Chapter 23 Electrical Potential

Transcription

1 hpte Electicl Potentil onceptul Polems [SSM] A poton is moved to the left in unifom electic field tht points to the ight. Is the poton moving in the diection of incesing o decesing electic potentil? Is the electosttic potentil enegy of the poton incesing o decesing? Detemine the oncept The poton is moving to egion of highe potentil. The poton s electosttic potentil enegy is incesing. An electon is moved to the left in unifom electic field tht points to the ight. Is the electon moving in the diection of incesing o decesing electic potentil? Is the electosttic potentil enegy of the electon incesing o decesing? Detemine the oncept The electon is moving to egion of highe electic potentil. The electon s electosttic potentil enegy is decesing. If the electic potentil is unifom thoughout egion of spce, wht cn e sid out the electic field in tht egion? Detemine the oncept If is constnt, its gdient is zeo; consequently the electic field is zeo thoughout the egion. If is known t only single point in spce, cn E e found t tht point? Eplin you nswe. Detemine the oncept No. E cn e detemined without knowing t continuum of points. 5 [SSM] Figue -9 shows point pticle tht hs positive chge Q nd metl sphee tht hs chge Q. Sketch the electic field lines nd equipotentil sufces fo this system of chges. 9

2 9 hpte Pictue the Polem The electic field lines, shown s solid lines, nd the equipotentil sufces (intesecting the plne of the ppe), shown s dshed lines, e sketched in the djcent figue. The point chge Q is the point t the ight, nd the metl sphee with chge Q is t the left. Ne the two chges the equipotentil sufces e sphees, nd the field lines e noml to the metl sphee t the sphee s sufce. 6 Figue - shows point pticle tht hs negtive chge Q nd metl sphee tht hs chge Q. Sketch the electic field lines nd equipotentil sufces fo this system of chges. Pictue the Polem The electic field lines, shown s solid lines, nd the equipotentil sufces (intesecting the plne of the ppe), shown s dshed lines, e sketched in the djcent figue. The point chge Q is the point t the ight, nd the metl sphee with chge Q is t the left. Ne the two chges the equipotentil sufces e sphees, nd the field lines e noml to the metl sphee t the sphee s sufce. ey f fom oth chges, the equipotentil sufces nd field lines ppoch those of point chge Q locted t the midpoint. 7 Sketch the electic field lines nd equipotentil sufces fo the egion suounding the chged conducto shown in Figue -, ssuming tht the conducto hs net positive chge.

3 Electic Potentil 9 Pictue the Polem The equipotentil sufces e shown with dshed lines, the field lines e shown in solid lines. It is ssumed tht the conducto cies positive chge. Ne the conducto the equipotentil sufces follow the conducto s contous; f fom the conducto, the equipotentil sufces e sphees centeed on the conducto. The electic field lines e pependicul to the equipotentil sufces. 8 Two equl positive point chges e septed y finite distnce. Sketch the electic field lines nd the equipotentil sufces fo this system. Pictue the Polem The equipotentil sufces e shown with dshed lines, the electic field lines e shown with solid lines. Ne ech chge, the equipotentil sufces e sphees centeed on ech chge; f fom the chges, the equipotentil sufce is sphee centeed t the midpoint etween the chges. The electic field lines e pependicul to the equipotentil sufces. 9 Two point chges e fied on the -is. () Ech hs positive chge q. One is t nd the othe is t. At the oigin, which of the following is tue? () E nd, () E nd kq/, () E ( kq )iˆ nd, () E ( kq )iˆ nd kq/, (5) None of the ove. () One hs positive chge q nd the othe hs negtive chge q. The positive point chge is t nd the negtive point chge is t. At the oigin, which of the following is tue? () E nd,

