Electric Fields. Chapter Outline Properties of Electric Charges 23.2 Insulators and Conductors 23.3 Coulomb s Law 23.4 The Electric Field


 Stuart Stewart
 2 years ago
 Views:
Transcription
1 P U Z Z L E R Soft contct lenses re comfortble to wer becuse the ttrct the proteins in the werer s ters, incorporting the comple molecules right into the lenses. The become, in sense, prt of the werer. Some tpes of mkeup eploit this sme ttrctive force to dhere to the skin. Wht is the nture of this force? (Chrles D. Winters) c h p t e r Electric Fields Chpter Outline 23.1 Properties of Electric Chrges 23.2 Insultors nd Conductors 23.3 Coulomb s Lw 23.4 The Electric Field 23.5 Electric Field of Continuous Chrge Distribution 23.6 Electric Field Lines 23.7 Motion of Chrged Prticles in Uniform Electric Field 708
2 T he electromgnetic force between chrged prticles is one of the fundmentl forces of nture. We begin this chpter b describing some of the bsic properties of electric forces. We then discuss Coulomb s lw, which is the fundmentl lw governing the force between n two chrged prticles. Net, we introduce the concept of n electric field ssocited with chrge distribution nd describe its effect on other chrged prticles. We then show how to use Coulomb s lw to clculte the electric field for given chrge distribution. We conclude the chpter with discussion of the motion of chrged prticle in uniform electric field Properties of Electric Chrges PROPERTIES OF ELECTRIC CHARGES A number of simple eperiments demonstrte the eistence of electric forces nd chrges. For emple, fter running comb through our hir on dr d, ou will find tht the comb ttrcts bits of pper. The ttrctive force is often strong enough to suspend the pper. The sme effect occurs when mterils such s glss or rubber re rubbed with silk or fur. Another simple eperiment is to rub n inflted blloon with wool. The blloon then dheres to wll, often for hours. When mterils behve in this w, the re sid to be electrified, or to hve become electricll chrged. You cn esil electrif our bod b vigorousl rubbing our shoes on wool rug. The electric chrge on our bod cn be felt nd removed b lightl touching (nd strtling) friend. Under the right conditions, ou will see sprk when ou touch, nd both of ou will feel slight tingle. (Eperiments such s these work best on dr d becuse n ecessive mount of moisture in the ir cn cuse n chrge ou build up to lek from our bod to the Erth.) In series of simple eperiments, it is found tht there re two kinds of electric chrges, which were given the nmes positive nd negtive b Benjmin Frnklin ( ). To verif tht this is true, consider hrd rubber rod tht hs been rubbed with fur nd then suspended b nonmetllic thred, s shown in Figure When glss rod tht hs been rubbed with silk is brought ner the rubber rod, the two ttrct ech other (Fig. 23.1). On the other hnd, if two chrged rubber rods (or two chrged glss rods) re brought ner ech other, s shown in Figure 23.1b, the two repel ech other. This observtion shows tht the rubber nd glss re in two different sttes of electrifiction. On the bsis of these observtions, we conclude tht like chrges repel one nother nd unlike chrges ttrct one nother. Using the convention suggested b Frnklin, the electric chrge on the glss rod is clled positive nd tht on the rubber rod is clled negtive. Therefore, n chrged object ttrcted to chrged rubber rod (or repelled b chrged glss rod) must hve positive chrge, nd n chrged object repelled b chrged rubber rod (or ttrcted to chrged glss rod) must hve negtive chrge. Attrctive electric forces re responsible for the behvior of wide vriet of commercil products. For emple, the plstic in mn contct lenses, etfilcon, is mde up of molecules tht electricll ttrct the protein molecules in humn ters. These protein molecules re bsorbed nd held b the plstic so tht the lens ends up being primril composed of the werer s ters. Becuse of this, the werer s ee does not tret the lens s foreign object, nd it cn be worn comfortbl. Mn cosmetics lso tke dvntge of electric forces b incorporting mterils tht re electricll ttrcted to skin or hir, cusing the pigments or other chemicls to st put once the re pplied. QuickLb Rub n inflted blloon ginst our hir nd then hold the blloon ner thin strem of wter running from fucet. Wht hppens? (A rubbed plstic pen or comb will lso work.)
3 710 CHAPTER 23 Electric Fields Rubber Rubber Figure 23.1 F F Glss () F F Rubber () A negtivel chrged rubber rod suspended b thred is ttrcted to positivel chrged glss rod. (b) A negtivel chrged rubber rod is repelled b nother negtivel chrged rubber rod. (b) Chrge is conserved Another importnt spect of Frnklin s model of electricit is the impliction tht electric chrge is lws conserved. Tht is, when one object is rubbed ginst nother, chrge is not creted in the process. The electrified stte is due to trnsfer of chrge from one object to the other. One object gins some mount of negtive chrge while the other gins n eul mount of positive chrge. For emple, when glss rod is rubbed with silk, the silk obtins negtive chrge tht is eul in mgnitude to the positive chrge on the glss rod. We now know from our understnding of tomic structure tht negtivel chrged electrons re trnsferred from the glss to the silk in the rubbing process. Similrl, when rubber is rubbed with fur, electrons re trnsferred from the fur to the rubber, giving the rubber net negtive chrge nd the fur net positive chrge. This process is consistent with the fct tht neutrl, unchrged mtter contins s mn positive chrges (protons within tomic nuclei) s negtive chrges (electrons). Figure 23.2 Rubbing blloon ginst our hir on dr d cuses the blloon nd our hir to become chrged. Chrge is untized Quick Quiz 23.1 If ou rub n inflted blloon ginst our hir, the two mterils ttrct ech other, s shown in Figure Is the mount of chrge present in the blloon nd our hir fter rubbing () less thn, (b) the sme s, or (c) more thn the mount of chrge present before rubbing? In 1909, Robert Millikn ( ) discovered tht electric chrge lws occurs s some integrl multiple of fundmentl mount of chrge e. In modern terms, the electric chrge is sid to be untized, where is the stndrd smbol used for chrge. Tht is, electric chrge eists s discrete pckets, nd we cn write Ne, where N is some integer. Other eperiments in the sme period showed tht the electron hs chrge e nd the proton hs chrge of eul mgnitude but opposite sign e. Some prticles, such s the neutron, hve no chrge. A neutrl tom must contin s mn protons s electrons. Becuse chrge is conserved untit, the net chrge in closed region remins the sme. If chrged prticles re creted in some process, the re lws creted in pirs whose members hve eulmgnitude chrges of opposite sign.
