Chapter 15. ELECTRIC POTENTIALS and ENERGY CONSIDERATIONS

Transcription

1 Ch. 15--Elect. Pt. and Enegy Cns. Chapte 15 ELECTRIC POTENTIALS and ENERGY CONSIDERATIONS A.) Enegy Cnsideatins and the Abslute Electical Ptential: 1.) Cnside the fllwing scenai: A single, fixed, pint chage Q sits in space with nthing aund it. A secnd psitive chage is bught in a distance units fm the fist chage. Once in psitin, is eleased and, being cmpletely fee, acceleates. a.) It shuld be bvius that 's acceleatin is due t its inteactin with the electic field geneated by the fixed chage. Althugh thee is nthing wng with this intepetatin, thee is anthe way we culd lk at the situatin. b.) Cnsideing the idea f enegy: On the assumptin that static electic fields (i.e., electic fields that d nt change with time) ae assciated with cnsevative fces (in fact, they ae), we culd claim that 's acceleatin was the cnseuence f its cnveting electical POTENTIAL ENERGY affded it by its pesence in Q's electic field t KINETIC ENERGY. c.) As we have seen in pevius chaptes, the ptential enegy functin f a paticula fce field is a mathematical cntivance designed t allw us t keep tack f the amunt f kinetic enegy a bdy can ptentially pick up if allwed t feefall in the fce field. i.) Put anthe way, if yu knw hw much ptential enegy a bdy has when at tw diffeent pints in a fce field, yu can detemine hw much wk the field des as the bdy mves fm the ne pint t the the. Knwing the net amunt f wk dne, yu can use the wk/enegy theem t detemine the bdy's change f kinetic enegy. d.) In the case f gavitatinal fce fields, the amunt f gavitatinal ptential enegy a bdy has is dependent upn the bject's 73

2 mass (U g mgh 1 ). In the case f electic fces, the amunt f electical ptential enegy an bject has depends upn the bject's chage. The lage the chage, the me electical ptential enegy the bdy will have. e.) The electic field is a vect whse magnitude tells us the fce pe unit chage available at a given pint in the field. Wuld it nt be useful t define a simila uantity elated t the cncept f electical ptential enegy? f.) Just such a uantity has, indeed, been defined. It is called the abslute electical ptential f a pint-f-inteest due t the pesence f an electic-field. It is a SCALAR uantity that tells yu hw much ptential enegy pe unit chage thee is available at a paticula pint in an electic field..) Mathematically, the abslute electical ptential V a at a Pint A in space is defined as: V A (U A /), whee U A is the amunt f ptential enegy a chage will have if placed at Pint A. a.) The units f the abslute electical ptential ae "enegy-pechage-unit", jules pe culmb. This unit is given a special name: it is called a VOLT. b.) The abslute electical ptential at an abitay Pint A is smetimes efeed t as "the abslute vltage at Pint A." c.) Example: A +4 culmb chage is fund t have 8000 jules f ptential enegy when placed at a paticula pint in an electic field. What is the abslute vltage (the abslute electical ptential) at that pint? i.) Fm the definitin: V A (U/) A (8000 j)/(4 C) 000 jules/c 000 vlts. 74

3 Ch. 15--Elect. Pt. and Enegy Cns. Nte: If yu knw V A and want t knw the amunt f ptential enegy a chage has when at A (i.e., U A ), the manipulated expessin U A V A will d the jb. Example: A x10-1 culmb chage is placed in an electic field at a pint whse electical ptential is 500 vlts. Hw much ptential enegy will the chage have? Slutin: U A V A (x10-1 C)(500 vlts) 5x10-9 jules. 3.) Cnside a psitive chage placed and eleased at Pint A in an electic field (the field lines ae shwn in Figue 15.1). The chage will acceleate, picking up kinetic enegy as it des, and sne late finds itself at Pint B. The fllwing bsevatins can be made abut Pints A and B. vltage at Pint A geate than vltage at Pint B Pint A Pint B a.) Just as massive bjects in a gavitatinal field natually feefall fm highe t lwe ptential enegy, psitive chages acceleate in electic fields fm highe psitive chage will acceleate feely fm A t B; negative chage will acceleate feely fm B t A. E. fld. FIGURE 15.1 vltage pints t lwe vltage pints. In the wds, V A must be geate than V B. Nte 1: With mass, ne neve has t wy abut bjects feefalling upwad fm a lwe ptential enegy level t a highe ptential enegy level. Afte all, thee is nly ne kind f mass and the eath's gavity always pulls it dwnwad. Unftunately, thee ae tw kinds f chages--psitive and negative nes--each f which eacts exactly the ppsite t the the when put in an electic field. Ou they has been set up n the assumptin that psitive chages acceleate in the diectin f electic field vects, which means psitive chages acceleate fm highe t lwe vltages. That means negative chages will d exactly the ppsite. (T see this, think abut the diectin an electn will acceleate if put at the lwe vltage, Pint B). Nte : It des nt matte whethe the electic field lines ae getting clse tgethe futhe apat (that is, whethe the electic field intensity is getting lage smalle)--the abslute vltage becmes less as yu pceed in the diectin f electic field lines. 75

4 B.) Wk and Vltage Diffeences: 1.) The amunt f wk W dne by a cnsevative fce field n a bdy that mves fm Pint A t Pint B in the field is intimately elated t the bdy's change f ptential enegy. That is: W - U..) In an electic field, the amunt f wk pe unit chage dne n a chaged bdy as it mves fm Pint A t Pint B in the field euals: W/ - U/ -(U B - U A )/ -[(U B /) - (U A /)] -(V B - V A ) W/ - V. a.) V is fmally called the electical ptential diffeence between Pints A and B. In eveyday usage, it is smetimes efeed t as "the vltage diffeence between Pints A and B." Nte: Althugh in sme texts thee is vey little ntatinal delineatin between the tw, thee is a BIG diffeence between a field's vltage at a pint and a vltage diffeence between tw pints. In this text, an abslute electical ptential will eithe be subscipted t designate the pint t which it is attached made evident thugh the cntext f the situatin. The will ccasinally be mitted when efeing t a vltage diffeence, mst ntably when dealing with a pwe supply like a battey in an electic cicuit. When in dubt, lk t the cntext f the pblem. If the vltage value is attached t a paticula pint, it is an abslute electical ptential (i.e., an abslute vltage). If the vltage value is elated t a vltage change between tw pints, yu ae lking at an electical ptential diffeence (i.e., a vltage diffeence). Example: A 6 vlt battey has a 6 vlt electical ptential diffeence between the + and - teminals. As such, the numbe tells yu hw much wk pe unit chage the battey will d n chage-caies as they tavel thugh a wie fm ne teminal t the the. 76

