprotons gtneutrons and gtelectrons the net charge q Npe Nee Np Nee Where NP and Ne are the number of protons and electrons contained in the object If N Ny the material is neutral there is no net charge If there are multiple point charges the forces add by superposition 1272011 If two point particles having charges q1 and q2 are at distance r r 51 unit newton N Coulomb39s law apart the particles 39 exert forces on each 1 other It Ht 7 r F k 7 Vi The magnitude of the force at every point in space is proportional to test charge q0 due to Coulomb s Law It is convenient to divide the force by q0 and define the force per charge at every point in space This quantity is referred as the electric e E F E 7 3970 SI unit NC In general we say that the electric field a distance r from a point T charge q has the magnitude Use this equation to find the magnitude The electric field of a positive charge and towardsa negative one Use this rule to find the direction 1272011 Just as electric forces can be supelposed electric elds can as well FIGURE 273 Electric fieids obey the principle of superposition Vivid ul39 souruc Chili39ng i and 3 1 i 5 NW IlL l vlccliic eld at hix 1mm The Electric Flux mule 7s 3 W uinmi if n m Lu dwmzi nmnb of in ii39quot1 39il39 i w l Definition of Electric Flux III I 39 EA cos i Si unit N mZC The change in potential energy depends on the sign of a charge as well as on its magnitude I n iquot i7 4 gt E E Tm pulcniui mlurg m mka Thcimnm Icih rg nluucgumc mumum he mlhcdirccmmuif ulmmicu gtogtnllwxllremml n C mm mm smug as H il t The chair mum mu Ilhc nugilin c pllilg mm m 2mm w umm w hum lhx ncgmnc mm Gauss s Law For any closed surface enclosing total charge Qmthe net electric ux through the surface is 11 Ed2i amp El Gauss39s law states that the electric ux through a closed surface is proportional to the charge enclosed by the surface Dummy u mm If a particle moves through a potential difference AV its electric potential energy changes by AU qtest AV We can write the conservation of energyequations 12 m39Vi2 qVi 12 m39Vf2 qu Therefore the electric potential of a point charge q source is 1m in witsm l m vnlrs kt Lon 1 win The potential m created by a point charge q source at any distance r from the charge 1 200 Hill II 200 Min 4110 full nvllllr 1272011 If more than two charges are present the potential energyis the sum of the potential energies due to all pairs of charges The electric potential of a group of point charges is the algebraic sum of the potentials of each charge By SUM we mean that the potential of a given charge may be positive or negative and SIGN is veLy important during the calculations of the total potential 1132011 Ga uss s Law Symmetry prelz39prlead Suppose we know only two things about eleciric field determines the electric fied matcauses her hair 1 An electric field points away from positive to stream outward using charges toward negative charges Guass39s law we can 2 An electric field exeris a force on a charged deduce electric fields parb39cl f e simp y rorn the shape of From this information we can not deduce 9 Charge d39Str39butlon39 information about the eleciric field of an infinitely long charged cylinder Chapter Goa quot T m u ndersia nd and a pply Gauss39s law We don t know the parameters of the h d 39 d Some char 8 l lll l c arge CY In er distributions have A manslatinnal E F We know only that the charge is quottabMal 0r re echve 39 positive and the charge distribution symmetry has a symmetry A charge dlSIl lhlIIIOI I We say that the charge distribution is symmetric if there is a group of geometric transformations that don t t be cyl f39dr cally 5 mmeh39lc cause the physlcal change y ill The symmehy nfan elech39ic eld must f quotimlvw RPMWm NW graftll Me symmehy ofMe clmye quot 1m m Mum ism39 uh39nn 1 L mm H 4 mm mmmn lli x r 39 39L If39l39lnl hllrgjlhiingml 7m 139 1 ul 3 C 5 lEUTlllll rm 1 mum In general the electric eld ofa 1132011 cylindrically symmetric charge distribution acannot have a component parallel to the cylinder axis bcannot have a component tangent to the circular cross section w m llllh ll H l 7 5 ll ll This is the only shape for the electric the symmetry of the charge distribut field that matches 39 n 9 4 If we know that 1 the charge distribution is cylindrically s metric and 2 the electric elds point away from positive charge We deduce the shape of electric eld Knowing the shape helps us to nd the strenglft of electric e d does it decrease like 1r 1r2 or constant new n4 l lnmlr q39mnlclr lam C lillllritnl smut ltnnhllwllllllch nlmerlrlll u nlm l rumum smw a The field is cullllllg uul lll39 czlt39h law of llll hml Thch l1llll he il1llllL ch lll th hm a Opaque bth We can t see what is inside the box Can you gure out what is in the box 1132011 L I held passing Hum I hm Suppose we surround a regIon of space L WIth a closed surface a surface that divides space into inside and outside regions A closed surface through which the electric eld passes is called a Gaussian surface implies here39s nu not charge in the ho 111is is an imaginary mathematical surface It should be in 3d but it is easier to draw We don t know how much charge is there A better box would be more informative sulrounxdlvxg a m c A lwodlmensmnal The Gaussian surface is more useful when muss secumv IS usually easler In draw it matches the shape of the eld 1Hllllllll v ll l MM 1 The electric eld doesn t flow like a uid use the metaphor The Latin word ow is flux The Electric Flux was m u The sum u mu n The amount of electric field passing through a surface is cal e a Electric flux b G ss39s Law c Electricity d Charge surface density e None of the above 1132011 a max electric flux A ii I v llull 2 Ir v Definition of Electric Flux I I EA cos I b o o 7 m SI unit N m fC Exercise The Elecztrl ux d measures the amount The electric flux through the surface shown in of electric field passmg through a surface figure is 25quot m2c of area A whose normal to the surface is What is the electric field strength tilted at angle Hfrom the field Ie ELA EAcos We can define the electric flux more concisely using the dotproduct tiquot If ml electric ux ol uconsianlclccu39ic eld lilcm k Ilium l l l Exercise I5 v I I 39 I i The electric field t 4 I gt I g t t t y 4 quot is constant over m 1 1 l i J each face of the 39 39 i cube Does the 1 l l l 1 box contain I 0 I l T o J 39 quot pOSItlve charge 1 l The total electrlc flux through this box IS negative charge f 7 a 6 Nm2C or no charge 5 l iuld slrvngllls b 4 Nm2C Explainl in NC c 2 Nm2C d 1 Nm2C e 0 Nm2C m 1132011 rlcmxr 7n 1 uh5d SuHaL c m an W W M TactIcs Evaluatlng surface integrals quothis Evaluating swim integral 1 n u InnM W 3 w W lt1 I u v uhuc mm m H mm m clcmu lqu M V mm m mum um I h Ilw mu h emmr mm IhrrHgly uw nutm 4 El in w my l lll1HA M b j 124 mm The electric flux through a closed TaCtiC5 Finding the flux through 3 cc is closed surface 1 M 3 Finding the uxlhmughadosedsurlam Er Human The electric flux is still the summation h tilt mutlsfnfx I mww nu of the fluxes through a huge number MW of tiny pieces pieces that now cover a closed surface n g Gauss s Law Using Gauss s Law For any closedsurface enclosing total charge 0quotthe net electric ux through the 1Gauss s law applies only to