4 23.2 Insultors nd Conductors 711 From our discussion thus fr, we conclude tht electric chrge hs the following importnt properties: Two kinds of chrges occur in nture, with the propert tht unlike chrges ttrct one nother nd like chrges repel one nother. Chrge is conserved. Chrge is untized. Properties of electric chrge INSULATORS AND CONDUCTORS It is convenient to clssif substnces in terms of their bilit to conduct electric chrge: Electricl conductors re mterils in which electric chrges move freel, wheres electricl insultors re mterils in which electric chrges cnnot move freel. Mterils such s glss, rubber, nd wood fll into the ctegor of electricl insultors. When such mterils re chrged b rubbing, onl the re rubbed becomes chrged, nd the chrge is unble to move to other regions of the mteril. In contrst, mterils such s copper, luminum, nd silver re good electricl conductors. When such mterils re chrged in some smll region, the chrge redil distributes itself over the entire surfce of the mteril. If ou hold copper rod in our hnd nd rub it with wool or fur, it will not ttrct smll piece of pper. This might suggest tht metl cnnot be chrged. However, if ou ttch wooden hndle to the rod nd then hold it b tht hndle s ou rub the rod, the rod will remin chrged nd ttrct the piece of pper. The eplntion for this is s follows: Without the insulting wood, the electric chrges produced b rubbing redil move from the copper through our bod nd into the Erth. The insulting wooden hndle prevents the flow of chrge into our hnd. Semiconductors re third clss of mterils, nd their electricl properties re somewhere between those of insultors nd those of conductors. Silicon nd germnium re wellknown emples of semiconductors commonl used in the fbriction of vriet of electronic devices, such s trnsistors nd lightemitting diodes. The electricl properties of semiconductors cn be chnged over mn orders of mgnitude b the ddition of controlled mounts of certin toms to the mterils. When conductor is connected to the Erth b mens of conducting wire or pipe, it is sid to be grounded. The Erth cn then be considered n infinite sink to which electric chrges cn esil migrte. With this in mind, we cn understnd how to chrge conductor b process known s induction. To understnd induction, consider neutrl (unchrged) conducting sphere insulted from ground, s shown in Figure When negtivel chrged rubber rod is brought ner the sphere, the region of the sphere nerest the rod obtins n ecess of positive chrge while the region frthest from the rod obtins n eul ecess of negtive chrge, s shown in Figure 23.3b. (Tht is, electrons in the region nerest the rod migrte to the opposite side of the sphere. This occurs even if the rod never ctull touches the sphere.) If the sme eperiment is performed with conducting wire connected from the sphere to ground (Fig. 23.3c), some of the electrons in the conductor re so strongl repelled b the presence of Metls re good conductors Chrging b induction
5 712 CHAPTER 23 Electric Fields () (b) (c) (d) Figure 23.3 Chrging metllic object b induction (tht is, the two objects never touch ech other). () A neutrl metllic sphere, with eul numbers of positive nd negtive chrges. (b) The chrge on the neutrl sphere is redistributed when chrged rubber rod is plced ner the sphere. (c) When the sphere is grounded, some of its electrons leve through the ground wire. (d) When the ground connection is removed, the sphere hs ecess positive chrge tht is nonuniforml distributed. (e) When the rod is removed, the ecess positive chrge becomes uniforml distributed over the surfce of the sphere. (e)
6 23.3 Coulomb s Lw 713 Chrged object Insultor Induced chrges QuickLb Ter some pper into ver smll pieces. Comb our hir nd then bring the comb close to the pper pieces. Notice tht the re ccelerted towrd the comb. How does the mgnitude of the electric force compre with the mgnitude of the grvittionl force eerted on the pper? Keep wtching nd ou might see few pieces jump w from the comb. The don t just fll w; the re repelled. Wht cuses this? Figure 23.4 () (b) () The chrged object on the left induces chrges on the surfce of n insultor. (b) A chrged comb ttrcts bits of pper becuse chrges re displced in the pper the negtive chrge in the rod tht the move out of the sphere through the ground wire nd into the Erth. If the wire to ground is then removed (Fig. 23.3d), the conducting sphere contins n ecess of induced positive chrge. When the rubber rod is removed from the vicinit of the sphere (Fig. 23.3e), this induced positive chrge remins on the ungrounded sphere. Note tht the chrge remining on the sphere is uniforml distributed over its surfce becuse of the repulsive forces mong the like chrges. Also note tht the rubber rod loses none of its negtive chrge during this process. Chrging n object b induction reuires no contct with the bod inducing the chrge. This is in contrst to chrging n object b rubbing (tht is, b conduction), which does reuire contct between the two objects. A process similr to induction in conductors tkes plce in insultors. In most neutrl molecules, the center of positive chrge coincides with the center of negtive chrge. However, in the presence of chrged object, these centers inside ech molecule in n insultor m shift slightl, resulting in more positive chrge on one side of the molecule thn on the other. This relignment of chrge within individul molecules produces n induced chrge on the surfce of the insultor, s shown in Figure Knowing bout induction in insultors, ou should be ble to eplin wh comb tht hs been rubbed through hir ttrcts bits of electricll neutrl pper nd wh blloon tht hs been rubbed ginst our clothing is ble to stick to n electricll neutrl wll. Quick Quiz 23.2 Object A is ttrcted to object B. If object B is known to be positivel chrged, wht cn we s bout object A? () It is positivel chrged. (b) It is negtivel chrged. (c) It is electricll neutrl. (d) Not enough informtion to nswer COULOMB S LAW Chrles Coulomb ( ) mesured the mgnitudes of the electric forces between chrged objects using the torsion blnce, which he invented (Fig. 23.5). Chrles Coulomb ( ) Coulomb's mjor contribution to science ws in the field of electrosttics nd mgnetism. During his lifetime, he lso investigted the strengths of mterils nd determined the forces tht ffect objects on bems, thereb contributing to the field of structurl mechnics. In the field of ergonomics, his reserch provided fundmentl understnding of the ws in which people nd nimls cn best do work. (Photo courtes of AIP Niels Bohr Librr/E. Scott Brr Collection)
7 714 CHAPTER 23 Electric Fields B A Figure 23.5 Coulomb constnt Suspension hed Fiber Coulomb s torsion blnce, used to estblish the inversesure lw for the electric force between two chrges. Chrge on n electron or proton Coulomb confirmed tht the electric force between two smll chrged spheres is proportionl to the inverse sure of their seprtion distnce r tht is, F e 1/r 2. The operting principle of the torsion blnce is the sme s tht of the pprtus used b Cvendish to mesure the grvittionl constnt (see Section 14.2), with the electricll neutrl spheres replced b chrged ones. The electric force between chrged spheres A nd B in Figure 23.5 cuses the spheres to either ttrct or repel ech other, nd the resulting motion cuses the suspended fiber to twist. Becuse the restoring torue of the twisted fiber is proportionl to the ngle through which the fiber rottes, mesurement of this ngle provides untittive mesure of the electric force of ttrction or repulsion. Once the spheres re chrged b rubbing, the electric force between them is ver lrge compred with the grvittionl ttrction, nd so the grvittionl force cn be neglected. Coulomb s eperiments showed tht the electric force between two sttionr chrged prticles is inversel proportionl to the sure of the seprtion r between the prticles nd directed long the line joining them; is proportionl to the product of the chrges 1 nd 2 on the two prticles; is ttrctive if the chrges re of opposite sign nd repulsive if the chrges hve the sme sign. From these observtions, we cn epress Coulomb s lw s n eution giving the mgnitude of the electric force (sometimes clled the Coulomb force) between two point chrges: (23.1) where k e is constnt clled the Coulomb constnt. In his eperiments, Coulomb ws ble to show tht the vlue of the eponent of r ws 2 to within n uncertint of few percent. Modern eperiments hve shown tht the eponent is 2 to within n uncertint of few prts in The vlue of the Coulomb constnt depends on the choice of units. The SI unit of chrge is the coulomb (C). The Coulomb constnt k e in SI units hs the vlue This constnt is lso written in the form F e k e 1 2 r 2 k e N m 2 /C 2 k e where the constnt 0 (lowercse Greek epsilon) is known s the permittivit of free spce nd hs the vlue C 2 /N m 2. The smllest unit of chrge known in nture is the chrge on n electron or proton, 1 which hs n bsolute vlue of e C Therefore, 1 C of chrge is pproimtel eul to the chrge of electrons or protons. This number is ver smll when compred with the number of 1 No unit of chrge smller thn e hs been detected s free chrge; however, recent theories propose the eistence of prticles clled urks hving chrges e/3 nd 2e/3. Although there is considerble eperimentl evidence for such prticles inside nucler mtter, free urks hve never been detected. We discuss other properties of urks in Chpter 46 of the etended version of this tet.
8 23.3 Coulomb s Lw 715 TABLE 23.1 Chrge nd Mss of the Electron, Proton, nd Neutron Prticle Chrge (C) Mss (kg) Electron (e) Proton (p) Neutron (n) free electrons 2 in 1 cm 3 of copper, which is of the order of Still, 1 C is substntil mount of chrge. In tpicl eperiments in which rubber or glss rod is chrged b friction, net chrge of the order of 10 6 C is obtined. In other words, onl ver smll frction of the totl vilble chrge is trnsferred between the rod nd the rubbing mteril. The chrges nd msses of the electron, proton, nd neutron re given in Tble EXAMPLE 23.1 The Hdrogen Atom The electron nd proton of hdrogen tom re seprted (on the verge) b distnce of pproimtel m. Find the mgnitudes of the electric force nd the grvittionl force between the two prticles. Solution From Coulomb s lw, we find tht the ttrctive electric force hs the mgnitude F e k e e 2 r N m2 ( C) 2 C 2 ( m) N Using Newton s lw of grvittion nd Tble 23.1 for the prticle msses, we find tht the grvittionl force hs the mgnitude F g G m em p r N m2 kg 2 ( kg)( kg) ( m) N The rtio F e /F g Thus, the grvittionl force between chrged tomic prticles is negligible when compred with the electric force. Note the similrit of form of Newton s lw of grvittion nd Coulomb s lw of electric forces. Other thn mgnitude, wht is fundmentl difference between the two forces? When deling with Coulomb s lw, ou must remember tht force is vector untit nd must be treted ccordingl. Thus, the lw epressed in vector form for the electric force eerted b chrge 1 on second chrge 2, written F 12, is F 12 k e 1 2 r 2 rˆ (23.2) where rˆ is unit vector directed from 1 to 2, s shown in Figure Becuse the electric force obes Newton s third lw, the electric force eerted b 2 on 1 is 2 A metl tom, such s copper, contins one or more outer electrons, which re wekl bound to the nucleus. When mn toms combine to form metl, the soclled free electrons re these outer electrons, which re not bound to n one tom. These electrons move bout the metl in mnner similr to tht of gs molecules moving in continer.
9 716 CHAPTER 23 Electric Fields r 2 F 12 1 rˆ F 21 () 1 F 21 (b) F 12 2 Figure 23.6 Two point chrges seprted b distnce r eert force on ech other tht is given b Coulomb s lw. The force F 21 eerted b 2 on 1 is eul in mgnitude nd opposite in direction to the force F 12 eerted b 1 on 2. () When the chrges re of the sme sign, the force is repulsive. (b) When the chrges re of opposite signs, the force is ttrctive. eul in mgnitude to the force eerted b 1 on 2 nd in the opposite direction; tht is, F 21 F 12. Finll, from Eution 23.2, we see tht if 1 nd 2 hve the sme sign, s in Figure 23.6, the product 1 2 is positive nd the force is repulsive. If 1 nd 2 re of opposite sign, s shown in Figure 23.6b, the product 1 2 is negtive nd the force is ttrctive. Noting the sign of the product 1 2 is n es w of determining the direction of forces cting on the chrges. Quick Quiz 23.3 Object A hs chrge of 2 C, nd object B hs chrge of 6 C. Which sttement is true? () F AB 3F BA. (b) F AB F BA. (c) 3F AB F BA. When more thn two chrges re present, the force between n pir of them is given b Eution Therefore, the resultnt force on n one of them euls the vector sum of the forces eerted b the vrious individul chrges. For emple, if four chrges re present, then the resultnt force eerted b prticles 2, 3, nd 4 on prticle 1 is F 1 F 21 F 31 F 41 EXAMPLE 23.2 Find the Resultnt Force Consider three point chrges locted t the corners of right tringle s shown in Figure 23.7, where C, C, nd 0.10 m. Find the resultnt force eerted on 3. Solution First, note the direction of the individul forces eerted b 1 nd 2 on 3. The force F 23 eerted b 2 on 3 is ttrctive becuse 2 nd 3 hve opposite signs. The force F 13 eerted b 1 on 3 is repulsive becuse both chrges re positive. The mgnitude of F 23 is F 23 k e N m2 ( C)( C) C 2 (0.10 m) N Note tht becuse 3 nd 2 hve opposite signs, F 23 is to the left, s shown in Figure 23.7.