5 Ch. 15--Elect. Pt. and Enegy Cns. C.) Electical Ptentials and Cnstant Electic Fields: 1.) F cnstant electic fields, we can calculate the wk pe unit chage as: Electical Ptential and a Cnstant Electic Field W/ (F e. d)/, whee F e is the fce n the chage due t the electic field E, and d is a displacement vect defining the sepaatin between the elative ientatin f the Pints A and B (see Figue 15.). Pint A V A d V B Pint B E E a.) Reaanging and manipulating yields: FIGURE 15. W/ (F e /). d E. d. b.) Cmbining this esult with W/ - V we get: E. d - V. c.) As can be seen, this elatinship links the electical ptential diffeence between tw pints in a cnstant electic field t: i.) The electic field vect, and; ii.) A displacement vect between the tw pints-in-uestin. iii.) This is a vey pweful, useful expessin that will cme in handy. UNDERSTAND IT WELL!.) Cnseuences f E. d - V: Cnside the electic field lines shwn in Figue 15.3 n the next page (igne gavity). Given that V A 1 vlts, V B 3 vlts, and the distance between Pints A and B, Pints A and C, and Pints A and D is metes: 77

6 a.) Detemine the magnitude f the electic field: i.) Mving fm A t B alng the E field lines, we can wite: E. d - V E d AB cs ξ -(V B - V A ), whee ξ is the angle between the line f E and the line f d AB. cnstant E. fld. with: d d d metes, A/B A/C A/D V 1 vlts, and A V 3 vlts. B Pint A E. fld. Pint B ii.) We knw d AB, θ, and the vltages at Pints A and B. All we d nt knw is the magnitude f the electic field. Slving f that uantity yields: E -(V B - V A )/(d AB cs θ) -(3 v - 1 v)/[( m)(cs 0 )] 4.5 vlts/mete Pint C Pint D Nte that the dts in this field ae nt epesenting chages-- they ae pints! FIGURE 15.3 Nte: A vlt pe mete is eual t a newtn pe culmb. b.) Detemine vltage V C : diectin f E i.) Once again, we knw: E. d - V E d AC cs θ - (V C - V A ). ii.) Lking at Figue 15.4, θ 10 (θ is suppsed t be the angle between the line f E and the line f d AC, whee d AC is a vect diected diectin f d Pint A Pint C 0 30 angle between E and d is 0 10 FIGURE

7 Ch. 15--Elect. Pt. and Enegy Cns. fm the stating Pint A t the finishing Pint C). Plugging in the knwn values and slving, we get: V C V A - E d AC cs θ (1 v) - (4.5 nt/c)( m) cs vlts. Nte: The fact that the vltage at Pint C is lage than the vltage at Pint A makes sense: the vltage shuld be lage "upsteam," s t speak, in the electic field. c.) Detemine V D : i.) Mving fm A t D, we have: E. d - V E d AD cs θ - (V D - V A ). ii.) Nting that the angle between the line f E and the line f d AD is θ 90 (althugh this is nt explicitly stated in Figue 15.3, this shuld nevetheless be bvius fm lking at the sketch), we slve f V D : V D V A - E d AD cs θ (1 vlts) - (4.5 vlts/m)( m) cs 90 1 vlts. Imptant Nte 1: The line between Pints A and D is pependicula t the line f the electic field. That means the dt pduct E. d is ze (the dt pduct f vects pependicula t ne anthe will always be ze), which in tun means V is ze. If the change f the electical ptential between tw pints is ze, the vltage at each pint must be the same. That is exactly what we have hee: V A V D 1 vlts. Imptant Nte : Thee is a gup f pints each f which has a vltage f 1 vlts. Cnnecting thse pints defines an euiptential line. Euiptential lines ae always PERPENDICULAR t electic field lines (in thee dimensinal situatins, yu can have whle euiptential sufaces). 79

8 Peviusly, we detemined the electic field lines geneated by tw eual, psitive chages. Euiptential lines f that situatin ae shwn in Figue 15.5 t the ight. euiptential lines (dashed in the sketch belw) geneated by tw eual pint chages +Q +Q E. fld. lines FIGURE 15.5 D.) The Electical Ptential f a Pint Chage: 1.) Just as it was pssible t deive a geneal algebaic expessin f the size f the electic field ceated by a SINGLE POINT CHARGE Q, it is als pssible t deive a geneal algebaic expessin f the abslute electical ptential (i.e., the amunt f ptential enegy pe unit chage) geneated by a field-pducing pint-chage Q as a functin f the distance fm the chage..) The geneal elatinship between a cnstant electic field and its electical ptential functin was peviusly deived as: V -E. d (Euatin 1), Cnstant E and d whee V V B - V A is the electical ptential diffeence between tw pints A and B in the electic field, E is the electic field vect, and d is a vect beginning at Pint A and ending at Pint B. Figue 15.6 e-caps the set-up. V A d V B E FIGURE