4 9 hpte () E nd kq/, () E ( kq )iˆ nd, () E ( kq )iˆ nd kq/, (5) None of the ove. Pictue the Polem We cn use oulom s lw nd the supeposition of fields to find E t the oigin nd the definition of the electic potentil due to point chge to find t the oigin. () Apply oulom s lw nd the E E q t E q t supeposition of fields to find the kq ˆ kq ˆ electic field E t the oigin: i i The potentil t the oigin is given y: () Apply oulom s lw nd the supeposition of fields to find the electic field E t the oigin: q t q t kq kq kq nd ( ) is coect. E E q t Eq t kq kq kq iˆ iˆ iˆ The potentil t the oigin is given q t q t y: kq k( q) nd ( ) is coect. The electosttic potentil (in volts) is given y (, y, z)., whee is constnt, nd is in metes. () Sketch the electic field fo this potentil. () Which of the following chge distiutions is most likely esponsile fo this potentil: () A negtively chged flt sheet in the z plne, () point chge t the oigin, () positively chged flt sheet in the plne, o () unifomly chged sphee centeed t the oigin. Eplin you nswe. Pictue the Polem We cn use E iˆ to find the electic field coesponding to the given potentil nd then compe its fom to those poduced y the fou ltentives listed.

5 Electic Potentil 95 () Find the electic field coesponding to this potentil function: If >, then [ ] If <, then [ ] nd: nd: E iˆ. [ ]iˆ E > E <. iˆ m. iˆ m [. ] iˆ A sketch of the electic field in this egion follows: y () ( ) is coect ecuse field lines end on negtive chges. [SSM] The electic potentil is the sme eveywhee on the sufce of conducto. Does this men tht the sufce chge density is lso the sme eveywhee on the sufce? Eplin you nswe. Detemine the oncept No. The locl sufce chge density is popotionl to the noml component of the electic field, not the potentil on the sufce. Thee identicl positive point chges e locted t the vetices of n equiltel tingle. If the length of ech side of the tingle shinks to one-fouth of its oiginl length, y wht fcto does the electosttic potentil enegy of this system chnge? (The electosttic potentil enegy ppoches zeo if the length of ech side of the tingle ppoches infinity.) Pictue the Polem Points A, B, nd e t the vetices of n equiltel tingle of side. The electosttic potentil enegy of the system of the thee equl positive point chges is the totl wok tht must e done on the chges to ing them fom infinity to this configution. q B A q q

6 96 hpte The electosttic potentil enegy is the wok equied to ssemle the thee chges t the vetices of the equiltel tingle: Plce the fist chge t point A. To ccomplish this step, the wok W A tht is needed is zeo: Bing the second chge to point B. The wok equied is W q, whee A is the potentil t point B due to the fist chge t point A distnce wy: B A U W W W () A W A kq WB qa q B kq Similly, W is given y: W q kq kq kq q Sustituting fo W A, W B, nd W in eqution () yields: U kq kq kq If the tingle is epnded to fou times its oiginl size, its electosttic potentil enegy U ecomes: kq kq U' U Hence the electosttic potentil enegy of this system chnges y fcto of. Estimtion nd Appoimtion Polems [SSM] Estimte mimum the potentil diffeence etween thundecloud nd Eth, given tht the electicl ekdown of i occus t fields of oughly. 6 /m. Pictue the Polem The field of thundecloud must e of ode. 6 /m just efoe lightning stike. Epess the potentil diffeence etween the cloud nd the eth s function of thei seption d nd electic field E etween them: Ed