10 23.3 Coulomb s Lw 717 Figure The force eerted b 1 on 3 is F 13. The force eerted b 2 on 3 is F 23. The resultnt force F 3 eerted on 3 is the vector sum F 13 F 23. F 23 3 F N m2 ( C)( C) C 2 2(0.10 m) 2 11 N The force F 13 is repulsive nd mkes n ngle of 45 with the is. Therefore, the nd components of F 13 re eul, with mgnitude given b F 13 cos N. The force F 23 is in the negtive direction. Hence, the nd components of the resultnt force cting on 3 re F 3 F 13 F N 9.0 N 1.1 N F 3 F N We cn lso epress the resultnt force cting on 3 in unitvector form s F 3 ( 1.1i 7.9j) N The mgnitude of the force eerted b 1 on 3 is F 13 k e 1 3 (!2) 2 Eercise force F 3. Answer Find the mgnitude nd direction of the resultnt 8.0 N t n ngle of 98 with the is. EXAMPLE 23.3 Where Is the Resultnt Force Zero? Three point chrges lie long the is s shown in Figure The positive chrge C is t 2.00 m, the positive chrge C is t the origin, nd the resultnt force cting on 3 is zero. Wht is the coordinte of 3? Solution Becuse 3 is negtive nd 1 nd 2 re positive, the forces F 13 nd F 23 re both ttrctive, s indicted in Figure From Coulomb s lw, F 13 nd F 23 hve mgnitudes F 13 k e 1 3 (2.00 ) 2 F 23 k 2 3 e 2 For the resultnt force on 3 to be zero, F 23 must be eul in mgnitude nd opposite in direction to F 13, or k e k e 1 3 (2.00 ) 2 Noting tht k e nd 3 re common to both sides nd so cn be dropped, we solve for nd find tht ( )( C) 2 ( C) Solving this udrtic eution for, we find tht m. Figure 23.8 (2.00 ) Wh is the negtive root not cceptble? 2.00 m F 23 3 F 13 1 Three point chrges re plced long the is. If the net force cting on 3 is zero, then the force F 13 eerted b 1 on 3 must be eul in mgnitude nd opposite in direction to the force F 23 eerted b 2 on 3. EXAMPLE 23.4 Find the Chrge on the Spheres Two identicl smll chrged spheres, ech hving mss of kg, hng in euilibrium s shown in Figure The length of ech string is 0.15 m, nd the ngle is 5.0. Find the mgnitude of the chrge on ech sphere. Solution From the right tringle shown in Figure 23.9, we see tht sin /L. Therefore, L sin (0.15 m)sin m The seprtion of the spheres is m. The forces cting on the left sphere re shown in Figure 23.9b. Becuse the sphere is in euilibrium, the forces in the
11 718 CHAPTER 23 Electric Fields horizontl nd verticl directions must seprtel dd up to zero: (1) F T sin F e 0 (2) F T cos mg 0 From Eution (2), we see tht T mg /cos ; thus, T cn be eliminted from Eution (1) if we mke this substitution. This gives vlue for the mgnitude of the electric force F e : (3) F e mg tn ( kg)(9.80 m/s 2 )tn N From Coulomb s lw (E. 23.1), the mgnitude of the electric force is F e k e 2 r 2 L Figure 23.9 θ () θ L = 0.15 m θ = 5.0 L T cos θ () Two identicl spheres, ech crring the sme chrge, suspended in euilibrium. (b) The freebod digrm for the sphere on the left. F e (b) θ mg T θ T sin θ where r m nd is the mgnitude of the chrge on ech sphere. (Note tht the term 2 rises here becuse the chrge is the sme on both spheres.) This eution cn be solved for 2 to give Eercise If the chrge on the spheres were negtive, how mn electrons would hve to be dded to them to ield net chrge of C? Answer 2 F er 2 ( N)(0.026 m) 2 k e N m 2 /C C electrons. QuickLb For this eperiment ou need two 20cm strips of trnsprent tpe (mss of ech 65 mg). Fold bout 1 cm of tpe over t one end of ech strip to crete hndle. Press both pieces of tpe side b side onto tble top, rubbing our finger bck nd forth cross the strips. Quickl pull the strips off the surfce so tht the become chrged. Hold the tpe hndles together nd the strips will repel ech other, forming n inverted V shpe. Mesure the ngle between the pieces, nd estimte the ecess chrge on ech strip. Assume tht the chrges ct s if the were locted t the center of mss of ech strip. Q Figure A smll positive test chrge 0 plced ner n object crring much lrger positive chrge Q eperiences n electric field E directed s shown. 0 E THE ELECTRIC FIELD Two field forces hve been introduced into our discussions so fr the grvittionl force nd the electric force. As pointed out erlier, field forces cn ct through spce, producing n effect even when no phsicl contct between the objects occurs. The grvittionl field g t point in spce ws defined in Section 14.6 to be eul to the grvittionl force F g cting on test prticle of mss m divided b tht mss: g F g /m. A similr pproch to electric forces ws developed b Michel Frd nd is of such prcticl vlue tht we shll devote much ttention to it in the net severl chpters. In this pproch, n electric field is sid to eist in the region of spce round chrged object. When nother chrged object enters this electric field, n electric force cts on it. As n emple, consider Figure 23.10, which shows smll positive test chrge 0 plced ner second object crring much greter positive chrge Q. We define the strength (in other words, the mgnitude) of the electric field t the loction of the test chrge to be the electric force per unit chrge, or to be more specific
12 23.4 The Electric Field 719 the electric field E t point in spce is defined s the electric force F e cting on positive test chrge 0 plced t tht point divided b the mgnitude of the test chrge: E F e 0 (23.3) Definition of electric field Note tht E is the field produced b some chrge eternl to the test chrge it is not the field produced b the test chrge itself. Also, note tht the eistence of n electric field is propert of its source. For emple, ever electron comes with its own electric field. The vector E hs the SI units of newtons per coulomb (N/C), nd, s Figure shows, its direction is the direction of the force positive test chrge eperiences when plced in the field. We s tht n electric field eists t point if test chrge t rest t tht point eperiences n electric force. Once the mgnitude nd direction of the electric field re known t some point, the electric force eerted on n chrged prticle plced t tht point cn be clculted from This drmtic photogrph cptures lightning bolt striking tree ner some rurl homes.