9 Ch. 15--Elect. Pt. and Enegy Cns. 3.) Hw des this expessin change when eithe the field the path vaies? a.) The W/ dne as a test chage mves fm Pint A t Pint B will always eual the change f electical ptential between the tw pints. As such, the left-hand side f Euatin 1 will nt change at all. b.) The E. d tem is eally (F/). d (i.e., the wk--f. d--pe unit chage ). If F vaies, d vaies, the angle between F and d vaies, the wk dne alng ne sectin f the path will nt be the same as the wk dne alng anthe sectin. c.) T deal with this pblem: ttal wk/unit chage is -(V - V ) which, in tun, euals B A E d i.) Define a diffeential path length d alng an abitaily selected sectin f the path (see Figue 15.7); V A V B ii.) Detemine the amunt f wk pe unit chage (i.e., E. d) dne n a chage mving ve d; path between Pints A and B d 0 E iii.) Then sum ve the entie path using integatin (i.e., E. d). FIGURE 15.7 d.) Summaizing: V -E. d becmes: V E d, whee d is an abitay, diffeential vect alng the path. 81

10 4.) With all this in mind, we want t deive a geneal algebaic expessin f the electical ptential a distance units fm a pint chage Q. a.) T begin with, we must decide whee the electical ptential shuld be ze. A ptential enegy functin is taditinally defined as ze whee its assciated fce functin is ze. As an electical ptential is nthing me than a mdified ptential enegy functin, we will define its ze whee its assciated electical field is ze. In the case f a pint chage, that is at infinity. b.) Having made that decisin, we must detemine the amunt f wk pe unit chage available due t the electic field-pduced by a field-pducing pint chage as we mve fm infinity t a pint units fm the pint chage (see Figue 15.8). hw much wk pe unit chage is dne as ne mves fm infinity t a distance units fm the field pducing chage Q? Q c.) Nting that the electical ptential is ZERO at infinity, (i.e., that V( ) 0) 1 and the electic field functin f a pint chage is 4πε is a unit vect in the adial diectin), the math yields: V E d 1 Q V ( ) V( ) d π [( ) )] 4 ε Q 1 V () d π 4 ε Q 1 4 π ε 1 Q. 4 π ε 5.) Thee ae a cuple f things t ntice abut this expessin: Q FIGURE 15.8 (whee a.) The euatin lks smething like the magnitude f the electic field euatin f a pint chage (i.e., it lks like kq/ ). The esem- 8

11 Ch. 15--Elect. Pt. and Enegy Cns. blance is supeficial. Electical ptentials ae enegy-elated scala uantities. That means they can be negative, but being s has nthing t d with diectin. Nte: Gavitatinal ptential enegy is meaningful nly in the cntext f detemining the wk dne by a cnsevative gavitatinal fce as a bdy mves fm ne pint t anthe in the field. It is pefectly acceptable f a ptential enegy value t be negative at sme pint--all that HAS TO BE TRUE is that W fld. - U. The same is tue f electical ptentials. It makes n diffeence what thei values ae, psitive negative, just as lng as the amunt f wk pe unit chage (W/) available t a chage as it mves thugh a given electic field euals minus the change in electical ptential (i.e., minus the vltage change, - V) between the beginning and end pints f the mtin. b.) V at A due t pt.chg.q kq/ wks eually well f psitive negative chages. But unlike the electic field euatin in which chage uantities ae always inseted as magnitudes, the sign f the chage must be included when slving f the electical ptential f a pint chage. In the wds, a psitive pint chage will pduce a psitive electical ptential and a negative pint chage will pduce a negative electical ptential. Nte: The vaiable tells yu the distance between the pint f inteest and the pint chage. When used in the electical ptential euatin f a pint chage, this value must always be psitive. Why? A chage at the igin f a cdinate axis is ging t pvide as much ptential enegy pe unit chage at x metes as it will at x - metes. E.) Electical Ptential f a System f Pint Chages: 1.) A gup f pint chages will pduce sme net electic ptential field in the space aund them. That field will be the SCALAR SUM f the electical ptentials geneated by the individual pint chages invlved in the cnfiguatin..) Example: A 3 micculmb ( µc) chage is placed at x -3 metes and a -3 µc chage is placed at x 3 metes (see Figue 15.9 t the ight). Q -Q (-3,0) (3,0) FIGURE

12 a.) What is the net electical ptential at x 0? V V due t Q1 + V due t Q k Q 1 / 1 + k Q / (9x10 9 )(3x10-6 C)/(3 m) + (9x10 9 )(-3x10-6 C)/(3 m) 0. Nte: At fist glance, this pbably seems bizae. Afte all, if yu eleased a psitive chage at x 0, it will cetainly acceleate twad the negative chage. That is t say, it will cetainly have ptential enegy. The espnse t this is, "Nt necessaily." If yu put a mass 1 mete abve the gund, yu culd define 1 mete abve the gund t be the ze ptential enegy level. Let the mass lse and it acceleates even thugh its ptential enegy has been defined as ze. Ptential enegy is meaningful nly in the cntext f ptential enegy changes. Only then ae they elated t the wk dne by the field as the bdy mves thugh it. The same is tue f electical ptentials. Yu can have ze electical ptential at sme spt as lng as the electical ptential dwnsteam (i.e., futhe dwn alng the electic field lines) is even lwe (in this case it wuld have t be negative). b.) What is the net electic ptential at x +1 mete? V 1 V due t Q1 + V due t Q k Q 1 / 1 + k Q / (9x10 9 )(3x10-6 C)/(4 m) + (9x10 9 )(-3x10-6 C)/( m) -6.75x10 3 vlts. Nte: The electic field alng the x-axis between the chages is t the ight. Ntice that makes x 1 "dwnsteam" elative t the igin. Ntice als that the electical ptential at x 1 is less than at the igin, just as was pedicted in the Nte abve. F.) Deiving the Electical Ptential Functin f an Extended, Chaged Object Using a Diffeential Chage Appach: 1.) Detemine a geneal expessin f the electical ptential alng the cental axis f a hp f adius R and upn which esides a net chage Q. 84