7 Assuming tht the thundecloud is t distnce of out km ove the sufce of the eth, the potentil diffeence is ppoimtely: Electic Potentil 97 6 (. /m)( m). 9 Note tht this is n uppe ound, s thee will e loclized chge distiutions on the thundecloud which ise the locl electic field ove the vege vlue. The specifictions fo the gp width of typicl utomotive spk plug is ppoimtely equl to the thickness of the cdod used fo mtchook coves. Becuse of the high compession of the i-gs mitue in the cylinde, the dielectic stength of the mitue is oughly. 7 /m. Estimte the mimum potentil diffeence coss the spk gp duing opeting conditions. Pictue the Polem The potentil diffeence etween the electodes of the spk plug is the poduct of the electic field in the gp nd the seption of the electodes. We ll ssume tht the seption of the electodes is. mm. Epess the potentil diffeence etween the electodes of the spk plug s function of thei seption d nd electic field E etween them: Sustitute numeicl vlues nd evlute : Ed 7 (. /m)(. m).k 5 The dius of poton is ppoimtely. 5 m. Suppose two potons hving equl nd opposite moment undego hed-on collision. Estimte the minimum kinetic enegy (in Me) equied y ech poton to llow the potons to ovecome electosttic epulsion nd collide. Hint: The est enegy of poton is 98 Me. If the kinetic enegies of the potons e much less thn this est enegy, then non-eltivistic clcultion is justified. Pictue the Polem We cn use consevtion of enegy to elte the initil kinetic enegy of the potons to thei electosttic potentil enegy when they hve ppoched ech othe to the given "dius." Apply consevtion of enegy to elte the initil kinetic enegy of the potons to thei electosttic potentil when they e septed y distnce : K U K U i i o, ecuse U i K f, K i U f f f

8 98 hpte Becuse ech poton hs kinetic enegy K: K e π K e 8π Sustitute numeicl vlues nd evlute K: K 8π Me 9 (.6 ) 5 (. m) N m.5 e J.6 9 J emks: Becuse the kinetic enegies of the potons is ppoimtely.8% of thei est enegy ( K E est.7 Me 98 Me.8%,) the noneltivistic clcultion ws justified. 6 When you touch fiend fte wlking coss ug on dy dy, you typiclly dw spk of out. mm. Estimte the potentil diffeence etween you nd you fiend just efoe the spk. Pictue the Polem The mgnitude of the electic field fo which dielectic ekdown occus in i is out. M/m. We cn estimte the potentil diffeence etween you nd you fiend fom the poduct of the length of the spk nd the dielectic constnt of i. Epess the poduct of the length of the spk nd the dielectic constnt of i: (. M/m)(.mm) 6.k 7 Estimte the mimum sufce chge density tht cn eist t the end of shp lightning od so tht no dielectic ekdown of i occus. Pictue the Polem The mimum electic field E m just outside the end of the lightning od is elted to the mimum sufce chge density σ m. Epess the mimum electic field just outside the end of the lightning od s function of the mimum sufce chge density σ m : σ m Em σ m Em

9 Electic Potentil 99 Sustitute numeicl vlues nd evlute σ m : σ m N m m μ /m 8 The electic-field stength ne the sufce of Eth is out /m. () Estimte the mgnitude of the chge density on the sufce of Eth. () Estimte the totl chge on Eth. (c) Wht is vlue of the electic potentil t Eth s sufce? (Assume the potentil is zeo t infinity.) (d) If ll Eth s electosttic potentil enegy could e hnessed nd conveted to electic enegy t esonle efficiency, how long could it e used to un the consume households in the United Sttes? Assume the vege Ameicn household consumes out 5 kw h of electic enegy pe month. Pictue the Polem () The chge density is the poduct of nd the mgnitude of the electic field t the sufce of Eth. () The totl chge on Eth is the poduct of its sufce chge density nd its e. (c) The electic potentil t Eth s sufce is given y whee Q is the chge on Eth nd is its dius. (d) We cn estimte how long Eth s electosttic enegy could un households in the United Sttes y dividing the enegy ville y the te of consumption of electicl enegy. () The mgnitude of the chge density σ on the sufce of Eth is popotionl to the electic field stength E t the sufce of Eth: σ E Sustitute numeicl vlues nd evlute σ: σ n/m N m m () The totl chge on Eth is given y: Q σ A πσ Sustitute numeicl vlues nd evlute Q: n Q π.656 m.5 M ( 67 km) (c) The electic potentil t the sufce of Eth is given y: whee is Eth s dius.