13 720 CHAPTER 23 Electric Fields TABLE 23.2 Source Tpicl Electric Field Vlues E (N/C) Fluorescent lighting tube 10 Atmosphere (fir wether) 100 Blloon rubbed on hir Atmosphere (under thundercloud) Photocopier Sprk in ir Ner electron in hdrogen tom >> 0 () Figure (b) () For smll enough test chrge 0, the chrge distribution on the sphere is undisturbed. (b) When the test chrge 0 is greter, the chrge distribution on the sphere is disturbed s the result of the proimit of 0. rˆ rˆ Figure r () E (b) 0 A test chrge 0 t point P is distnce r from point chrge. () If is positive, then the electric field t P points rdill outwrd from. (b) If is negtive, then the electric field t P points rdill inwrd towrd. P 0 P E Eution Furthermore, the electric field is sid to eist t some point (even empt spce) regrdless of whether test chrge is locted t tht point. (This is nlogous to the grvittionl field set up b n object, which is sid to eist t given point regrdless of whether some other object is present t tht point to feel the field.) The electric field mgnitudes for vrious field sources re given in Tble When using Eution 23.3, we must ssume tht the test chrge 0 is smll enough tht it does not disturb the chrge distribution responsible for the electric field. If vnishingl smll test chrge 0 is plced ner uniforml chrged metllic sphere, s shown in Figure 23.11, the chrge on the metllic sphere, which produces the electric field, remins uniforml distributed. If the test chrge is gret enough ( 0 W 0 ), s shown in Figure 23.11b, the chrge on the metllic sphere is redistributed nd the rtio of the force to the test chrge is different: (F e / 0 F e / 0 ). Tht is, becuse of this redistribution of chrge on the metllic sphere, the electric field it sets up is different from the field it sets up in the presence of the much smller 0. To determine the direction of n electric field, consider point chrge locted distnce r from test chrge 0 locted t point P, s shown in Figure According to Coulomb s lw, the force eerted b on the test chrge is F e k e 0 r 2 rˆ where rˆ is unit vector directed from towrd 0. Becuse the electric field t P, the position of the test chrge, is defined b E F e / 0, we find tht t P, the electric field creted b is E k e (23.4) r 2 rˆ If is positive, s it is in Figure 23.12, the electric field is directed rdill outwrd from it. If is negtive, s it is in Figure 23.12b, the field is directed towrd it. To clculte the electric field t point P due to group of point chrges, we first clculte the electric field vectors t P individull using Eution 23.4 nd then dd them vectorill. In other words, t n point P, the totl electric field due to group of chrges euls the vector sum of the electric fields of the individul chrges. This superposition principle pplied to fields follows directl from the superposition propert of electric forces. Thus, the electric field of group of chrges cn
14 23.4 The Electric Field 721 This metllic sphere is chrged b genertor so tht it crries net electric chrge. The high concentrtion of chrge on the sphere cretes strong electric field round the sphere. The chrges then lek through the gs surrounding the sphere, producing pink glow. be epressed s (23.5) where r i is the distnce from the ith chrge i to the point P (the loction of the test chrge) nd rˆi is unit vector directed from i towrd P. Quick Quiz 23.4 E k e i A chrge of 3 C is t point P where the electric field is directed to the right nd hs mgnitude of N/C. If the chrge is replced with 3 C chrge, wht hppens to the electric field t P? i r i 2 rˆi EXAMPLE 23.5 Electric Field Due to Two Chrges A chrge C is locted t the origin, nd second chrge C is locted on the is, 0.30 m from the origin (Fig ). Find the electric field t the point P, which hs coordintes (0, 0.40) m. Solution First, let us find the mgnitude of the electric field t P due to ech chrge. The fields E 1 due to the 7.0 C chrge nd E 2 due to the 5.0 C chrge re shown in Figure Their mgnitudes re E 1 k 1 e r N m2 ( C) C 2 (0.40 m) m N/C E 2 k 2 e r N m2 ( C) C 2 (0.50 m) N/C The vector E 1 hs onl component. The vector E 2 hs n component given b E 2 cos 3 nd negtive component given b E 2 sin 4 5 E 2. 5 E 2 Hence, we cn epress the vectors s Figure E 1 P 1 φ θ E E m 0.50 m The totl electric field E t P euls the vector sum E 1 E 2, where E 1 is the field due to the positive chrge 1 nd E 2 is the field due to the negtive chrge 2. θ 2
15 722 CHAPTER 23 Electric Fields E j N/C E 2 ( i j) N/C The resultnt field E t P is the superposition of E 1 nd E 2 : E E ( i E 2 j) N/C From this result, we find tht E hs mgnitude of N/C nd mkes n ngle of 66 with the positive is. Eercise Find the electric force eerted on chrge of C locted t P. Answer N in the sme direction s E. EXAMPLE 23.6 Electric Field of Dipole An electric dipole is defined s positive chrge nd negtive chrge seprted b some distnce. For the dipole shown in Figure 23.14, find the electric field E t P due to the chrges, where P is distnce W from the origin. Solution At P, the fields E 1 nd E 2 due to the two chrges re eul in mgnitude becuse P is euidistnt from the chrges. The totl field is E E 1 E 2, where E 1 E 2 k e r 2 The components of E 1 nd E 2 cncel ech other, nd the components dd becuse the re both in the positive direction. Therefore, E is prllel to the is nd hs mgnitude eul to 2E 1 cos. From Figure we see tht cos Therefore, /r /( 2 2 ) 1/2. k e E 2E 1 cos 2k e ( 2 2 ) 2 k e ( 2 2 ) 3/2 Becuse W, we cn neglect 2 nd write E k e ( 2 2 ) 1/2 Thus, we see tht, t distnces fr from dipole but long the perpendiculr bisector of the line joining the two chrges, the mgnitude of the electric field creted b the dipole vries s 1/r 3, wheres the more slowl vring field of point chrge vries s 1/r 2 (see E. 23.4). This is becuse t distnt points, the fields of the two chrges of eul mgnitude nd opposite sign lmost cncel ech other. The 1/r 3 vrition in E for the dipole lso is obtined for distnt point long the is (see Problem 21) nd for n generl distnt point. The electric dipole is good model of mn molecules, such s hdrochloric cid (HCl). As we shll see in lter chpters, neutrl toms nd molecules behve s dipoles when plced in n eternl electric field. Furthermore, mn molecules, such s HCl, re permnent dipoles. The effect of such dipoles on the behvior of mterils subjected to electric fields is discussed in Chpter 26. Figure r θ P θ θ θ E 1 E 2 E The totl electric field E t P due to two chrges of eul mgnitude nd opposite sign (n electric dipole) euls the vector sum E 1 E 2. The field E 1 is due to the positive chrge, nd E 2 is the field due to the negtive chrge ELECTRIC FIELD OF A CONTINUOUS CHARGE DISTRIBUTION Ver often the distnces between chrges in group of chrges re much smller thn the distnce from the group to some point of interest (for emple, point where the electric field is to be clculted). In such situtions, the sstem of
16 23.5 Electric Field of Continuous Chrge Distribution 723 chrges is smered out, or continuous. Tht is, the sstem of closel spced chrges is euivlent to totl chrge tht is continuousl distributed long some line, over some surfce, or throughout some volume. To evlute the electric field creted b continuous chrge distribution, we use the following procedure: First, we divide the chrge distribution into smll elements, ech of which contins smll chrge, s shown in Figure Net, we use Eution 23.4 to clculte the electric field due to one of these elements t point P. Finll, we evlute the totl field t P due to the chrge distribution b summing the contributions of ll the chrge elements (tht is, b ppling the superposition principle). The electric field t P due to one element crring chrge is where r is the distnce from the element to point P nd rˆ is unit vector directed from the chrge element towrd P. The totl electric field t P due to ll elements in the chrge distribution is pproimtel where the inde i refers to the ith element in the distribution. Becuse the chrge distribution is pproimtel continuous, the totl field t P in the limit i : 0 is E k e lim i :0 i E k e r 2 rˆ E k e i i r 2 rˆi i i r 2 rˆi k e i d r 2 rˆ (23.6) where the integrtion is over the entire chrge distribution. This is vector opertion nd must be treted ppropritel. We illustrte this tpe of clcultion with severl emples, in which we ssume the chrge is uniforml distributed on line, on surfce, or throughout volume. When performing such clcultions, it is convenient to use the concept of chrge densit long with the following nottions: A continuous chrge distribution E r P Figure The electric field t P due to continuous chrge distribution is the vector sum of the fields E due to ll the elements of the chrge distribution. Electric field of continuous chrge distribution rˆ If chrge Q is uniforml distributed throughout volume V, the volume chrge densit is defined b Q V Volume chrge densit where hs units of coulombs per cubic meter (C/m 3 ). If chrge Q is uniforml distributed on surfce of re A, the surfce chrge densit (lowercse Greek sigm) is defined b Q A Surfce chrge densit where hs units of coulombs per sure meter (C/m 2 ). If chrge Q is uniforml distributed long line of length, the liner chrge densit is defined b Q Liner chrge densit where hs units of coulombs per meter (C/m).
17 724 CHAPTER 23 Electric Fields If the chrge is nonuniforml distributed over volume, surfce, or line, we hve to epress the chrge densities s dq dv dq where dq is the mount of chrge in smll volume, surfce, or length element. da dq d EXAMPLE 23.7 The Electric Field Due to Chrged Rod A rod of length hs uniform positive chrge per unit length nd totl chrge Q. Clculte the electric field t point P tht is locted long the long is of the rod nd distnce from one end (Fig ). Solution Let us ssume tht the rod is ling long the is, tht d is the length of one smll segment, nd tht d is the chrge on tht segment. Becuse the rod hs chrge per unit length, the chrge d on the smll segment is d The field d E due to this segment t P is in the negtive direction (becuse the source of the field crries positive chrge Q ), nd its mgnitude is d. de k d e 2 k e d 2 Becuse ever other element lso produces field in the negtive direction, the problem of summing their contributions is prticulrl simple in this cse. The totl field t P due to ll segments of the rod, which re t different distnces from P, is given b Eution 23.6, which in this cse becomes 3 E k e d where the limits on the integrl etend from one end of the rod ( ) to the other ( ). The constnts k e nd cn be removed from the integrl to ield 2 E k e d 2 k e 1 k e 1 1 k e Q ( ) where we hve used the fct tht the totl chrge Q. If P is fr from the rod ( W ), then the in the denomintor cn be neglected, nd E k e Q / 2. This is just the form ou would epect for point chrge. Therefore, t lrge vlues of /, the chrge distribution ppers to be point chrge of mgnitude Q. The use of the limiting techniue (/ : ) often is good method for checking theoreticl formul. de P Figure d d = λdλ The electric field t P due to uniforml chrged rod ling long the is. The mgnitude of the field t P due to the segment of chrge d is k e d/ 2. The totl field t P is the vector sum over ll segments of the rod. EXAMPLE 23.8 The Electric Field of Uniform Ring of Chrge A ring of rdius crries uniforml distributed positive totl chrge Q. Clculte the electric field due to the ring t point P ling distnce from its center long the centrl is perpendiculr to the plne of the ring (Fig ). Solution The mgnitude of the electric field t P due to the segment of chrge d is de k e d r 2 This field hs n component de de cos long the is nd component de perpendiculr to the is. As we see in Figure 23.17b, however, the resultnt field t P must lie long the is becuse the perpendiculr components of ll the 3 It is importnt tht ou understnd how to crr out integrtions such s this. First, epress the chrge element d in terms of the other vribles in the integrl (in this emple, there is one vrible,, nd so we mde the chnge d The integrl must be over sclr untities; therefore, ou must epress the electric field in terms of components, if necessr. (In this emple the field hs onl n component, so we do not bother with this detil.) Then, reduce our epression to n integrl over single vrible (or to multiple integrls, ech over single vrible). In emples tht hve sphericl or clindricl smmetr, the single vrible will be rdil coordinte. d).