13 Ch. 15--Elect. Pt. and Enegy Cns. a.) Define a diffeential amunt f chage d n the hp. Als, define the distance between the diffeential chage and an abitay pint (x, 0) n the cental axis (all f this is shwn in Figue 15.10). b.) The diffeential electical ptential dv at (x, 0) due t the diffeential chage d is: 1 d dv 4 π ε 1 d 4 π ε ( R + x ) 1/. d R hp 1/ (R + x ) (x, 0) FIGURE c.) The nly vaiable in this expessin is d (all else is fixed in the pblem--even x has been defined as a specific cdinate). As such, we can pull ut the cnstants and integate, yielding: V dv d 4π ε( R + x ) 1 π R + / 4 ε ( x ) Q 4 π ε ( R + / x ) 1/ 1 1. d.) Cnside nw a flat disk f adius R with a net chage Q n its suface. What is the electical ptential at an abitay pint (x, 0) alng its cental axis (see Figue n the next page)? a.) If we assume a diffeential chage d esides within a hp f adius a and thickness da, we can use the expessin deived abve f the electical ptential f a hp by substituting d f Q and a f R. Ding s yields a diffeential electical ptential dv f: 85

14 d dv 1 4 π ε ( a + x ) 1/ b.) The pblem? The vaiable d vaies as a vaies. T accmmdate, we have t expess d in tems f a. c.) A disk with chage Q unifmly distibuted ve its suface will have a suface chage density σ eual t: σ Q/(πR ).. da R a 1/ (a + x ) flat disk (x, 0) d da (Q/πR )[(πa)da] d.) Knwing the FIGURE adius a and diffeential thickness da f the hp, and having defined σ, we can wite: d σda, whee da is the diffeential aea f the hp. This diffeential aea will be the pduct f the hp's cicumfeence and thickness, : da (πa)da. e.) Putting it all tgethe, we get: d σda Q ada π R ( π ) Q R ( ada ) [ ] Q R ada dv ( ) 1 π 1/ 4 ε ( a + x ) Q ( a) π ε R ( a + x ) 1/ da. 86

15 Ch. 15--Elect. Pt. and Enegy Cns. f.) With this, the net electical ptential f the entie disk alng its cental axis becmes: V dv Q R ( a) da π R a 0 a + 1/ ε ( x ) Q R a x 1/ R + π [( ) ] ε a 0 Q R R + x 1/ x π [( ) ]. ε G.) Me Fun With Extended, Chaged Objects: 1.) S fa, we have dealt with extended bjects by detemining the diffeential electical ptential due t a diffeential pint chage n the bject, then by integating t detemine the ttal electical ptential due t all chages n the bject. Cntinuing n in this fashin, cnside a d f length L with a ttal chage Q distibuted unifmly upn its suface (see d da Figue 15.1). Detemine the electical ptential at pint (b, 0) da n the axis..) We must begin by defining a diffeential chage d lcated sme distance a units up the vetical axis. Assuming the sectin length upn which d esides has a diffeential length da, and defining λ as the chage pe unit length n the d, we can wite the diffeential chage as d (λ)da. Nte: The chage pe unit length in this case will eual Q/L. a Q (b, 0) FIGURE

16 3.) T detemine the electical ptential at (b, 0): a.) The electical ptential due t a pint chage is: V() 1 4πε, whee the chage is witten with the sign f its chage intact and is the distance between the field-pducing chage and the pint f inteest. b.) In u case, the pint chage is the diffeential chage d. Additinally, (a + b ) 1/. c.) Putting it all tgethe, we can wite the diffeential electical ptential at (b, 0) as: dv 1 4πε d (a + b ) 1/. d.) Due t the symmety f the pblem, we can integate fm a 0 t a L, then duble that esult. Substituting in f d and λ and ding this integatin, the net electical ptential at (b, 0) is: V dv 1 d 1/ 4π ε ( a + b ) 1 L ( λ) da ( ) 4π a 0 1/ ε ( a + b ) 1 L ( Q/ L) da π a 0 a + 1/ ε ( b ) Q L 1 4 π L a 0 a + 1 ε / ( b ) da. Nte: Yu may need a bk f integals t slve this, depending upn hw much expeience yu have had with Calculus. Cntinuing: 88

17 Ch. 15--Elect. Pt. and Enegy Cns. 1/ [ [ ]] Q ln a+ ( a + b ) 4 π ε L 1/ 1/ [ [ ]] Q ln + + π [ L ( L b ) ] ln 0+ ( 0 + b ) 4 ε L Q L+ ( L + b ) ln 4 π ε L b 1/ L a 0. H.) Abslute Electical Ptentials f a Spheically Symmetic Chage Cnfiguatin: 1.) Thee ae times when it is nt cnvenient t define a diffeential bit f chage, detemine dv f that chage at a pint f inteest, then integate t detemine the net electical ptential at the pint. In such cases, we must evet t a me fundamental appach..) Hw did we deive the geneal expessin f the abslute electical ptential f a pint chage back in Sectin D-3-c? The pcedue is pesented belw: a.) We calculated the amunt f wk pe unit chage (i.e., E. d) available as we mved fm infinity t sme abitay pint in the pint chage's field. b.) We nticed that W/ - U/ - V -[V() - V( )], whee V( ) was defined as ze. c.) We cmbined Pat a and Pat b t geneate the euatin: whee E pt. chg. Q/(4πε ). V() E pt.chg. d, Nte: If this isn't clea, lk back at the sectin in which this deivatin was actually dne. 3.) The pcedue utlined abve was used t detemine the electical ptential at sme pint units fm a pint chage, but the pcedue itself can be used n any chage cnfiguatin. 89