a closed ll d surface Is surface ca e a Gaussian surface 2A Gaussian surface is not a physical Qiquot surface It is an imaginary mathematical surface in the space surrounding one or more charges 3We can t nd the electric field from I W eed CDE39IIA39 GU This result for the electric flux is known as auss39sLaw Gauss s law a one to apply Gauss s law states Mat Me elecm39c flux Gauss s law in situations where f wrong1 a closed surface isprnpnr nnal to symmetry and superposi on we already Me charge enclosed by Me surface can guess the 515 0f the e 1132011 ional cross sections Problem Solving Strategy Ga uss s eres and a Law These mens through three Imensional closed sph cube Rank order from largest to smallest the electrlc fluxes 1 a to me through surfaces 3 lb 8 i u Amimnilhc lmgcJimibuuou 39Ilvcynwnmiry libcluuim mm mlmw mm urlAu mm m mmu mum m m mulnm mum Im 39 minnurm Ir ch 5 t mnulilmfmm quotwrinceleilvcrlangmlmorpcrpuniitulnr mllw iccmu Iiuihl m Tin m mn mprmcumlmu mm m GAugt39gt 1 4 7 fl 439 s 3 quot f u mm new 5 1 mm 13 m mlquot the Wm Mug1 m 9 me V 9 9 Exercise Exercise Draw 2d cross sections of 3d close W the net electric ux through a cylinder surfaces through which the electric ux is a IIItao b Stilt0 c 0 d Io G 2 lQ nC 1 nC inside i 100 nC Exercise A surface endoses the charges q 32 quotC q Consider the surface S shown in the sketch 59 up q3 41 quotc Find the electric flux through Is the electric flux through this surface a negative b positive or c zero H1is surface According to Gauss law the shape of the urface oes notmatte oan the amount of charge it encloses j 1132011 km in wpdiwih i h I WHWhvw V mulhnManw wmu pm m z m mmu m mime Chm 39 mm Twmimmmm I k 39n mmquot Pam mquot numdrw mem mnnmmnm mu u v u mqu Zillifliili M 3333 3531 The net Electric flux is 390 thmugh a closed surface that does not contain any charges This exercise is a very good example of how Exercise powerful Gauss39s law can be Use Gauss s law to nd the electric eld of an in nite plane of charge with surface charge density n Finding the electric d of a plane of charge using the superposl Ion principle was difficult With Gauss39s law the problem is simple enough Wh we use the su er osition atall Blc the Gauss39s law is effective only in a limited numberof situations where the field is highly symmetric The superposition always work 1202011 Au 7 11H 1 m mu 4 Electric Potential and l l 39 Electric Ptential Ener quot I l 39 gt Electric force is also conservative 9 Gram7i f gt There must be an Electric cons ajrva We orce potential energy U 5 2271 s niggeg ljl t d 39th th I t 39 I assoc39a e WI e e ec quotc to the forge of gravity 3 H JI I a 1 r 5 If the charge All 1 A r i 39 l is moved A quot1 39 l w upward 1 H f V w distanced 17 The work has been done by u I x the electric force W qOEd V 39 mourn the change In potential AU 2 W qud u 3i energy 4 Eectric potential energy of the positive charge If the charge is negative 9 increases just as the gravitational potential the electric force is upward and does positive energy of a ball increases when it is raised work as the charge moves through the distance d against the force of gravity to a higher altitude 9 change in potential energy is negative AU qud w 4 I Au 7 my E Pi LI In 1202011 We can define the elecMcpaten aW as electric potential energy per charge Just as it was useful to define the electric eld E FM The electric potential V like the electric field is a property of the source charges and Is independent of the test charge Definition of Electric Potential V 1 lu c 51 unit joulecxmlomb 2 volt V battery 1800 gtKilovolts kV gtMicrovolts uV gtMillivolts mV This unit is named for Alessandro Volta who invented the electric potential erence Charged particles move t diff The potential energy of a charged particle is determined by the electric potential qV Charged particles speed up or slow down as they move through a region of changing f The potential difference between two points is often called the voltage hrough a potential i a particle moves through a potential difference AV its electric potential energy changes by AU q AV We can write the conservation afenergyequations 2 m39Vi2 qVi 2 m1 qu A proton and an electron are in a constant electric eld created by oppositely charged es You release the proton from the positive side and the electron from the negative side Which feels the larger electric force 1 proton 2 electron 3 both feel the same force 4 neither there is no force 5 theyfeel the same magnitude force but in opposite directions 4 W le V eleclmn O O pmmn V 1202011 A proton and an electron 1 proton are in a constant electric 2 demoquot eld created by oppositely charged Plates39 4 neitherthere is no You release the proton acceleration from the positive side 5 they feel the same and the electron from magnitude acceleration but the negative side in opposite directions 3 both feel the same acceleration 1 proton A proton and an electron are 2 electron in a constant electric field created by oppositely charged plate 3 both acquire the same KE 4 neither there is no on release the proton from change of KE 9 05m me and the 5 they both acquire the electron from the am E but with negative side lt n opposite signs Which potential graph describes the electric eld at the left Which has the larger 17 Vm 39 2 When it strikes the opposite 77 V Qquot acceleration electron plate which one has more elem Wm Wquot Rank in order from largest to smallest the potentials Va to V8 at the points a to e i i 1 1 1 1 l 1 r l J 39 l r r l l a 1 1 7 A proton is 100 V J V llmv l l released from I l l rest at point B I l i i where the At Bo Ct potential is 0 V AftenNard the proton Amoves toward Awith a steady speed Bmoves toward A with an increasing speed Cmoves toward C with a steady speed Dmoves toward C with an increasing speed E remains at rest at B The electric eld depends on the rate of change of the electric potential with position quot IV AV 39 L E 1 A3 Vm AV is negative when E and As are in the same direction both positive or both negative Exercise A uniform electric field with a magnitude of 1200 N C points in the negative x direction A particle with a mass of 359 and a charge of 0045 uC is released form rest at point a In which direction will this charge move b What speed will it have after moving throughout distance of 5 cm 1202011 Exercise A computer monitor accelerates electrons and directs them to the screen in order to create an image If the accelerating plates are 105 cm apart and have a potential difference of 25500V what is the magnitude of the uniform electric field between them The Electric Potential Energy of Point Charges If the positive test charge is released the repulsive force between q and q0 will cause it to accelerate away from the origin The difference in potential energy between pointsA and B kquq klM Therefore the electric potential of a point charge q source is ill in Mixm l39in ulh InnL will i Illlll The potential created by a point charge q source at any distance r w from the charge iEl lquotl rm 100 200 mn 4011 mi now ll mm The electric potential V like the electric field is a property of the source charges and is independent of the test charge The unit of electric potential is the joule per coulomb which