18 If we consider the disk s set of concentric rings, we cn use our result from Emple 23.8 which gives the field creted b ring of rdius nd sum the contrivrious chrge segments sum to zero. Tht is, the perpendiculr component of the field creted b n chrge element is cnceled b the perpendiculr component creted b n element on the opposite side of the ring. Becuse r ( 2 2 ) 1/2 nd cos /r, we find tht de de cos k e d r 2 r All segments of the ring mke the sme contribution to the field t P becuse the re ll euidistnt from this point. Thus, we cn integrte to obtin the totl field t P : Figure d 23.5 Electric Field of Continuous Chrge Distribution 725 k e ( 2 2 ) 3/2 d () r θ P de de de This result shows tht the field is zero t 0. Does this finding surprise ou? Eercise E k e ( 2 2 ) 3/2 d k e ( 2 2 ) 3/2 d k e ( 2 2 ) 3/2 Q Show tht t gret distnces from the ring ( W ) the electric field long the is shown in Figure pproches tht of point chrge of mgnitude Q. A uniforml chrged ring of rdius. () The field t P on the is due to n element of chrge d. (b) The totl electric field t P is long the is. The perpendiculr component of the field t P due to segment 1 is cnceled b the perpendiculr component due to segment (b) θ de 1 de 2 EXAMPLE 23.9 The Electric Field of Uniforml Chrged Disk A disk of rdius R hs uniform surfce chrge densit. Clculte the electric field t point P tht lies long the centrl perpendiculr is of the disk nd distnce from the center of the disk (Fig ). Solution Figure r R dr A uniforml chrged disk of rdius R. The electric field t n il point P is directed long the centrl is, perpendiculr to the plne of the disk. d P butions of ll rings mking up the disk. B smmetr, the field t n il point must be long the centrl is. The ring of rdius r nd width dr shown in Figure hs surfce re eul to 2 r dr. The chrge d on this ring is eul to the re of the ring multiplied b the surfce chrge densit: d 2 r dr. Using this result in the eution given for E in Emple 23.8 (with replced b r), we hve for the field due to the ring k e de ( 2 r 2 ) 3/2 (2 r dr) To obtin the totl field t P, we integrte this epression over the limits r 0 to r R, noting tht is constnt. This gives E k e R 2r dr 0 ( k e 2 r 2 ) 3/2 R ( 2 r 2 ) 3/2 d(r 2 ) 0 k e ( 2 r 2 ) 1/2 1/2 R 0 2 k e ( 2 R 2 ) 1/2
19 726 CHAPTER 23 Electric Fields This result is vlid for ll vlues of. We cn clculte the field close to the disk long the is b ssuming tht R W ; thus, the epression in prentheses reduces to unit: 0 1/(4 k e ) where is the permittivit of free spce. As we shll find in the net chpter, we obtin the sme result for the field creted b uniforml chrged infinite sheet. E 2 k e 2 0 A Figure Electric field lines penetrting two surfces. The mgnitude of the field is greter on surfce A thn on surfce B. B ELECTRIC FIELD LINES A convenient w of visulizing electric field ptterns is to drw lines tht follow the sme direction s the electric field vector t n point. These lines, clled electric field lines, re relted to the electric field in n region of spce in the following mnner: The electric field vector E is tngent to the electric field line t ech point. The number of lines per unit re through surfce perpendiculr to the lines is proportionl to the mgnitude of the electric field in tht region. Thus, E is gret when the field lines re close together nd smll when the re fr prt. These properties re illustrted in Figure The densit of lines through surfce A is greter thn the densit of lines through surfce B. Therefore, the electric field is more intense on surfce A thn on surfce B. Furthermore, the fct tht the lines t different loctions point in different directions indictes tht the field is nonuniform. Representtive electric field lines for the field due to single positive point chrge re shown in Figure Note tht in this twodimensionl drwing we show onl the field lines tht lie in the plne contining the point chrge. The lines re ctull directed rdill outwrd from the chrge in ll directions; thus, insted of the flt wheel of lines shown, ou should picture n entire sphere of lines. Becuse positive test chrge plced in this field would be repelled b the positive point chrge, the lines re directed rdill w from the positive point () Figure (b) The electric field lines for point chrge. () For positive point chrge, the lines re directed rdill outwrd. (b) For negtive point chrge, the lines re directed rdill inwrd. Note tht the figures show onl those field lines tht lie in the plne contining the chrge. (c) The drk res re smll pieces of thred suspended in oil, which lign with the electric field produced b smll chrged conductor t the center. (c)
20 23.6 Electric Field Lines 727 chrge. The electric field lines representing the field due to single negtive point chrge re directed towrd the chrge (Fig b). In either cse, the lines re long the rdil direction nd etend ll the w to infinit. Note tht the lines become closer together s the pproch the chrge; this indictes tht the strength of the field increses s we move towrd the source chrge. The rules for drwing electric field lines re s follows: The lines must begin on positive chrge nd terminte on negtive chrge. The number of lines drwn leving positive chrge or pproching negtive chrge is proportionl to the mgnitude of the chrge. No two field lines cn cross. Rules for drwing electric field lines Is this visuliztion of the electric field in terms of field lines consistent with Eution 23.4, the epression we obtined for E using Coulomb s lw? To nswer this uestion, consider n imginr sphericl surfce of rdius r concentric with point chrge. From smmetr, we see tht the mgnitude of the electric field is the sme everwhere on the surfce of the sphere. The number of lines N tht emerge from the chrge is eul to the number tht penetrte the sphericl surfce. Hence, the number of lines per unit re on the sphere is N/4 r 2 (where the surfce re of the sphere is 4 r 2 ). Becuse E is proportionl to the number of lines per unit re, we see tht E vries s 1/r 2 ; this finding is consistent with Eution As we hve seen, we use electric field lines to ulittivel describe the electric field. One problem with this model is tht we lws drw finite number of lines from (or to) ech chrge. Thus, it ppers s if the field cts onl in certin directions; this is not true. Insted, the field is continuous tht is, it eists t ever point. Another problem ssocited with this model is the dnger of gining the wrong impression from twodimensionl drwing of field lines being used to describe threedimensionl sitution. Be wre of these shortcomings ever time ou either drw or look t digrm showing electric field lines. We choose the number of field lines strting from n positivel chrged object to be C nd the number of lines ending on n negtivel chrged object to be C, where C is n rbitrr proportionlit constnt. Once C is chosen, the number of lines is fied. For emple, if object 1 hs chrge Q 1 nd object 2 hs chrge Q 2, then the rtio of number of lines is N 2 /N 1 Q 2 /Q 1. The electric field lines for two point chrges of eul mgnitude but opposite signs (n electric dipole) re shown in Figure Becuse the chrges re of eul mgnitude, the number of lines tht begin t the positive chrge must eul the number tht terminte t the negtive chrge. At points ver ner the chrges, the lines re nerl rdil. The high densit of lines between the chrges indictes region of strong electric field. Figure shows the electric field lines in the vicinit of two eul positive point chrges. Agin, the lines re nerl rdil t points close to either chrge, nd the sme number of lines emerge from ech chrge becuse the chrges re eul in mgnitude. At gret distnces from the chrges, the field is pproimtel eul to tht of single point chrge of mgnitude 2. Finll, in Figure we sketch the electric field lines ssocited with positive chrge 2 nd negtive chrge. In this cse, the number of lines leving 2 is twice the number terminting t. Hence, onl hlf of the lines tht leve the positive chrge rech the negtive chrge. The remining hlf terminte Figure () (b) () The electric field lines for two point chrges of eul mgnitude nd opposite sign (n electric dipole). The number of lines leving the positive chrge euls the number terminting t the negtive chrge. (b) The drk lines re smll pieces of thred suspended in oil, which lign with the electric field of dipole.
Answer, Key Homework 8 David McIntyre 1
Answer, Key Homework 8 Dvid McIntyre 1 This printout should hve 17 questions, check tht it is complete. Multiplechoice questions my continue on the net column or pge: find ll choices before mking your
More informationPHY 222 Lab 8 MOTION OF ELECTRONS IN ELECTRIC AND MAGNETIC FIELDS
PHY 222 Lb 8 MOTION OF ELECTRONS IN ELECTRIC AND MAGNETIC FIELDS Nme: Prtners: INTRODUCTION Before coming to lb, plese red this pcket nd do the prelb on pge 13 of this hndout. From previous experiments,
More informationOr more simply put, when adding or subtracting quantities, their uncertainties add.