18 4.) T see this, cnside the fllwing example: A thick spheical shell f inside adius R 1 and utside adius R (see Figue 15.13) is sht full f chage such that its vlume chage density is ρ ka, whee a is an abitay distance fm the cente f the sphee t a pint inside the shell, and k has a magnitude f ne with the apppiate units. Detemine the electical ptential functin V() f > R, R > > R 1, and f R 1 >. chaged spheical shell R 1 ka 5.) F > R : R FIGURE a.) What is the fist thing that has t be dne? Decide whee the electical ptential is t be ze. In this case, the electic field geneated by the chage cnfiguatin is ze at infinity, s that is whee we will define u ze electical ptential pint. b.) We want t use the elatinship: V E d. c.) By setting that expessin int the cntext f u pblem--that is, by evaluating it between the ze electical ptential pint and an abitay pint, we can wite: V() V( ) E d. Nte: Why is this desiable? Because by u wn definitin, V( ) 0. That means that the evaluatin f the integal will leave us with a geneal expessin f V(), which is exactly what we want. d.) Min pblem: T d this integal, we need t have a geneal expessin f the electic field E IN THE REGION OVER WHICH THE INTEGRAL IS TO BE TAKEN. T get that, we will use GAUSS'S LAW. Defining a spheical Gaussian suface whse adius is > R, we get: 90

19 Ch. 15--Elect. Pt. and Enegy Cns. S E utside ds ρdv ε E ut (4π ) R ar 1 [ ] ( ka) (4πa )da ε 4πk R 3 Eut π ( a ) da 4 ε a R1 kr ( 4 R 4 1 ) 1. 4ε e.) Knwing the electic field in the egin utside the sphee, we can nw pceed t detemine the electical ptential functin f this chage cnfiguatin: ( ) kr 4 R V () V( ) ( d) 4ε ( 4 ) kr 4 R1 V () 4ε kr 4 R 4ε ( 4 1 ) kr 4 R 4ε ( 4 1 ) ( 4 1 ) kr 4 R 4ε. 1 d 1 [ ] ) F R > > R 1 : a.) This is cnsideably me inteesting because it intduces a pblem with which we have nt yet had t cpe. We want t detemine V between infinity and a pint units fm the sphee's cente, whee is inside the sphee. The difficulty is that the electic field functin inside the sphee is diffeent fm the electic field functin utside the sphee. S what E d we use in - E. d? 91

20 b.) We must use E f the egin in which it is applicable. That means: i.) T detemine the vltage diffeence between and R (this will be V(R ) - V( )), we must use the electic field functin deived f the field utside the shell (i.e., E utside ); ii.) T detemine the vltage diffeence between R and an abitay pint in between R 1 and R (this will be V() - V(R )), we must deive an electic field functin f the field inside the shell (call this E inside ). iii.) The net electical ptential diffeence between (i.e., whee the electical ptential is ze) and an abitay psitin units fm the cente and inside the sphee will be: [V() - V(R )] + [V(R ) - V( )] V(). c.) T detemine E in, we must use Gauss's Law. Shwn in tuncated fm belw, this yields: S E inside ds ρdv ε E in (4π ) ar 1 4πk 3 Ein π ( a ) da 4 ε a R1 4 k ( R 4 1 ) 4ε k kr 4ε 4ε 4 1 [ ] ( ka) (4πa )da. ε d.) We can nw use u appach t detemine V(): 9

21 Ch. 15--Elect. Pt. and Enegy Cns. V () [ VR ( ) V( ) ]+[ V () VR ( ) ] R ut d in d + E E R 4 4 R kr ( R ) 1 1 ( d) 4 k ( ) kr ( [ ] + 4ε ) [( d) ] R 4ε 4ε 4 4 kr ( R R + k + 1 ) d d R d R 4ε 4ε R 4 4 R 3 kr ( R1 ) 1 k 4 1 R R ε ε R 4 4 kr ( R 1 ) 1 4ε R k R R. 4ε 3 3 R 7.) F R 1 > : a.) Using the appach utlined abve, we can wite: V() [V() - V(R 1 )] + [V(R 1 ) - V(R )] + [V(R ) - V( )]. That is, the vltage jump fm infinity t R plus the vltage jump between R and R 1 plus the vltage jump fm R 1 t will give yu the net vltage jump between infinity (whee V 0) and inside the hllw. b.) We aleady knw the electic field functin f the ute and inside-shell ptins. What is the electic field functin inside the hllw? A Gaussian suface that esides inside the hllw will have n chage enclsed within its bunds. That means the electic field in that egin must be ze. c.) Ze electic field in a egin means n electical ptential diffeence fm pint t pint in the field. This, in tun, implies a cnstant electical ptential thughut the egin. d.) Electical ptential functins, being enegy elated, must be cntinuus functins (electic field functins, n the the hand, d nt have t be cntinuus). As such, the electical ptential at the edge f the hllw will be the same as the electical ptential anywhee inside the hllw. Knwing that, we can wite: 93

22 V ( ) [ VR ( ) V( ) ]+[ VR ( 1) VR ( ) ]+[ V ( ) VR ( 1) ] R ut d in d [ ] + R1 E + E 0 R 4 4 R kr ( R ) 1 1 4ε R 4 1 k ( ) kr ( [ ] + ) ( d) [ ] ( d) R 4ε 4ε 4 4 kr ( R ) k R + 4ε R + 1 R R. 4ε 3 3 R1 R Nte: This is imptant. The electical ptential will be the same at evey pint in a egin in which the electic field is ze. THIS INCLUDES REGIONS INSIDE CONDUCTORS. Althugh yu haven't seen a pblem with a cnduct in it yet, yu will. I.) Abslute Electical Ptential f a Cylindically Symmetic Chage Cnfiguatin: 1.) The appach we have used f the deteminatin f the electical ptential functin f a cmplex chage cnfiguatin with spheical symmety will als wk f chage cnfiguatins that have cylindical symmety (with a twist that will becme evident shtly). The fllwing example will illustate this..) A cnducting cylindical shell f adius R has a cnstant suface chage density σ n it (see Figue 15.14). Dwn its cental axis uns a wie f adius R 1 whse linea chage density -λ is cnstant. Detemine a functin f the abslute electical ptential at, whee < R 1. Css-sectin f Caxial System --Cylindical Symmety-- R 1 a.) Assuming the electical ptential is ze at infinity, we need t execute the fllwing peatin: R wie with linea chage density - cylindical shell with suface chage density FIGURE