is called the volt V 1 volt 1V 1JC 1202011 If more than two charges are present the potential energy is the sum of the Once the potential has been determined The positive charge is Tolifixllmjn l 39 it s very easy to find the potential energy U the end VIeW Of a t W gm L positively charged glass rod A negatively charged particle moves in a circular arc around the lamest to smallest the potential differences Isl2 Isl3 and Isl23 between points 1 and 2 points 1 and 3 and points 2 and 3 glass rod potential energies due to all pairs of Is the work done on the quot charges charged particle by the End view m rod s electric field dwng mu positive negative or mm a A Positive B Negative C Zero Rank in order from 391 Rank in order from largest to smallest the potential energies IIa to lid 0 these faurpairsof charges Each symbol represents the same amount of charge Find the change in electric potential energy AU as a charge of a 220 x10395C or b 110x10396 C moves from point A to B given that the change in electric potential between these two points is AV VB VA 240V a quota quotb gt quotc quota a AK3gtAV12gtAV23 b ublldgtllallc hAV13AV23 12 c U cgt b a 3 AV120bcthese d udgtu gtubgtu 2 3 23 poinls are at the same 9 quotd gt quotb quotc gt quota e AVE gt AV gt AV distance 4 mm H 1 n u The Electric Potential of Many Exerclse Charges 111e electric potential Vat a point in space is the sum of the potentials due to each charge V21iy 4760 r where I is the distance from charge q to the point in space where the potential is being calculated In other words the electric potential like the electric field obeys Me printable of superposition 1202011 Superposition of the Electric Potential The electric potential of a group of point charges is the algebraic sum of the potentials of each charge By SUM we mean that the potential of a given charge may be positive or negative and SIGN is veg importantduring the calculations of the total potential Exercise Find the electric potential at point P Exercis Find the electric potential enemy for this system of charges 1 The total electric um potential enemy of two or more charges is the sum of H1e potenlial energiesdue to each pair of charges Exercise uc work must be done to move H1e three charges in Figure infinitely farfrom one another The Electric Potential of a Charged In praclice you are more likely to work with a charged sphere of radius Rand tolal charge Q than with a pointc arge Oulsi uniformly charged sphere the electric potential is identical to thatof a pointcharge Qat the center That is 1 2 V 47 r sphere ul39churgc r 2 R 776 Or in a more useful form the potential outside a sphere that is charged to potential lEI is sphere charged to potential V This eq tells us that the potential of a sphere is Vo on the surface and decreases inversely with the distance 1202011 mm iiit iiiullmmuhn n limwlihillnu ix v e 5 v The electric potential of a Linuiwiinniiiuiiinminim lyulwiilrIIMHHNNIhilIlmiiillciinnlrill LullheiiJ nnnriiniiiu ilmcpulili minimum Wiles continuous distribution of charge Mm in Win em Ag inn in mm n c prewar inmhing a t IHINI39dUHHV1quotle WHWIHWU Hch H11lHI iii 39HL CMKN mIpi39U39CII g Au n is is III C lliuil step in milking iii uniisiunn iruni In sum ii an ini nIHK CAHKjUH intii roliitliimlt leuruliixIlw LIL m VAiHilllc I All dixlmuux um i Iv awnmi in imni iii Ilw mnniiniits Hitu l H Hllvkrilli I mt iiihiu Win heuwx HIL39 uilirdluilh mi quotth l39ur ilii i iiiihlt will hoan Carry mix lllk llINgl lull 39AKHAUH ruriiicnicieiiuirepreseniaimn Wuhwmm m Wm VGWMIW n Uni pi um Alwldl lhhutnunlmmc dpc 1ullHtuml dllhlkSyl n0ulm a ILk39Hmyllw noianmw c ivnuimnllnc nlpl MIRre ll 9 iv tic iiic mini chaigi g Ai nemuH liiccts ni ci incii yml ill quot u pmni LILIAgm iiiurnim iiirii imii in ire minim lH y m ulLuInL39lllk dleclric Ululllml s 2 it as mi mien hIM no mrii bimi huw In dclcnnmc 1Thiv dli lxinn 4 i i iiiriiiiiv are electric fields inside the human body the body is not a perfect conductor so there are also potential differences An electrocardiograph ploE ear 39 electrical activity 3 o 162011 The Electric Field Electric forces like gravity are long range forces No contact is required for one charged particle to exert a force on another charged article This concept troubled many of scientists of Me on s days Force should have some mechanism by which it is exerted mm ruhmmu ww v w M m m vvn Jun 1 Mum mm H Allcmlmv mm 1 The alteration of space around a mass is called the gravitational field The s ace around a chare is altered to create the electric eld Let39s introduce a liedmodellhat describes how charges interact Charges which we will call the source charges alter the space around them by creating an electric lie d Hmvxzzhzn rm mum inhibit10 le iuvgc W In x I X a I 1 m mm m u ulunrmmh Consider a positive pointcllarge q source chargeat the origin of a coordinate system Let s place a positive quottest d15rye qa at a distance r from the or 39 111e test chargequot qo experiences the force 111e magnitude of the force at every point in space is proportional to test charge 110 due to Coulomb s Law It is convenient to divide the force by 110 and define the force per marge at every point in space 111is quantity is referred as the electric In 5 unit NC 162011 E 3 1quot SI Lmit NC gt Electric field is a vector field gtThe electric fiel s a property of the source chame it Is Independent of the test charge 110 111e electric field at distance rfrom a point charge q39 r E clccu ic Iicld ol39n point charge where the unit vector for rpoints away from the charge to the point at which we want to know the field 111is unit vector expresses the idea away from q distance r from a point c arge q has the magnitude H Use this equation to find the magnitude 1 quot The electric field of a z wax 39 positive charge and tawardsa negative Use this rule to find the direction B tw th d d e een e re an 1 o the brown charge which of them experiences the 2 e 3 the same for both Between the red and the 1 them experiences the 21 greater electric farce due to the blue charge a the samefor both 162011 Electric field lines 1 Point in the direction of the field vector at every point 2 Startat positive charges orian ty 3 End at negative charges or iannIty 4 Are more dense where the field is stronger l 1 ll Drawing and uslng elemlc field lines 111e gure m 1 05mm eld Iimsnm continuum curvesdmwn alumni in represents the lhc elm k Ila vcc rs 0Ie Hue clwu39t dquot electric eld ofa vecloralmiy poinl is langcnl lo the 1L I line al lhul poml dipole as a eld Closcly uccd eld linen reprcscm larger licld sli cnglli vector diagram H with longer eld vectors Widely spaced lines indicnlc a V 39 Slt litllel39lleltl sucnglh quot I I quot gt lt 39 f 1 I un Whatare the signs of the charges ll 9 0 whose electric zl e 0 fields are shown at 3 Q 0 Ight H G 0 5 no waytntell 162011 Which of the charges has H1e greater magnitude 1 2 a sum the same Just as electric forces can be superposed eleclric elds can as well FIGURE 273 Electric fields obey the principle of superposition t 1 and 1 l iclah ui mum ch l thu Ith ulcctm t d ii this pumi At the position of What Is the electrlc eld 20 the do me at the center of the square 5En electric eld points a Up b Down c Left d Right e The electric