Propgtion of Uncertint through Mthemticl Opertions Since the untit of interest in n eperiment is rrel otined mesuring tht untit directl, we must understnd how error propgtes when mthemticl opertions re
More informationr 2 F ds W = r 1 qe ds = q
Chpter 4 The Electric Potentil 4.1 The Importnt Stuff 4.1.1 Electricl Potentil Energy A chrge q moving in constnt electric field E experiences force F = qe from tht field. Also, s we know from our study
More information5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one.
5.2. LINE INTEGRALS 265 5.2 Line Integrls 5.2.1 Introduction Let us quickly review the kind of integrls we hve studied so fr before we introduce new one. 1. Definite integrl. Given continuous relvlued
More informationAREA OF A SURFACE OF REVOLUTION
AREA OF A SURFACE OF REVOLUTION h cut r πr h A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundr of solid of revolution of the tpe discussed in Sections 7.
More informationCypress Creek High School IB Physics SL/AP Physics B 2012 2013 MP2 Test 1 Newton s Laws. Name: SOLUTIONS Date: Period:
Nme: SOLUTIONS Dte: Period: Directions: Solve ny 5 problems. You my ttempt dditionl problems for extr credit. 1. Two blocks re sliding to the right cross horizontl surfce, s the drwing shows. In Cse A
More informationNewton s Three Laws. d dt F = If the mass is constant, this relationship becomes the familiar form of Newton s Second Law: dv dt
Newton s Three Lws For couple centuries before Einstein, Newton s Lws were the bsic principles of Physics. These lws re still vlid nd they re the bsis for much engineering nlysis tody. Forml sttements
More informationTheory of Forces. Forces and Motion
his eek extbook  Red Chpter 4, 5 Competent roblem Solver  Chpter 4 relb Computer Quiz ht s on the next Quiz? Check out smple quiz on web by hurs. ht you missed on first quiz Kinemtics  Everything
More informationExperiment 6: Friction
Experiment 6: Friction In previous lbs we studied Newton s lws in n idel setting, tht is, one where friction nd ir resistnce were ignored. However, from our everydy experience with motion, we know tht
More informationChapter G  Problems
Chpter G  Problems Blinn College  Physics 2426  Terry Honn Problem G.1 A plne flies horizonlly t speed of 280 mês in position where the erth's mgnetic field hs mgnitude 6.0µ105 T nd is directed t n
More informationSection 74 Translation of Axes
62 7 ADDITIONAL TOPICS IN ANALYTIC GEOMETRY Section 74 Trnsltion of Aes Trnsltion of Aes Stndrd Equtions of Trnslted Conics Grphing Equtions of the Form A 2 C 2 D E F 0 Finding Equtions of Conics In the
More informationApplications to Physics and Engineering
Section 7.5 Applictions to Physics nd Engineering Applictions to Physics nd Engineering Work The term work is used in everydy lnguge to men the totl mount of effort required to perform tsk. In physics
More informationOperations with Polynomials
38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: Write polynomils in stndrd form nd identify the leding coefficients nd degrees of polynomils Add nd subtrct polynomils Multiply
More informationPhysics 2102 Lecture 2. Physics 2102
Physics 10 Jonthn Dowling Physics 10 Lecture Electric Fields ChrlesAugustin de Coulomb (17361806) Jnury 17, 07 Version: 1/17/07 Wht re we going to lern? A rod mp Electric chrge Electric force on other
More informationCONIC SECTIONS. Chapter 11
CONIC SECTIONS Chpter 11 11.1 Overview 11.1.1 Sections of cone Let l e fied verticl line nd m e nother line intersecting it t fied point V nd inclined to it t n ngle α (Fig. 11.1). Fig. 11.1 Suppose we
More informationPROBLEMS 13  APPLICATIONS OF DERIVATIVES Page 1
PROBLEMS  APPLICATIONS OF DERIVATIVES Pge ( ) Wter seeps out of conicl filter t the constnt rte of 5 cc / sec. When the height of wter level in the cone is 5 cm, find the rte t which the height decreses.
More informationHelicopter Theme and Variations
Helicopter Theme nd Vritions Or, Some Experimentl Designs Employing Pper Helicopters Some possible explntory vribles re: Who drops the helicopter The length of the rotor bldes The height from which the
More informationGraphs on Logarithmic and Semilogarithmic Paper
0CH_PHClter_TMSETE_ 3//00 :3 PM Pge Grphs on Logrithmic nd Semilogrithmic Pper OBJECTIVES When ou hve completed this chpter, ou should be ble to: Mke grphs on logrithmic nd semilogrithmic pper. Grph empiricl
More informationAAPT UNITED STATES PHYSICS TEAM AIP 2010
2010 F = m Exm 1 AAPT UNITED STATES PHYSICS TEAM AIP 2010 Enti non multiplicnd sunt preter necessittem 2010 F = m Contest 25 QUESTIONS  75 MINUTES INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD
More informationUnit 6: Exponents and Radicals
Eponents nd Rdicls : The Rel Numer Sstem Unit : Eponents nd Rdicls Pure Mth 0 Notes Nturl Numers (N):  counting numers. {,,,,, } Whole Numers (W):  counting numers with 0. {0,,,,,, } Integers (I): 
More informationPhysics 43 Homework Set 9 Chapter 40 Key
Physics 43 Homework Set 9 Chpter 4 Key. The wve function for n electron tht is confined to x nm is. Find the normliztion constnt. b. Wht is the probbility of finding the electron in. nmwide region t x
More informationDouble Integrals over General Regions
Double Integrls over Generl egions. Let be the region in the plne bounded b the lines, x, nd x. Evlute the double integrl x dx d. Solution. We cn either slice the region verticll or horizontll. ( x x Slicing
More informationMathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100
hsn.uk.net Higher Mthemtics UNIT 3 OUTCOME 1 Vectors Contents Vectors 18 1 Vectors nd Sclrs 18 Components 18 3 Mgnitude 130 4 Equl Vectors 131 5 Addition nd Subtrction of Vectors 13 6 Multipliction by
More informationt 3 t 4 Part A: Multiple Choice Canadian Association of Physicists 1999 Prize Exam
Cndin Assocition of Physicists 1999 Prize Exm This is three hour exm. Ntionl rnking nd prizes will be bsed on student s performnce on both sections A nd B of the exm. However, performnce on the multiple
More informationA.7.1 Trigonometric interpretation of dot product... 324. A.7.2 Geometric interpretation of dot product... 324
A P P E N D I X A Vectors CONTENTS A.1 Scling vector................................................ 321 A.2 Unit or Direction vectors...................................... 321 A.3 Vector ddition.................................................
More informationLECTURE #05. Learning Objective. To describe the geometry in and around a unit cell in terms of directions and planes.
LECTURE #05 Chpter 3: Lttice Positions, Directions nd Plnes Lerning Objective To describe the geometr in nd round unit cell in terms of directions nd plnes. 1 Relevnt Reding for this Lecture... Pges 6483.
More informationIntroduction to Integration Part 2: The Definite Integral
Mthemtics Lerning Centre Introduction to Integrtion Prt : The Definite Integrl Mr Brnes c 999 Universit of Sdne Contents Introduction. Objectives...... Finding Ares 3 Ares Under Curves 4 3. Wht is the
More informationSection A4 Rational Expressions: Basic Operations
A Appendi A A BASIC ALGEBRA REVIEW 7. Construction. A rectngulr opentopped bo is to be constructed out of 9 by 6inch sheets of thin crdbord by cutting inch squres out of ech corner nd bending the
More informationSection 2.3. Motion Along a Curve. The Calculus of Functions of Several Variables
The Clculus of Functions of Severl Vribles Section 2.3 Motion Along Curve Velocity ccelertion Consider prticle moving in spce so tht its position t time t is given by x(t. We think of x(t s moving long
More informationThe Velocity Factor of an Insulated TwoWire Transmission Line
The Velocity Fctor of n Insulted TwoWire Trnsmission Line Problem Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ 08544 Mrch 7, 008 Estimte the velocity fctor F = v/c nd the
More informationVectors. The magnitude of a vector is its length, which can be determined by Pythagoras Theorem. The magnitude of a is written as a.
Vectors mesurement which onl descries the mgnitude (i.e. size) of the oject is clled sclr quntit, e.g. Glsgow is 11 miles from irdrie. vector is quntit with mgnitude nd direction, e.g. Glsgow is 11 miles
More informationAnswer, Key Homework 10 David McIntyre 1
Answer, Key Homework 10 Dvid McIntyre 1 This printout should hve 22 questions, check tht it is complete. Multiplechoice questions my continue on the next column or pge: find ll choices efore mking your
More informationUse Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions.