23 Ch. 15--Elect. Pt. and Enegy Cns. [ 1 ] + [ 1 ] + [ ] V () V () VR ( ) VR ( ) VR ( ) VR ( ) V( ) R R inne d middle d ute d. R + 1 R + E E E 1 3.) T execute the peatin, we need t use Gauss's Law t detemine the vaius electic fields. Ding s yields: a.) F < R 1 : i.) Because we ae dealing with a cnducting wie, fee chage n the wie distibutes itself alng the wie's ute edge leaving n chage inside that suface. As such, the electic field is ze. b.) F R 1 < < R : i.) Thee is a negative chage pe unit length alng the cental axis. If we define a Gaussian cylinde f adius R 1 < < R and length L, we get: c.) F > R : S E middle ds ε encl L Emiddle π L λ ( ) ε λ. Emiddle π ε i.) If we define a Gaussian cylinde f adius ( > R ) and length L, we get: S E ut ds ε encl L R L Eut π L λ + σ( π ) ( ) ε λ σr Eut +. π ε ε 95

24 4.) We ae nw eady t use u techniue. [ 1 ] + [ 1 ] + [ ] V () V () VR ( ) VR ( ) VR ( ) VR ( ) V( ) R R inne d middle d ute d R + 1 R + E E E 1 R 1 λ 1 λ 1 σ 1 [ 0]+ πε R R [ ] + + π ε ε ( d) [ ] ( d) R λ R 1 1 λ σ [ 0]+ πε + R d R R 1 πε ε d λ R R R [ ]+ [ ] R π + 1 λ σ 0 ln [ ln ] ε πε ε λ R [ R R ] π + λ σ ln 1 ln [ ln R ln ] ε πε ε λ R R R 1 λ σ ln + ln π R π. ε ε ε 5.) We haven't dne anything technically wng, but we have cme up with an expessin that makes n sense (ln (R / ) ln (0)... this desn't exist). What happened? a.) The electic field functin we deived using Gauss's Law (i.e., E α 1/) is actually the field f an infinitely lng wie. Such a wie wuld have an infinite amunt f chage n it, which autmatically puts it utside the ealm f pssibility. F a finite length wie, Gaussian symmety wks (i.e., E is in a adial diectin ut fm the wie) as lng as ne desn't get t fa fm the wie. At infinity, the electic field expessin isn't gd. If we had used the diffeential pint chage appach t detemine E, we wuld have ended up with an electic field expessin that was gd f all space aund a finite wie. That functin tuns ut t be E α 1/ 3. Using u appach with this electic field functin wuld have yielded a geneal electical ptential functin whse value at infinity wuld have been ze, but that wuldn't have explded f finite. b.) Was this execise a waste f time? Npe, nt if it made yu think abut the appach used t deive V() fm E. 96

25 Ch. 15--Elect. Pt. and Enegy Cns. J.) Deiving an Electic Field Functin fm an Electical Ptential Functin: 1.) S fa, we have used the fact that the change f electical ptential is elated t the aea unde an electic field vesus psitin gaph. That is, knwing the electic field f a chage cnfiguatin, we have detemined V by being cleve with the elatinship V - E. d. We nw want t g the the way, detemining E knwing smething abut its assciated electical ptential functin V()..) Cnside the elatinship between the diffeential ptential change dv, the electic field E that pduces the ptential field, and a diffeential displacement d ve which the change ccus. a.) Assuming E and d ae in the same diectin (let's assume it is in the x diectin), we can wite: dv -(E x )dx. b.) It shuld be bvius fm examinatin that if the abve expessin is cect, the x cmpnent f the electic field must eual: E x -dv/dx. c.) Additinally, simila expessins shuld be tue f the y and z diectins. 3.) Put in wds, the electic field cmpnent at a paticula pint in a paticula diectin euals the ate at which the electical ptential changes as ne mves sme diffeential distance in that diectin. Put anthe way, the electic field must eual the slpe f the electical ptential vesus psitin gaph, evaluated at a pint f inteest. 4.) The electic field functin is a vect. The peat that executes a ate f change with psitin calculatin and makes the esult int a vect is the del peat. In sht, E V. K.) Electic Fields and Pint Chage Cnfiguatins: 1.) Take an easy example fist. We knw that the electical ptential f a psitive pint chage is: 97

26 V 4πε, whee is the distance between the field-pducing chage and the pint f inteest. What is the electic field functin assciated with this electical ptential functin?.) We have a small pblem hee as the functin is in adial symmety. The del peat in spheical cdinates is (n, yu needn't knw this): V V + 1 V θ Θ+ 1 V sin θ ϕ Φ. a.) Using this with u ptential functin, we get: E V 4 ε π ( ) ( 1/ ) π 4 1 π 4 ε. 4 π ε ε ( ) 1/ 1 1/ 1 1/ + Θ+ θ sin θ ϕ + 0Θ+ 0Φ ( ) b.) This is the mst geneal fm f the electic field due t a pint chage, cmplete with adial symmety. The pblem? Yu ae nt suppsed t knw the del peat in spheical cdinates. T cicumvent the difficulty, we culd as well have assumed the pint f inteest was alng the x-axis and used ectangula cdinates. Ding s wuld yield an electical ptential functin f: Φ V 1 4πε x. and an electic field expessin f: 98

27 Ch. 15--Elect. Pt. and Enegy Cns. E V V V V i + j + k 4 π ε x y z ( 1/ x) ( 1/ x) ( 1/ x) i + j + k 4 π ε x y z 1 i. 4 π ε x c.) Again, this is the electic field functin f a pint chage (thugh in esticted fm as E f a pint chage isn't slely a functin f x). 3.) Cnside the tw eual pint chages shwn in Figue a.) The electical ptential at pint (x, 0) due t the tp chage is: V 4πε, whee (a + x ) 1/. b.) The electical ptential at pint (x, 0) due t the bttm chage is: V 4πε, a a 1/ (a + x ) eual pint chages (x, 0) FIGURE whee (a + x ) 1/. c.) The ttal electical ptential at (x, 0) is the scala sum f thse tw electical ptentials, : 99