field is zero ProblemSolving Strategy The electric field of multiple point charges vmunn Fur Ilw pimurml ruprcxunlmnm I Esluhlisl 1c00nlinulc lcmmuslin Ilu lumliunx I he L hill gt s ldcnlil39y m poinl pm 39hit39 an wnnl mvuluululc m clcclric new I Dnm IIIC cluclric I lcldul39 lL39ll L39I ill I I UNmymmtnj nulrunninu m mmlnmmmm Ewanmm mm Tn Innlhcnlulicul mprmumiun i 7 l Funwa clnugculmrumm in IMuucc I rmn I Lunl Ihc ullylc Hi 1 Hum he 1 Hit new 3mngm or may harch mm mm II vculur E m mmpmmnl lmm I Sum UIL vccmr mnumncuh hulk lrllnillc I ll needed lclcrnmw lllc mugniuulc ch All wan of 1 162011 Exercise What are the strength and direction of the electric field at the position indicated by the dot Specify the direction as an angle above or below the horizontal Exercise What are the strength and direction of the electric field at the position indicated by the dot in the figure Specify the direction as an angle above or below horizontal V 5 cm lUnC ln CH cm A very long thin rod with linear charge density A has an electric field E I 1 MI I IQI I BIA 39 lm i Z m L39 4176 r 3 1L y 417EHI L2 4175 I Where ris the radial distance away from the rod The field of an infinitely long charged wire decreases as 1r 0 MC 39 51 um I cm 39 3 111C The Electric Field of a Continuous Charge Distribution The linear charge densityof an object of length Land charge Q m V 1 is defined as Linear charge density which has units of Cm is the amount of charge per meterof length IQ 39 l 2W l IQI I 475 I LZf 475110 4775 r A ul n 1 m 1mm lmmm hm m lwgv 1 1 y A 1 qu u 162011 Which of the following actions wi in the electric eld strength at the position of the dot Charged rod 3 Make the rod ionger without changing the charge o Make me rod fatter without changing the charge c Make me rod shorter without changing the charge d Remove charge from the ro e Make the rod narrower without changing the charge May 111e electric eld strength 50cm from a very long charged wire is 2000 NC What is the electric field strength 10 cm from t e wire A Plane of Charge The surface charge density I1 of a two dimensional distribution e iiii h of area A is defined as Surface charge density with units Clmz is the amount of charge per square meter 111e electric eld of an in nite plane of charge with surface charge density qis E l COI Hl lnl piiiiii r c 25 For a positively charged plane with I gt 0 the electric field points away am he plane on both sides of the plane For a negatively charged plane with I lt 0 the electric field points towardsthe plane on both sides of the plane nww n M iwu v mm M i piaiir at Rank in order from largest to smallest a the electric field h l strengths g to g at r 4 these five points near I a plane 0 charge aEgtEgtEgtEgtEd bEEampEdE cEgtEampgtEdE d5ampEdEgtE 04gtQgt gtggtg 162011 mi km wwmm A parallelplate capacitor opposite charges In this ideal case the electric eld A m hm mm nmuh Pl mum m clculnnlm mm In cup mm 1 mu I m puuchl m m mm Lugs Sitic we of alumnae 94 vl0lE Llclhcmpuuhu I u39 uc ul plmiu w Nu m mm m nouns 2722 1116 eiecwc 1d ofa The figure PM shows that the electric field is my idem mnmcnnr I 139 the samein L 1121 strength and mumx direction at every point in mde mmm mmm 1mm m m space This is called a m wuxmu m 111 my uniform electric eld 111e electric eld inside a capacitoris 1 T7 pncillvr L L 1mm positive In ncgmm u Q fx39mn pusluvc lo negative 501 where A is the surface area of each electrode Outside the capacitor plates where E and E have equal magnitudes but opposite directions the electric field is lero a Rank in order from largest to smallest the ME to E at the ve points Explain b Rank in order from largest to smallest the force F1 to F5 a proton would experience if placed at point 1 to Sin this parallel 39 39 Explain 7 1111 1111 m A parallel plate capacitor is formed from two 4cm x 4cm electrodes spaced 2mm apart 111e electric field strength inside the capacitor is 1x105 NC What is the charge in nC on each electrode 162011 The onaxis electric field of a charged disk mm H U M Charge of radius R centered on the origin with axis 11 u39wu1m 39 parallel to z and surface charge density n QINR2 is Elihu T 25 1 3u R3 NOTE This expression is only valid for z gt 0 The field for z lt 0 has the same magnitude but points in the opposite direction Let39s move very far from the disk A Sphere of Charge At distant 2 gtgt R the disk appears to be a point Charge Q A sphere of charge Qand radius R be it a uniformly charged sphere orjust a 77 spherical shell has an electric field outside I M Ell T the sphere that is exactly the same as that of a point charge Q located at the center of the sphere w Q If ZgtgtR Esphcrc I fort2R Motion of a charged particle in an The direction of the Electric Field force depends on the sig of the J charge I gtA n r poSItIve charge I quot experiences a force cw 3 It39s is nh1m in a direction of E gt A negative charge experiences a force in the 7 39 opposite direction F 7 7E of E If we know the electric field we can calculate the force on any charge 162011 The electric field exerts a force An electron Is e 130 2 IE placed at the E E pOSItIon marked by the dot on a charged particle If this is the only force acting on I it causes the charged particle to agt Hgt The force on the E E electron IS accelerate with 1 I t I III I In a uniform field the acceleration is constant A10 the right Bto the left qE Czero quot 77 mumquot DThere s not enough information to tell Exercise Exercise A 209 ball hangs frm a str39ng The ball An object of mass m 1g is attached to a string Wives In th negatwe x cl39reFt39on When a and placed in a uniform electric eld of 1000NC un39form hor39zontal eleCtr39C eld Of The object is in static equilibrium Find a the magnitude 5510 NC is applied in the sign and the magnitude of the charge Q b the positive x direction After the ball comes to tension in the string Draw a force diagram rest the strin is at anle 10 with the 937 vertical a What are the sign and the magnitude of the charge on the ball b What is the tension in the string Draw a 1 force diagram Di oles in an Electric Field There is no net force but the electric field does p affect the dipole the electric field exerts a I w munquot quotFM cm torque on a dipole and causes the dipole to rotate kII139LlL nllll1l lepulu E T F ssin6qE pEsin0 F mun 17 0 The mrrgue m y mpme r l 111111 1 x 39 uuw There is no net force on a dipole in an I Ii r r 7 uniform electric field h a h I were isteange H r H F qE the dipole makes with W F 39qE 9 Fuel o the electric field I 7 u xxln ludmlgv mmlwxmutuugu nu mi xurlmu m um mum All the dipoles rotate until they are align with the electric field this is a 39 39 162011 The Magnetic Field gtWhat is magnetism gtHow are magnetic fields created gtWhat are their properties Chapter Goal To learn how to calculate and use the magnetic eld gtIfa bar magnet is placed on a piece of cork and allowed to float in a dish of water it always align itself in north south direction gtMag nets always come with two poles magnetic monopoles do not exist EEl v 1713 HE gtLike pols repel and opposite poles attract Just as an electric charge creates an electric field quot so the magnet does create a Magnetic 8 The magnetic eld can be visualized using magneticI39eldlins