Lerning Objectives Loci nd Conics Lesson 3: The Ellipse Level: Preclculus Time required: 120 minutes In this lesson, students will generlize their knowledge of the circle to the ellipse. The prmetric nd
More informationMechanics Cycle 1 Chapter 5. Chapter 5
Chpter 5 Contct orces: ree Body Digrms nd Idel Ropes Pushes nd Pulls in 1D, nd Newton s Second Lw Neglecting riction ree Body Digrms Tension Along Idel Ropes (i.e., Mssless Ropes) Newton s Third Lw Bodies
More informationNet Change and Displacement
mth 11, pplictions motion: velocity nd net chnge 1 Net Chnge nd Displcement We hve seen tht the definite integrl f (x) dx mesures the net re under the curve y f (x) on the intervl [, b] Any prt of the
More informationCOMPONENTS: COMBINED LOADING
LECTURE COMPONENTS: COMBINED LOADING Third Edition A. J. Clrk School of Engineering Deprtment of Civil nd Environmentl Engineering 24 Chpter 8.4 by Dr. Ibrhim A. Asskkf SPRING 2003 ENES 220 Mechnics of
More information9.3. The Scalar Product. Introduction. Prerequisites. Learning Outcomes
The Sclr Product 9.3 Introduction There re two kinds of multipliction involving vectors. The first is known s the sclr product or dot product. This is soclled becuse when the sclr product of two vectors
More informationSPECIAL PRODUCTS AND FACTORIZATION
MODULE  Specil Products nd Fctoriztion 4 SPECIAL PRODUCTS AND FACTORIZATION In n erlier lesson you hve lernt multipliction of lgebric epressions, prticulrly polynomils. In the study of lgebr, we come
More informationPolynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )
Polynomil Functions Polynomil functions in one vrible cn be written in expnded form s n n 1 n 2 2 f x = x + x + x + + x + x+ n n 1 n 2 2 1 0 Exmples of polynomils in expnded form re nd 3 8 7 4 = 5 4 +
More informationWeek 11  Inductance
Week  Inductnce November 6, 202 Exercise.: Discussion Questions ) A trnsformer consists bsiclly of two coils in close proximity but not in electricl contct. A current in one coil mgneticlly induces n
More informationModule 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur
Module Anlysis of Stticlly Indeterminte Structures by the Mtrix Force Method Version CE IIT, Khrgpur esson 9 The Force Method of Anlysis: Bems (Continued) Version CE IIT, Khrgpur Instructionl Objectives
More informationIntegration. 148 Chapter 7 Integration
48 Chpter 7 Integrtion 7 Integrtion t ech, by supposing tht during ech tenth of second the object is going t constnt speed Since the object initilly hs speed, we gin suppose it mintins this speed, but
More informationP.3 Polynomials and Factoring. P.3 an 1. Polynomial STUDY TIP. Example 1 Writing Polynomials in Standard Form. What you should learn
33337_0P03.qp 2/27/06 24 9:3 AM Chpter P Pge 24 Prerequisites P.3 Polynomils nd Fctoring Wht you should lern Polynomils An lgeric epression is collection of vriles nd rel numers. The most common type of
More information6.2 Volumes of Revolution: The Disk Method
mth ppliction: volumes of revolution, prt ii Volumes of Revolution: The Disk Method One of the simplest pplictions of integrtion (Theorem ) nd the ccumultion process is to determine soclled volumes of
More informationSection 1: Crystal Structure
Phsics 927 Section 1: Crstl Structure A solid is sid to be crstl if toms re rrnged in such w tht their positions re ectl periodic. This concept is illustrted in Fig.1 using twodimensionl (2D) structure.
More information1. 1 m/s m/s m/s. 5. None of these m/s m/s m/s m/s correct m/s
Crete ssignment, 99552, Homework 5, Sep 15 t 10:11 m 1 This printout should he 30 questions. Multiplechoice questions my continue on the next column or pge find ll choices before nswering. The due time
More informationScalar and Vector Quantities. A scalar is a quantity having only magnitude (and possibly phase). LECTURE 2a: VECTOR ANALYSIS Vector Algebra
Sclr nd Vector Quntities : VECTO NLYSIS Vector lgebr sclr is quntit hving onl mgnitude (nd possibl phse). Emples: voltge, current, chrge, energ, temperture vector is quntit hving direction in ddition to
More informationVectors 2. 1. Recap of vectors
Vectors 2. Recp of vectors Vectors re directed line segments  they cn be represented in component form or by direction nd mgnitude. We cn use trigonometry nd Pythgors theorem to switch between the forms
More informationVersion 001 Summer Review #03 tubman (IBII20142015) 1
Version 001 Summer Reiew #03 tubmn (IBII20142015) 1 This printout should he 35 questions. Multiplechoice questions my continue on the next column or pge find ll choices before nswering. Concept 20 P03
More informationChapter 6 Solving equations
Chpter 6 Solving equtions Defining n eqution 6.1 Up to now we hve looked minly t epressions. An epression is n incomplete sttement nd hs no equl sign. Now we wnt to look t equtions. An eqution hs n = sign
More informationA5682: Introduction to Cosmology Course Notes. 4. Cosmic Dynamics: The Friedmann Equation. = GM s R 2 s(t).
4. Cosmic Dynmics: The Friedmnn Eqution Reding: Chpter 4 Newtonin Derivtion of the Friedmnn Eqution Consider n isolted sphere of rdius R s nd mss M s, in uniform, isotropic expnsion (Hubble flow). The
More informationSection 54 Trigonometric Functions
5 Trigonometric Functions Section 5 Trigonometric Functions Definition of the Trigonometric Functions Clcultor Evlution of Trigonometric Functions Definition of the Trigonometric Functions Alternte Form
More informationReview guide for the final exam in Math 233
Review guide for the finl exm in Mth 33 1 Bsic mteril. This review includes the reminder of the mteril for mth 33. The finl exm will be cumultive exm with mny of the problems coming from the mteril covered
More informationPhys 207. Announcements. Hwk3 is posted on course website Quizzes & answers will be posted on course website Formula sheets.
Phys 07 Announcements Hwk3 is posted on course website Quizzes & nswers will be posted on course website ormul sheets Newton s 3 lws Tody s Agend How nd why do objects move? Dynmics 1 Dynmics Isc Newton
More informationBasic Analysis of Autarky and Free Trade Models
Bsic Anlysis of Autrky nd Free Trde Models AUTARKY Autrky condition in prticulr commodity mrket refers to sitution in which country does not engge in ny trde in tht commodity with other countries. Consequently
More informationFactoring Polynomials
Fctoring Polynomils Some definitions (not necessrily ll for secondry school mthemtics): A polynomil is the sum of one or more terms, in which ech term consists of product of constnt nd one or more vribles
More informationTheories of light and Interference S BALASUBRAMANYA SGL IN PHYSICS SARVODAYA PU COLLEGE, TUMKUR Important paints
Theories of light nd nterference S BALASUBRAMANYA SGL N PHYSCS SARVODAYA PU COLLEGE, TUMKUR mportnt pints Theories of Light Newton s Corpusculr theory (1675) Christin Huygen s Wve theory (1678) Mxwell
More informationDiffraction and Interference of Light
rev 12/2016 Diffrction nd Interference of Light Equipment Qty Items Prt Number 1 Light Sensor CI6504 1 Rotry Motion Sensor CI6538 1 Single Slit Set OS8523 1 Multiple Slit Set OS8523 1 Liner Trnsltor
More informationSCHOOL OF ENGINEERING & BUILT ENVIRONMENT. Mathematics. Basic Algebra
SCHOOL OF ENGINEERING & BUILT ENVIRONMENT Mthemtics Bsic Alger. Opertions nd Epressions. Common Mistkes. Division of Algeric Epressions. Eponentil Functions nd Logrithms. Opertions nd their Inverses. Mnipulting
More information4.0 5Minute Review: Rational Functions
mth 130 dy 4: working with limits 1 40 5Minute Review: Rtionl Functions DEFINITION A rtionl function 1 is function of the form y = r(x) = p(x) q(x), 1 Here the term rtionl mens rtio s in the rtio of two
More informationEcon 4721 Money and Banking Problem Set 2 Answer Key
Econ 472 Money nd Bnking Problem Set 2 Answer Key Problem (35 points) Consider n overlpping genertions model in which consumers live for two periods. The number of people born in ech genertion grows in
More informationAlgebra Review. How well do you remember your algebra?
Algebr Review How well do you remember your lgebr? 1 The Order of Opertions Wht do we men when we write + 4? If we multiply we get 6 nd dding 4 gives 10. But, if we dd + 4 = 7 first, then multiply by then
More informationSo there are two points of intersection, one being x = 0, y = 0 2 = 0 and the other being x = 2, y = 2 2 = 4. y = x 2 (2,4)
Ares The motivtion for our definition of integrl ws the problem of finding the re between some curve nd the is for running between two specified vlues. We pproimted the region b union of thin rectngles
More informationPure C4. Revision Notes
Pure C4 Revision Notes Mrch 0 Contents Core 4 Alger Prtil frctions Coordinte Geometry 5 Prmetric equtions 5 Conversion from prmetric to Crtesin form 6 Are under curve given prmetriclly 7 Sequences nd
More informationAn OffCenter Coaxial Cable
1 Problem An OffCenter Coxil Cble Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ 08544 Nov. 21, 1999 A coxil trnsmission line hs inner conductor of rdius nd outer conductor
More information1. 0 m/s m/s m/s m/s
Version PREVIEW Kine Grphs PRACTICE burke (1111) 1 This printout should he 30 questions. Multiplechoice questions m continue on the next column or pge find ll choices before nswering. Distnce Time Grph
More informationaddition, there are double entries for the symbols used to signify different parameters. These parameters are explained in this appendix.