28 V 4 πε π ε ( a + x ) 1/ Nte: What is s nice abut using electical ptential functins is that they ae nt vects. We can add them like scalas--we dn't have t hassle with beaking them int cmpnents befe using them.. d.) Accding t they, the electic field shuld be: E V a + x π ε x 1 [( ) ] + [( ) ] 1/ + a x i π ε x 1 3/ a + x x π ( ) ε ( )( ) i x π a + x 3 / i. ε ( ) [( ) ] + [( ) ] 1 1 a + x a + x i j k y z / / / e.) L and behld, this is exactly the expessin we deived using the definitin f the electic field f a pint chage and the vect appach intduced in the Electic Field chapte. f.) UNFORTUNATELY, WE'VE DETERMINED THE RIGHT ANSWER BUT WE HAVE DONE SOMETHING DIRTY IN THE PROCESS. T see the difficulty, cnside the next pblem. 4.) Cnside the tw ppsite pint chages shwn in Figue n the next page. Detemine the electic field at pint (xb, 0), using the electical ptential functin f a pint chage. a.) The electical ptential at pint (x, 0) due t the tp chage is: 100

29 Ch. 15--Elect. Pt. and Enegy Cns. V 4πε, whee (a + x ) 1/. b.) The electical ptential at pint (x, 0) due t the bttm chage - is: V 4πε, whee (a + x ) 1/. c.) The ttal electical ptential at (x, 0) is the scala sum f thse tw electical ptentials, ZERO! a a - 1/ (a + x ) ppsite pint chages (x, 0) FIGURE d.) This isn't a pblem as fa as the electical ptential is cncened. As lng as the ptential enegy chaacteistics ae maintained, finding a ze electical ptential pint is pefectly acceptable. The pblem is in tying t use that functin in cnjunctin with the electic field expessin E - V. Why wn't it wk? The pblem is ted in the fact that E - V euies a GENERAL expessin f the electical ptential (i.e., V(x, y)) t wk ppely. We didn't use such a functin in the tw psitive chages pblem dne in the pevius sectin, and we gt away with it due t the pblem's chage symmety. Withut that symmety, we wuld have lst. 5.) Cnside the tw ppsite pint chages shwn in Figue Detemine the electic field at pint (x b, 0), using the electical ptential functin f a pint chage. a.) We must begin by detemining the electical ptential at an abitay pint (x, y). F the tp chage, 1/ [(y-a) + x ] 1 a a - x x y + a 1/ [(y+a) + x ] Pint (x, y) y - a FIGURE

30 the electical ptential at (x, y) is: V 4 π ε 4 π ε ( y a) + x 1 [ ] 1 /. is: b.) The electical ptential at pint (x, y) due t the bttm chage - V 4πε 4π ε ( y+ a) + x [ ] 1 /. c.) The net electical ptential at (x, y) is the scala sum f thse tw electical ptentials, : V V + V + 4 π ε ( y a) + x 4π ε ( y+ a) + x 1 / 1 / 1/ 1/ ( y a) + x ( y a) x. 4 π ε [ ] [ ] [ ] [ + + ] [ ] d.) The electic field at (x, y) is: E V [ [( y a) + x ] [( y+ a) + x ] ] [( y a) + x ] ( y+ a) + x i + 4 π ε x y x 4 π ε [ ( y a) + x ] 1 [ [ ] ] / / / / x + [( y+ a) + x ] + ( y a) i ( y a) + x [ ] ( y+ a) + [( y+ a) + x ] 3 / 3/ 3/ 3/ j. j e.) This expessin is geneal f any pint in the x-y plane. 10

31 Ch. 15--Elect. Pt. and Enegy Cns. f.) It wuld be inteesting t see if the field is cect f a psitin alng the x-axis. T find ut, substitute x b, y 0 int u expessin. Ding s yields: E V x x + 4 π ε [ ( y a) + x ] ( y+ a) + x 4 π ε b 4 π ε a + b a 4 π ε a + b [ ] i + ( y a) ( y a) + x [ ] + ( y+ a) ( y+ a) + x [ ] 3/ 3/ 3/ 3/ b b + [ ( 0 a) + b ] ( 0+ a) + b [ ] a + ( 0 ) i ( 0 a) + b [ ] + ( 0 + a) ( 0 + a) + b [ ] 3/ 3/ 3/ 3/ + i [ ] [ + ] + 3/ 3 / [ + ] 3 / [ + ] 3 / a b a b a b [ ] j 3 /. b a + a j Electic Field due t Pt. Chages j j g.) Des this make sense? Yu bet. Lking at Figue 15.18, the chage cnfiguatin is shwn. Due t symmety, the x cmpnents f the electic fields add t ze while the y cmpnents (afte being added tgethe) eual: a a - 1/ [a + b ] 0 E [/(4π )] Pint (b, 0) 0 E y E sin 0 1/ E [ a/(a +b ) ] FIGURE E ( ) 4 π ε ( a + b ) / [ ] 1 cs θ( j ) a ( ) π [ + ] + 4 ε ( a b ) ( a b ) a j. π ( + 3/ 4 ε a b ) 1/ 1 / [ ] ( j ) 103