similar to the electric field mwmm mm mw 2132011 make compasses in half then pick up some The most 2132011 Wmquot M mm 1 mm Lam lug Hrc mm mm resemhls that of in pols of compass Tlu my ng imwllu Mm Tactics Righthand rule for fields To find the direction of the magnetic field due to a currentcarrying wire A l39humh ul gm hqu mummy in Autumn sz39fs Nghrhan mlelmll l i E9 1 cu v 7 u em Ol mnl yum mm cl ll u It unmnm nmllw Izm cul in rhc L my m Wm uit u we Ilin duH W Mg m mmung e m mulch mm m um mslmn in m mug t ncld rum muml me m 39 P What is the direction of the 1 399 gt magnetic field at 2 right the center point P a zero of the square loop 4 into the page The magnetic field at the position P points of current 5 out ofthe page a Into the page b u c Down d om of the page 2132011 The magnetic field has 1 39es 39 i it t Amagnetlc fleldls 39 s lllt39rlghlrlmlill hilt Tllrlch 1 created at all points In uillu39l inruni ham llit tint space surrounding a currentcarrying wire A magnetic field at each point is a vector It has both a e magnitude the magneti leld strength and a direction livl Because a current in a wire genemts a magnetic 1 lan ff xl jiws eld and a current is a collection of moving 39A magnetquot eld all rl With the lltllrl ChaI39QS 9 exe s ff 39ces 0 I H m HM moving diaryes are the source oltlie magnetic magnetlc poles l mu w u w Iiled Mighttit hell H1l il hl 4llll ul lllc mm g h lnt hm Wt mittl M lllhl ll r Pliinl flitll r39i ti Mmrnn litm l l wit UHIN clung llmiilnt The magnetic field of a charged particle q The tesla is a Mun of the magnetic moving with velocity vis given by the Biot Strength Iaw Another commonly used unit of magnetism is the 39 gauss r H quotl wizneiriet it a L t n it lgrltn 1 9m 16 10 7 W l V39llyliuhy em Magllctl dml 6 where ris the distance from the charge and His the angle between Vand r The constant p is called the permeability constant nu 4n x 10397 TmA 1257 x 10397 TmA inn tutl 2mm in mint The 51 unit Desa 1 tesla IT 1 NAm 2132011 Fhe magnetic field of a moving charge is in addition to the charge39s electric eld The fundamental distinc on between electric and magnetic fields Charges create elechic elds but only moving charges create magnetic elds The positive charge is moving straight out of the a e What is the direction of the magnetic field at the position of the dot F a Left 39 d b Rig ht c Down Vout of page d Up The magnetic 39elds like electri 39elds obey the principle of superposih39on If there are n moving charges the net magnetic field is given by the vector sum Blolal Bl 32 Bn In practice we are more interested in the neh39c eld nfa current a collection of moving charges than in a very small magnetic fields of individual charges The magnetic field 05 long straight wire carrying current I at a distance dfrom the wire is Exercise What are the magnetic field strength and c direction at polnts a to a 20 cm is gt H A b 40 cm 10A E 20 cm 2132011 Motors loudspeakers metal detectors A circular loop of wire with a circulating current is called a current loop lb All 107 The magnetic field at the enter of a ooil of III turns and radius R carrying a current I is 3 y R museum 2 generate magnetic fields with coils of wire Tactics Finding the magnetic field Irection of a current loop W1quot Filldluu lIe magnetic eld dllectinu 01 wrmu luuu Eu Inlxloluullht u m mtuldumum 3 c mummy nmumnwm tlmunum 11 W mam unlltmmm m nnthnlevHunmuquu1mmtu anllu39n pmnnuwmhulr n hu tluluux unltln Im HIM Ju39t lmml1ht nu tlnmlmuthiIMrihmml mm m Wu What is the current 1 direction in this loop I And which side of the loop is the north pole a Current counterclockwise north pole on bottom b Current clockwise north pole on bottom c Current counterclockwise north pole on top d Current clockwise north pole on top A current loop is a magnetic dipole A current loop is a magnet just like a permanent magnet My mum W A solenoid is a series of current loops formed into the shape of a cylinder 2132011 A solenoid is a magnet and we can use RHR 39 39 d to Identi y the northpole en w 39 r The magnetic dipole Pat39ent 395 momentof a current loop enclosing an area A is defined as 2 r w leI12lml l ML39KI I WYIUHIV ltl The 51 units of the magnetic dipole moment are A m2 The onaxis field of a magnetic dipole is It 7 mm mu me im m i mayin mlmlm Amp re s law m The value of the line integral around the closed Whenever total W 3 Ugtquot39Rquot39 F9 E39gt W path is 138x10395 Tm What are the direction in current 1m mlh passes or out of the page and magnitude of 13 through an area hounded by a closed curve the line int ml Id 3 39 m L1 2IA 2132011 Current Loops and Solenoids Hf quotquot ff quot j The strength of the magnetic field of a current loop is similar to the f m39 e c magnetic field of a bar magnet NM 397 fleld Inside a In the center ofthe loop 7 39 olenold ls f 139 A r w 5min Hquot nun twin ml u IN A innlim llll39llllllllllli39lll39i39 where n Nlis the number of turns per unit length The Magnetic Force on a Moving Charge left to rig ht Let s switch our attention from how Part39c39e f magnet Ids are generated to how 39quotm39es magnetic elds exert forces and torques A current consists of moving charges Let s consider the force a magnetic field exerts on a moving electric charge Magnitude orthe Magnetic Force F SI tlllil lwu39lun N I MW W H define the magnetic field SI unit nmvlun N De nillor of the Magnltude of the Magnetk Field B l a Mr sill ll Slnmt 1125b 1T lNMm mu 5m Typlcll magnenr lleld sllenmlls Field location Ficlll swung T Sll 5 X lo J R w x lll I 5 I U l Sllpclumdllclillg mngncl llt moving use a rightha positive charge nd 2132011 An electron moves perpendicular to a 1 magnetic field What is the direction of B h Into the page c Out of the page d U e Down A mime e enters a uniform magnetic field as shown Wh t s the direction of the magnetic force If a uniform magnetic field points into the page out oflhe page a o 1 out of the page A positive charge enters a uniform magnetic field as shown What is the direction of the magnetic force 1 outof the page 2 into the page 3 downward 4 upward 5 tothe left A positive charge enters a uniform magnetic field as shown What is the direction of the mag netic force 1 out of the page 2 into the page 3 zero 4 to the rig ht 5 to the left A positive charge enters a uniform magnetic field as shown What is the direction of the magnetic force 1 out of the age 2 into the page ero 3 z 4 to the rig ht 5 to the left 2132011 A posilively charged particle in an electric eld experiences a force in the direction of the eld in a magnetic eld the force is perpendicular to the eld 111is leads to very different motions Because the magnetic force is always perpendicular to the direction of motion the path of a particle is circular Also while an electric field can do work on a r 7 particle a magnetic bl K field cannot the particle39s speed remains constant For a particle of mass mand quot charge 4 moving at a speed Vin a magnetic field B the radius of the circle it travels is 7717 r 7 lqlB One of the gppli ations of circular motion in a magnetic eld is in a device known a mass speclrometer Exercise What is the acceleration of a proton moving with a speed of 95 ms at right angles to a magnetic field of 16T Magnitude of the Magnetic