APPENDIX A: The ellipse August 15, 1997 Becuse of its importnce in both pproximting the erth s shpe nd describing stellite orbits, n informl discussion of the ellipse is presented in this ppendix. The
More informationEQUATIONS OF LINES AND PLANES
EQUATIONS OF LINES AND PLANES MATH 195, SECTION 59 (VIPUL NAIK) Corresponding mteril in the ook: Section 12.5. Wht students should definitely get: Prmetric eqution of line given in pointdirection nd twopoint
More informationEinstein. Mechanics. In Grade 10 we investigated kinematics, or movement described in terms of velocity, acceleration, displacement, and so on.
Cmbridge University Press 9780521683593  Study nd Mster Physicl Sciences Grde 11 Lerner s Book Krin Kelder More informtion MODULE 1 Einstein Mechnics motion force Glileo Newton decelerte moment of
More informationHomework #4: Answers. 1. Draw the array of world outputs that free trade allows by making use of each country s transformation schedule.
Text questions, Chpter 5, problems 15: Homework #4: Answers 1. Drw the rry of world outputs tht free trde llows by mking use of ech country s trnsformtion schedule.. Drw it. This digrm is constructed
More informationSolutions to Section 1
Solutions to Section Exercise. Show tht nd. This follows from the fct tht mx{, } nd mx{, } Exercise. Show tht = { if 0 if < 0 Tht is, the bsolute vlue function is piecewise defined function. Grph this
More informationTreatment Spring Late Summer Fall 0.10 5.56 3.85 0.61 6.97 3.01 1.91 3.01 2.13 2.99 5.33 2.50 1.06 3.53 6.10 Mean = 1.33 Mean = 4.88 Mean = 3.
The nlysis of vrince (ANOVA) Although the ttest is one of the most commonly used sttisticl hypothesis tests, it hs limittions. The mjor limittion is tht the ttest cn be used to compre the mens of only
More informationCurve Sketching. 96 Chapter 5 Curve Sketching
96 Chpter 5 Curve Sketching 5 Curve Sketching A B A B A Figure 51 Some locl mximum points (A) nd minimum points (B) If (x, f(x)) is point where f(x) reches locl mximum or minimum, nd if the derivtive of
More informationAnswer, Key Homework 4 David McIntyre Mar 25,
Answer, Key Homework 4 Dvid McIntyre 45123 Mr 25, 2004 1 his printout should hve 18 questions. Multiplechoice questions my continue on the next column or pe find ll choices before mkin your selection.
More informationSolar and Lunar Tides
Solr nd Lunr Tides evised, Corrected, Simplified Copyriht Steve Olh, Irvine, CA, 009 solh@cox.net Keywords: Tide, Lunr Tide, Solr Tide, Tidl Force Abstrct Solr nd lunr tides were prt of life since there
More informationExample A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding
1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. 1 Exmple A rectngulr box without lid is to be mde
More informationVolumes of solids of revolution
Volumes of solids of revolution We sometimes need to clculte the volume of solid which cn be obtined by rotting curve bout the xxis. There is strightforwrd technique which enbles this to be done, using
More informationLecture 3 Gaussian Probability Distribution
Lecture 3 Gussin Probbility Distribution Introduction l Gussin probbility distribution is perhps the most used distribution in ll of science. u lso clled bell shped curve or norml distribution l Unlike
More informationto the area of the region bounded by the graph of the function y = f(x), the xaxis y = 0 and two vertical lines x = a and x = b.
5.9 Are in rectngulr coordintes If f() on the intervl [; ], then the definite integrl f()d equls to the re of the region ounded the grph of the function = f(), the is = nd two verticl lines = nd =. =
More informationThe Laws of Motion. chapter
chpter The Lws of Motion 5 5.1 The Concept of Force 5.2 Newton s First Lw nd Inertil Frmes 5.3 Mss 5.4 Newton s econd Lw 5.5 The Grvittionl Force nd Weight 5.6 Newton s Third Lw 5.7 Anlysis Models Using
More information14.2. The Mean Value and the RootMeanSquare Value. Introduction. Prerequisites. Learning Outcomes
he Men Vlue nd the RootMenSqure Vlue 4. Introduction Currents nd voltges often vry with time nd engineers my wish to know the men vlue of such current or voltge over some prticulr time intervl. he men
More informationWarmup for Differential Calculus
Summer Assignment Wrmup for Differentil Clculus Who should complete this pcket? Students who hve completed Functions or Honors Functions nd will be tking Differentil Clculus in the fll of 015. Due Dte:
More informationSect 8.3 Triangles and Hexagons
13 Objective 1: Sect 8.3 Tringles nd Hexgons Understnding nd Clssifying Different Types of Polygons. A Polygon is closed twodimensionl geometric figure consisting of t lest three line segments for its
More informationReview Problems for the Final of Math 121, Fall 2014
Review Problems for the Finl of Mth, Fll The following is collection of vrious types of smple problems covering sections.,.5, nd.7 6.6 of the text which constitute only prt of the common Mth Finl. Since
More informationPROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY
MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY Contents 1. ACT Compss Prctice Tests 1 2. Common Mistkes 2 3. Distributive
More informationTITLE THE PRINCIPLES OF COINTAP METHOD OF NONDESTRUCTIVE TESTING
TITLE THE PRINCIPLES OF COINTAP METHOD OF NONDESTRUCTIVE TESTING Sung Joon Kim*, DongChul Che Kore Aerospce Reserch Institute, 45 EoeunDong, YouseongGu, Dejeon, 35333, Kore Phone : 824286231 FAX
More informationAP QUIZ #2 GRAPHING MOTION 1) POSITION TIME GRAPHS DISPLACEMENT Each graph below shows the position of an object as a function of time.
AP QUIZ # GRAPHING MOTION ) POSITION TIME GRAPHS DISPLAEMENT Ech grph below shows the position of n object s function of time. A, B,, D, Rnk these grphs on the gretest mgnitude displcement during the time
More informationExponential and Logarithmic Functions
Nme Chpter Eponentil nd Logrithmic Functions Section. Eponentil Functions nd Their Grphs Objective: In this lesson ou lerned how to recognize, evlute, nd grph eponentil functions. Importnt Vocbulr Define
More information9 CONTINUOUS DISTRIBUTIONS
9 CONTINUOUS DISTIBUTIONS A rndom vrible whose vlue my fll nywhere in rnge of vlues is continuous rndom vrible nd will be ssocited with some continuous distribution. Continuous distributions re to discrete
More information1+(dy/dx) 2 dx. We get dy dx = 3x1/2 = 3 x, = 9x. Hence 1 +
Mth.9 Em Solutions NAME: #.) / #.) / #.) /5 #.) / #5.) / #6.) /5 #7.) / Totl: / Instructions: There re 5 pges nd totl of points on the em. You must show ll necessr work to get credit. You m not use our
More informationPythagoras theorem and trigonometry (2)
HPTR 10 Pythgors theorem nd trigonometry (2) 31 HPTR Liner equtions In hpter 19, Pythgors theorem nd trigonometry were used to find the lengths of sides nd the sizes of ngles in rightngled tringles. These
More informationBayesian Updating with Continuous Priors Class 13, 18.05, Spring 2014 Jeremy Orloff and Jonathan Bloom
Byesin Updting with Continuous Priors Clss 3, 8.05, Spring 04 Jeremy Orloff nd Jonthn Bloom Lerning Gols. Understnd prmeterized fmily of distriutions s representing continuous rnge of hypotheses for the
More informationAe2 Mathematics : Fourier Series
Ae Mthemtics : Fourier Series J. D. Gibbon (Professor J. D Gibbon, Dept of Mthemtics j.d.gibbon@ic.c.uk http://www.imperil.c.uk/ jdg These notes re not identicl wordforword with my lectures which will
More informationProjectile Motion CHAPTER 1
CHAPTER 1 PHYSICS ESSENTIALS STAGE 2 Projectile Motion Subject Outline In the bsence of ir resistnce nd moing under the ction of constnt grittionl force, projectile hs constnt ccelertion in the direction
More informationPicture Match Words Fusion Density Isotope Neutron Atomic Number Structure Components Function Atomic Mass Orbit
Picture Mtch Words Fusion Density Isotope Neutron Atomic Number Structure Components Function Atomic Mss Orbit Mterils copyrighted by the University of Louisville. Eductors re free to use these mterils
More informationElectric Circuits. Simple Electric Cell. Electric Current
Electric Circuits Count Alessndro olt (74587) Georg Simon Ohm (787854) Chrles Augustin de Coulomb (736 806) André Mrie AMPÈRE (775836) Crbon Electrode () Simple Electric Cell wire Zn Zn Zn Zn Sulfuric
More information1. In the Bohr model, compare the magnitudes of the electron s kinetic and potential energies in orbit. What does this imply?
Assignment 3: Bohr s model nd lser fundmentls 1. In the Bohr model, compre the mgnitudes of the electron s kinetic nd potentil energies in orit. Wht does this imply? When n electron moves in n orit, the
More informationUnderstanding 22. 23. The frictional force acting to the left is missing. It is equal in magnitude to the applied force acting to the right.
Chpter 3 Review, pges 154 159 Knowledge 1. (c) 2. () 3. (d) 4. (d) 5. (d) 6. (c) 7. (b) 8. (c) 9. Flse. One newton is equl to 1 kg /s 2. 10. Flse. A norl force is perpendiculr force cting on n object tht
More information