32 h.) Bttm line: Technically, using E - V t detemine E euies a geneal expessin f V. BUT, if all the chage in the cnfiguatin is the same kind (i.e., it is all psitive all negative), and if yu want E at a pint f symmety (i.e., alng the x-axis, f instance, in the tw psitive chage pblem fm the pevius sectin), and assuming yu ae nt sueamish abut expliting a mathematical anmaly that allws yu t d a pblem wng but get the answe ight, then yu can get away with setting the pblem up as was iginally dne in Pat 3 f Sectin K. QUESTIONS 15.1) The diamete f a typical hydgen atm is appximately metes acss. The chage n an electn is the same as the chage n a ptn (ne elementay chage unit--1.6x10-19 culmb) and the mass f an electn is 9.1x10-31 kilgams. If we assume that the electn fllws a cicula path aund the ptn: a.) What is the electic ptential at the edge f the atm's bunday due t the pesence f the ptn at the atm's cente? b.) Hw much electical ptential enegy des the electn have due t the pesence f the ptn? c.) Assuming it is mving with velcity.5x10 6 m/s, what is the electn's ttal enegy? 15.) The aeas A and B shwn in Figue I ae V A V B fund t have electical ptential values f V A -1 vlts and V B +0 vlts: FIGURE I a.) Re-dawing the sketch lage, daw in the electic field lines f the egin between and aund A and B; and b.) Daw the egin's euiptential lines at eight-vlt intevals. 15.3) A paticle made up f tw ptns and tw neutns is called an α paticle (the mass f eithe a ptn a neutn is 1.67x10-7 kilgams--the 104

33 Ch. 15--Elect. Pt. and Enegy Cns. chage n a ptn is 1.6x10-19 culmbs; and thee is n chage n a neutn). An α paticle is acceleated thugh an 18 Megavlts electical ptential diffeence (ne megavlt euals 10 6 vlts). If the acceleatin takes place alng a 1 mete lng tack: a.) Assuming the α paticle has n ptential enegy by the end f its un, hw much ptential enegy des it have at the beginning f its un? b.) Hw much wk pe unit chage des the field d n the α paticle duing its un? c.) Assuming it stats fm est, what is the paticle's velcity magnitude at the end f the un (appach this using cnsevatin f enegy, nt the wk/enegy theem)? 15.4) The fllwing infmatin is knwn abut the cnstant electic field shwn in Figue II t the ight and belw: the electic field intensity is 80 ntspe-culmb; the vltage V A 340 vlts; the vltage V E 30 vlts; distance d AB.5 metes; the distance d DE.50 metes and is pependicula t the electic field; and Pint C's vetical psitin is half-way between A and B. a.) Is V A geate less than V B? b.) Detemine the distance d AD sme the way than just eyeballing it. c.) Detemine V B. B d.) Thee ae a numbe f ways t detemine V C. Pick tw ways and d it. A e.) Hw much ptential enegy will be available t a 6 µc chage when placed at Pint A? f.) Hw much wk pe unit chage is dne by D the field as a 6 µc chage mves fm Pint A t Pint E? g.) Hw much wk is dne by the field n a 6 µc chage that mves fm Pint A t Pint B? h.) If V A had been 340 vlts and V B had been 90 vlts: i.) What wuld the electic field's diectin have been? ii.) What wuld the electic field's magnitude have been? electic field lines FIGURE II C E 15.5) Tw eual pint-chages ( C each) ae placed at ppsite cnes f a suae whse edges ae.4 metes lng. Assuming we neglect gavity, if C is placed at the exact cente f the suae and given a 105

34 slight nudge twad ne f the unccupied cnes, it will acceleate. Letting 1 's mass be m 7x10 - kilgams, what will 1 's velcity be by the time it eaches the unccupied cne twad which it acceleates? D the pblem algebaically fist, then put in the numbes. 15.6) A d f length L has a ttal chage Q distibuted unifmly upn its suface (see Figue IV). a.) Detemine the electical ptential at an abitay pint (x, y) in the field. b.) Knwing the electical ptential functin, hw culd yu detemine the electic field at (x, y)? (Yu dn't need t actually d it). L (x, y) Q's wth f chage unifmly distibuted ve the d's length 15.7) An electical ptential functin: V k 1 e kx + k /y 3 FIGURE IV exists within a egin. Assuming the k tems ae cnstants, what is the electic field functin f the egin? 15.8) An electic field functin: E kx 3 ke i+ ( k / y) j 1 exists within a egin that excludes y 0. If the k tems ae cnstants: a.) At what x and y cdinate must the electical ptential be ze? b.) What is the electical ptential functin f the egin? 15.9) A vey lage (essentially infinite) flat cnduct has n its suface a suface chage density eual t culmbs pe suae mete. Euiptential sufaces that diffe by 1 vlts ae pltted in the vicinity f the cnduct's face. a.) Relative t the cnduct's face, hw ae the euiptential sufaces iented? b.) Hw fa apat ae the sufaces? 106

35 Ch. 15--Elect. Pt. and Enegy Cns ) Electic field functins ae nt cntinuus. Electical ptential functins ae cntinuus. Explain hw yu knw this must be tue ) A cnducting d f adius R 1 is suunded by a caxially psitined pipe f adius R. A battey whse vltage is V is cnnected such that the d is attached t the gund f a battey (assume the electical ptential f the gund is ze) while the pipe is attached t the high vltage side f the pwe suce. Deive a geneal expessin f the electical ptential between the d and the pipe. In the wds, what is V() f R > > R 1? css-sectin f a lng caxial cable R 1 R V 0 V V assume the battey teminals put negative chage pe unit length (- ) n the inside d and + n the utside pipe 15.1) A chage Q exists (it is hung fm an incnseuential thead) at the cente f a spheical cnducting shell f inside adius R 1 and utside adius R. Knwing that the shell has Q's wth f fee chage placed n it, plt the electic field vesus psitin gaph and the electical ptential vesus psitin gaph f this cnfiguatin. Afte having used the they t detemine the field expessins f the vaius egins, substitute the fllwing values int yu expessins: R 1 metes, R 3 metes, and [Q/(4πε ) ] 1 (that is, assume Q is the ight size f this t be tue). Nte: The abve pblem culd as easily have been a spheical cnfiguatin in which thee wee suface vlume chage densities, it culd have been a cylindical cnfiguatin in which thee wee linea, suface, vlume chage densities. Be pepaed t deal with any cntingency. TRANSLATION: Review GAUSS'S LAW f bth spheical and cylindical symmety befe yu next test! 107

36 108