Force F F quB sin 0 Si unit newton N Fma Exercise A 125 pC particle with a mass of 28x10 5 kg moves perpendicular to a 101T magnetic field in a circular path of radius of 268m a How fast is the particle moving b How long will it take the particle to complete one orbit 111739 11le t 21trV 2132011 long If a particle s quot velocity makes an A it 14x10397N Find a the magnitude of mag 39el r a a the location of the particle Pah llflelWIHHToVe In I The Magnetic Force Exerted on a 51quot gammy WNW CurrentCaring Wire m m The force on a segment of a currentcarrying wire in a magnetic field is given by39 Magnetic Force on a numericCarrying Wire LB sin Ii SI Ilul ncwtun N 7 i 2123 ii i311 Wm iii Himiii iiiiiiiiiiiiiiii iiiiiiiiiiw Mi A horizontal wire carries a A honzontal WIre names a 1 left current and is inavertiral 396 current and Is In a vertlral 2 rlghl 2 right 3 zero magnetlc fleld What Is the 3 magnetlc fleld What Is 4 h d Cr h f Zero 391 t Ire Iono eorueon the direction of the force Iquot o 6 page 4 39quot 9 Page 5 out of the the wire 5 out of the on the wire Page age i 5 WM 2132011 Since a currentcarrying wire experiences a force in a Emir Cl magnetic field and a magnetic field is created by a A 045m copper rod WIth a mass of 017kg carrles currentcarrying wire a current 0f 11A In the PEER x d39rECtIOP39what It follows that the one currentcarrying wire will exerta are the magnltude and dIrectIon of the ml Imum force on another 39 39 39 7 magneuc flew neEdEd to lewtate the redquot There is a force between currentcarrying wires Iquot Cr 1 m 7 ymlru lml Magnetic Force on a Current Carrying Wire F ILB sin 51 unit newton N Hull lull I F I I LB I L L 7 7 End 2m 1 xrmvi 2211 Two straight wires run 1 toward each other up 1 x it y parallel to each other 2 away from each lll i39lfllfrlil l 4 each carryinga current 0 er 4 J in the direction shown 3 there 395 quot me below Ill I If the currents are in the same direction The two wires experience the force is attractive If the currents are in the opposite direction the force is repulsive a force in which direction Exercise Two long straight wires are separated by a distance of 122 cm One wire carries a currentof 275 A the other carries a currento 3 A Loops of Current and Magnetic Torque Consider a rectangular loop of height h and width w aFind the force per meter exerted on the first wire a bIs the force per meter exerted on the second wire can39VIng if current greater than less than or the same as the force per The loop Is placed In a meter exerted on the first wire uniform magnetic field B H l I W MAL Only the vertical sides I IHl lyl 39I 271 37d experlence forces that are 1 equal in magnitu e opposite in direction F IhB i 1 r 39 The forces do not operate at the same point so they create a torquearound the vertical axis of the loop 2132011 u If the plane ofthe loop is at an angle to the magnetic The Data Dorqu is the sum of A ed d I the torques from Each fame 4 Torque Exene on a lie angular Leap of Area A 7 7 llb sllit 1 unit rm iiinml lllBltigt mimic mum ElrsinceAhW 139 13 n 39 Eray mrc 2am MMquot autumn m Elm To increase the torque a long wire may be Consider a current loop in a uniform magnetic field B 0 MT 1 95A and d 0 wr ed In a loop many times or turns If the 4 Find the magnitude ofthe torque exerted on the loop number of turns Is N we have about the Vertical axis of rotation Torque Exerted on a General Loop ot Area A and NTurns NH iiri ll 9 iini39r39 N m The Won a current W IfN 50 m5 loop is proportional to the i it w i f the basis of a variety of I I useful electrical 2 H 213 instruments Here is a vll JEJJ Ferromagnetic materials are strongly attracted Materials respond differently to the force by magma me of a magnetic feld Th I t F k I N d h It gtA magnet VIIIquot strongly attract 5333 g iE mc e l aquot m a ferromagnehc materials These metals are strong ly gt weakly attract pawnagnehc matenals and attracted because i 9 t l t l f 39 thei dividualatoms have Itlllltl ahigherdegreeof I I I I I I I I magnetism duetotheir can gumtion oleectmns their atoms line u in the same magnetic direction gt weakly repel lliamagneh39c materials 2132011 Temperature effect Ferromagnetic materials like nickel or iron lose all their magnetic properties if they are heated to a high enough temperature The atoms become too excited by the heat to remain pointing in one direction for long The temperature at which a metal loses its magnetism is called the Curie temperature and it is different for every metal The Curie temperature for nickel for example is about 350 C Paramagnetic mateias a re metals that are weakly attracted to magnets Aluminum and copper are such metals These materials can become very weak magnets but their attractive force can only be measured with sensitive instruments Temperature can affect the magnetic properties of a material Paramagnetic materials like aluminum uranium and platinum become more magnetic when they are very cold Certain materials are diamagnetic which weakly repel a strong magnet Diamagnetic materials include water diamond wood and living organisms The funniest application of diamagnetic materials is magnetic levitation MRI Another important application of diamagnetic materials is magnetic resonance imaging MRI When carbonbased atoms in the body are ex osed to a stron manetic field the are slightly repelled by the field This movement of the atoms can be detected and used for analysis 212011 Electric Current and DC Circuits A closed path through which charge can flow returning to its starting point is called an electric circuit gtThe circuit in which the current always flows in the same direction is called directcurrent circuits DC gtCircuits with current that periodically reverse the direction is called alternatingcurrent circuits AC You push the sea of electrons the sea of electrons will quickly slow down and stop unless you continue pushing Rulzmlmglvm nI Juclncullhmlh ClL L illlllN 13 Which is the correct way 17 to light the lightbulb 4 aquota quot Fquot 9 with the battery 5 are quot t r f Bunny Win Rpm Bulb J T O 39 quot 4 i 7 we 1 Jillmiml Cuputllnr Suiwli v You Sui atbgfk will 4 m Maxim WI The actual motion of mm 7H Mbquot Hr gt m 1 Hum V39LV across ter a eh 39 electrons along a WIre cons an orce as IV is quite slow 7 to be applled 439 the electrons spend 7ltL 7 most of their time 39 bouncing around 7 randomly and there average speed drift if speed is small 212011 FIGURE 315 The Plenan rmrpm Under normal Circumstances Wires present some resistance to the motion of electrons my 1 1 MCLMHMWHI 1 mm mud n u lhrmiyil A mm In order to cause electrons to move against the resistance of a wire it is necessary to apply the poten al difference between its Elcuroiis Electrons do NOT ow in wires unless the wires are connected to a source of electrical energy rm clunmii mm i m mmum 1 i1 emu WM unugh nu W mum ur xlw m m Wm 111e electrons move in a closed path from the negative 1 terminal of the battery through the 1 circuit to the positive terminal A close analogy is provided by the water in quotquotquotquot quot Wm1 a garden hose szmmmi um m vim W F mun1nim 7 1 1 m n quot 111e direction of the current in an electric 39 c i is the direction in which 2 sit ve test charge would move m mm w 01 Wm M 0 Elecm39c current I is the ow of electric 39 charge from one place to another If Qis the tolal amountof charge that has moved past a point in a wire we define the current Iin the wire to be the rate of charge flow I I I E T39 Clll39l Cnl is he rule ill which charge ows 1 i 111e ampere A amp for short Echise A picture tube in a particular television draws a current of 15A How many electrons strike the screen every second 212011 A battepy uses chemical 1 I39eactiansto produce a l l potential difference between its terminalsAVbat The battery creates a potential difference AVwie AVbat between the ends of wire The potential difference AVwie causes an electric field in the wire The electric field establishes the current in the The magnitude of the current is determined by the battery and the wire39s resistance IAV R wire r FIGURE 3119 lhe currem I IS related to the potential difference M1 lh lmruuml lulmcn L39 trum39 m claim mm mml llh mmluum AV l lqnmuumm mmuw u jn39ll39L mllulIM In m c lL39CU39L mm Be careful Ohm s law is NOT a law of nature It is limited to those materials whose resistance R remains constant or very nearly so during use The materials to which Ohm s law applies are called ahmic mum u xw n 39 r 1 14 mm lw m Im um nunml m uu39lmw mm nle 1 i I Muir m woulmx it The current through an ohmic material is directly proportional to the potential difference Doubling the potential difference doubles the current Example Metals and other conductors are onmIc aeVIces Some materials and devices are nonohmic the current through the device is not directly proportional to the potential difference Exampl diodes are nonohmic devices You double the voltageacross a certain conductor and you observe the currentincreases three times What can you conclude 1 Ohm s law is obeyed since the current still increases when V increases 2 Ohm s law is not obeyed 3 this has nothing to do with Ohm s law 212011 The current is proportional the potential difference bw the ends of a conductor RAVl resistance of a conductor may quotm m m 1 m 101me w The were Simmm between insulaturs The resistance is a property of a specific I Wm H n A m co ductor c it depends on semiconductors and alumnus quot1101 me numm conductor s length diameter and the Cfquotdr m39s quotthe I resitivity of the material from which it is n eff Y quot39quot equot mad I resIstIvItIes pL R A The umts 0f res39Stance39 VOIts per ampere39 111e resistance of a metals is small a are called ohms cIrcuIts made from the metal WIres would 1 Q 1 VA have enormous currents and would quickly deplete the battery Two wires of the same length and diameter It is useful to limit the current in a circuit will have different resistances if they are with nhmicdevices called resistors made of different materials Resistors are made with poorly conducting materials ectro component designed to oppose an electric current by producing a voltage wnp twee A resistnris an el nic be n as x Nf ifiti f JIII 111111 terminals in proportion to the current I A Eumvs USA anm m quot chiwu quot rc 212011 W mm m 2 Exercise j 5 A bird lands on a bare copper wire caring a current of 32 A The wire is 8 gauge a E which means that is crosssectional area I is 013 cmz a Find the difference in potential between the bird s feet assuming they are separated by a distance of 60 cm b Will your answer to part a increase or decrease if the separation between the bird s feet increases S Band cam Code Islnigil 2nd Digit m Digit 0 Resistors connected end to end are said to we find that the equivalent resistance is be in series 1 W Equivalent Resistance for Resistors in Series Since the current through the series resistors R R R R H R must be the same in each I I1 I2 13 I I 2 quot 2 The total potential difference is the sum of the 51 unit Ohm Q potential differences across each resistor AV AV1 AV2 AV3 Assume that the voltage 1 12 V 1 12v of the battery is 9 V and 2 zero In the Clrcult below 2 zero that the three reSIstors 3 3 V what is the voltage 3 6 v are Identlcal across R1 4 8V Wh t 39 th t t39 I 4 4 V a Is e M 5 you need to 5 4 V difference across each know the actual R4 9 R2 2 a resistor value of R 12V 212011 Resistors are in parallel when What is the potential at points a to e they quot395 the same i ll 1 1 potential difference t Using the fact that the potential difference across each resistor is the same AV AVl 3 the total current is the sum of the currents in each resistor I II I Ia they can be replaced bya single equivalent 1 10 A resistance In the circuit below 2 19quot Equivalent Resistance for Resistors in Parallel What is the current 3 5 A v 391 throu h R 4 2A i i i 2 7 9 J le R1 R2 R3 km 5 7 A 51 unit ohm I w v If a circuit is more complex Acumbmalmnmveslsmvs gtstart with combinations of resistors that are either purely in series or in parallel Haunt n u The Lou me 1 mauled m Single fquw lem v gtreplace these with their equivalent Lug 212011 Exercise What is equivalent resistance between points a and b in figures I ltgtlHiJ 40 IH h vvvv qu gt lt ul Which of these diagrams represent the same circuit E3 Aa and b Bb and c Ca and c Da b and d Ea b and c Exercise Three resistors 110 53 Q and R are connected in series with a 12 V battery The total current flowing through the battery is 016A 3 Find the value of resistance R b Find the potential difference across each resistor An ideal batteries sepa rate the difference charges and create the potential Real batteries sepa rate the difference and provide a slight resistance internal resistance charges and create the potential LI 1 all the current continues to ow through the bulb current ows 2 half the current flows through 39 the wire the other ha f through a llght continues through the bulb bulb39 If a me Is 3 all the current flows through now connected the wire across the bu 4 none of the above what happens 4Q Two light bulbs A and B are connected in series to a constant voltage source When a wire is connected across B bulb A will 1 glow brighter than before 2 glow just the same as before 3 glow dimmer than before 4 go out completely 5 explode 212011 Energy and Power In Elecmc c39rcu39ts In materials for which ohm39s law holds the power when 3 ch ge moves across a potential difference can also be written is potential energy changes39 AU 4 HOW I IV MIR FR Therefore the power it takes to do this is V V2 T e W V iv AU AQW Eieftvicai Power K R Al Al 7 SI nil watt W This power mos y becomes heat inside the resistive material Ililii i i iilquot 1 identical with the 2 mm 2032932939 if i same resistance R 3 both the same identical bulbs quot A to D Which circuit 4 it depends on R roduces more quotWquot quotMquot x E D s a Q light brightness g AECg power EAgtCgtEgtD izv in Example More complex circuils cannot be broken down When the electric company sends you a bill your into series and pa ralle39 pieces usage is q ote I kilowatthours kWh They For these circuits Kirchholf srulesare useful are charging you for energy use and kWh are a gtThe junction rule is a consequence of charge measure of energy conservation gtthe loop rule is a consequence of energy conservation 1 kilnvt39nltrlmtlt illiltl Wiquotwhili lllllllJsuf lhlllls39 z 31 x iim wn ilV i JilV Mi 7 n 212011 The junction rule At anyjunction the current This basic conservation statement that the sum entering the junction must equal the current of the currents into a junction equals the sum of leaving it the currents leaving is called irclilwff s junction law For a junction the law of conservation of current 21in Elm i l 1 I r r if 2 m 2 out I1 I2 I3 I F J l T 11 2 13 0 where the Z symbol means summation The loop rule The algebraic sum of the potential quot quotAIml a closed loop must be zero it Whatfll e the l must return to its original value at the original magnltude and point the direction of 7g 7 u 7 e j the current in r f 391 l t AV 397 n 414 AM the fifth wire L r Mvhhnll39x imp MW A 15 A into the junclion B 15 A out of the junction C 1 A into the junction D1 A out of the junction E Not enough data to determine r a Imu lmvp Amgu muquot m my mum I 3mm n Tactics Using Kirchhoff s loop law 13139 Using Kivrhhn 39i twp lww 59 0 mm a rirmllillnmmlxlwl11H Mumquotva nnkmmuqummm n lullu lumumn 1 wurru l m and lululuuncmnnuu mum yum mum n gnu mm m lellllldlllk l ducclmnmllmm H munu llmi mm my line Mm nunm Imuiuu m uhmry lllUlLk 39lllll1ll will m lmrrmi n W a wmng u um um nlue m mm eml up mum m n mu 1 1 m m1 u v 212011 rlmmr 3 1 Amlym ul lln lmzu umul mug Kulelvull s loop law W m m Ilw will yu ilvruw illw 1mmquot v lmvmmllwm ilwuilwnluuhli Ollhm an 39 6 HM Mm an um Nu u me anmumm erm hquotIllllvnllmil hmllwwumm mmumuumm l 5 NM 111 um Mum JV39M 4 mum um Foramdmlhnllmw milwpmmvcu 7 If I mva mm 1 Harammur m 7m l lrltnlulxlu39mm gt 9 Appl mammal 2mm u 39 alummm u m n m m lrnu m Exerci Exercise Whatare the magnitude and direction of the Find the iurrent through eaCh VESismr Sing current in the 309 resistor KiI ClIlIOffS rules 3 ohm 2 o B V 39 A yl n D 11 12 13 4 ohm f i gt I H o 4 1 V1 m m Find the current through each resistor using a rules for series and parallel resistors and b Kirchhoff39s rues Capacitors AAAA vvvv l IZX vvvv AAAA vvvr A capacitor is a device that has capacity to store the electrical charge and the energy A simple type of a capacitor is the parallelplate capacitor It consists of two plates of area A separated by a distance d 212011 When the plates of capacitor are connected to the terminals of battery they become charged one plate Q the other Q The greater the charge Q for a given voltage V the greater the capacitance C The capacitance relates the charge to the potential difference I I I l I SI units coulomb volt farad F The way to increase the capacitance of a capacitor is to insert an insulating material dielectric between its plates 1 pF 103912F 1pF 10396F The Dielectric materials mm can be made from Ceramic Mica Polypropylene Polyester Electrolytic PM Tantalum and even air met ofdieleumc When a dielectric placed between the plates of a capacitor it gives a lower potential difference with the same charge due to the polarization of the material This increases the capacitance 212011 The polarization of the dielectric results in a lower electric eld within it the new field is given by dividing the original field by the dielectric constant Ir E E K This decreases the potential difference AV E a Q Ca F Therefore the capacitance of a parallelplate capacitor filled with a dielectric becomes The value of capacitance is determined by gtThe size of the plates gt The distance between them gt The type of dielectric material used For an insulating materials K gt1 TABLE 20 Dielectm comlm u The dielectric constant K is a Subxtance Dielectric conutanl K property of the WdlLr 5m material here are Nl39uprvm some exa mples quotquotl b 7 mm gin 10 Mira 4 mm 3 Mylar 11 Talon 2 il39 10m RH Vacuum c K 8quot A d Application 9 Dimmi Plales The keys on most computer keyboards are capacitors switches Pressing the key pushes two capacitor plates closer together increasing there capacitance Circuits Containing Capacitors Capacitors can be connected in seriesor in parallel When capacitors are 1 avg L connected in parallel I the potential In difference across each one is the same quotL In Therefore the equivalent capacitance is the sum of the individual capacitances Equivalent Capacitance for Capacitors in Parallel Cm C C3 a 7 2C 51 unit famd F 212011 Exercise Two capacitors 75 pF and 15 pF are connected in parallel across V battery a Find the equivalentcapacitance of the two capacitors c Find the charge stored on each capacitor Equivalent Capatita e lo Capacitors in Parallel 2C L Cl C Sl um 11le F b Which capacitor stores more charge Explain Capacitors connected in series do not have the same potential difference across them but they do all carry the same cha The total potential w difference is the sum of the potential k differences across each T 111erefore the equivalent capacitance is Equivalent Capacitance for capacitors in Series 1 1 i 7 T a i v r CmL C L C L 51mm fared F rc se Two capacitors 75 pF and 15 pF are connected in series across a 12V batter a Find the equivalentcapacitance of the two ca pac39 c Find the charge stored on each capacitor 1 x L k x n Equivalnnt apnclianle l apldlnrx u Swim I r 1 i A l I V elm mm 1 they do all carry the same charge ltors b Which capacitor stores more charge Explain What is the equivalent 3 20 capacitance Cm of the Z 2 43C combination below c C 3C d c 13c e c 12c How does the voltage Vlacross a VI V2 the firstcapacitor b V gt V2 c V lt V2 Q com pare to the voltage l2 across the second capacitor CZ d all voltages are zero q1ou17 w V c1ou17 c1om 212011 How does the charge Qlon the a Q 02 firstcapacitor b 0 gt 02 cl compare to c o lt 02 the charge 02 on d all charges are zero the second capacitor Q 6 10m 1 0V c1ouF 63101417 RC Circuits In a circuit containing only batteries and capacitors charge appears almost instantaneously on the capacitors when the circui is connected However if the circuit contains resistors as well this Is not the case 111e resistors Ii it the rate at which charge can ow The simplest example of such a circuit is called IEcircuit m 1 1 w l n r 2 Here ris the time constantof the circuit 5 l MWW t l r RC e time constant for the m discharge of this capacitor is 4 s E 1he capacitor doesn39t discharge because the resistors cancel eac other An application of RC circuits is the heart pacemaker the device uses the RC circuit to deliver precisely timed pulses directly to the heart 212011 Ammeters and Voltmeters An ammeberis a device for measuring current and a voltmeter measures voltages The current in the circuit must flow through the ammeter therefore the ammeter s ould have as aw a resistance as ossible for the least disturbance A lumeta measures the potential drop between two pains in a circuit It is connected in parallel in order to minimize the e ect on the circuit it should have as 39ble large a resistance as possl y QC 2 AAA z AAA v n C Getting Grounded Grounding a circuit services two functions 1 It provides a common reference potential ihe lrtultanalyses arcEduc we have dlstussed SD lar al mly f1 Dll39 tulhes an arise if you want to connect two d erent mum mquot 2 It is an important safety feature to prevent injury or death from a defective circuits Applications belies resistors MRlkrlixw MM ww 1wm r u r F amiiel leiisxnls N i I quotN iiiK 39 IiJ R Applications
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