10 hpte Multiplying nd dividing y nd sustituting fo yields: E Sustitute numeicl vlues nd evlute : m (d) Epess the ville enegy: eu e( Q ) Evil ( 67 km).9g whee e is the efficiency of enegy convesion. Assuming n efficiency of / yields: Epess the te t which enegy must e supplied to households in the United Sttes: ( Q ) Q Evil 6 P P eq eq pe household N households Assuming 8 million households in the United Sttes, sustitute numeicl vlues nd evlute the lifetime of the electicl enegy deived fom Eth s electosttic enegy: E P vil eq P eq pe household. h 6 Q N households (.5 M)(.9G) 6 s kwh h month 65 household month. d d h 8 6 households Electosttic Potentil Diffeence, Electosttic Enegy nd Electic Field 9 A point pticle hs chge equl to. μ nd is fied t the oigin. () Wht is the electic potentil t point. m fom the oigin ssuming tht t infinity? () How much wok must e done to ing second point pticle tht hs chge of. μ fom infinity to distnce of. m fom the.-μ chge? Pictue the Polem The oulom potentil t distnce fom the oigin eltive to t infinity is given y kq/ whee q is the chge t the oigin. The wok tht must e done y n outside gent to ing chge fom infinity to

11 Electic Potentil position distnce fom the oigin is the poduct of the mgnitude of the chge nd the potentil diffeence due to the chge t the oigin. () The oulom potentil of the chge is given y: kq Sustitute numeicl vlues nd evlute : 9 ( N m / )(.μ).m.9 k.9 k () The wok tht must e done is given y: W qδ Sustitute numeicl vlues nd evlute W: W (. )(.9 k).5mj μ The fcing sufces of two lge pllel conducting pltes septed y. cm hve unifom sufce chge densities tht e equl in mgnitude ut opposite in sign. The diffeence in potentil etween the pltes is 5. () Is the positive o the negtive plte t the highe potentil? () Wht is the mgnitude of the electic field etween the pltes? (c) An electon is elesed fom est net to the negtively chged sufce. Find the wok done y the electic field on the electon s the electon moves fom the elese point to the positive plte. Epess you nswe in oth electon volts nd joules. (d) Wht is the chnge in potentil enegy of the electon when it moves fom the elese point plte to the positive plte? (e) Wht is its kinetic enegy when it eches the positive plte? Pictue the Polem Becuse the electic field is unifom, we cn find its mgnitude fom E Δ/Δ. We cn find the wok done y the electic field on the electon fom the diffeence in potentil etween the pltes nd the chge of the electon nd find the chnge in potentil enegy of the electon fom the wok done on it y the electic field. We cn use consevtion of enegy to find the kinetic enegy of the electon when it eches the positive plte. () Becuse the electic foce on test chge is wy fom the positive plte nd towd the negtive plte, the positive plte is t the highe potentil. () Epess the mgnitude of the electic field etween the pltes in tems of thei seption nd the potentil diffeence etween them: Δ E Δ 5.m 5.k/m

12 hpte (c) elte the wok done y the electic field on the electon to the diffeence in potentil etween the pltes nd the chge of the electon: W qδ 8. 9 (.6 )( 5) 7 J onveting 8. 7 J to e 7 W ( J) yields: e J 5e (d) elte the chnge in potentil enegy of the electon to the wok done on it s it moves fom the negtive plte to the positive plte: ΔU W 5e (e) Apply consevtion of enegy to otin: ΔK ΔU 5e A unifom electic field tht hs mgnitude. k/m points in the diection. () Wht is the electic potentil diffeence etween the. m plne nd the. m plne? A point pticle tht hs chge of. μ is elesed fom est t the oigin. () Wht is the chnge in the electic potentil enegy of the pticle s it tvels fom the. m plne to the. m plne? (c) Wht is the kinetic enegy of the pticle when it ives t the. m plne? (d) Find the epession fo the electic potentil () if its vlue is chosen to e zeo t. Pictue the Polem () nd () We cn use the definition of potentil diffeence to find the potentil diffeence (. m) () nd (c) consevtion of enegy to find the kinetic enegy of the chge when it is t. m. (d) We cn find () if () is ssigned vious vlues t vious positions fom the definition of potentil diffeence. () Apply the definition of finite potentil diffeence to otin:. m (.m) ( ) E d Ed (. kn/)(.m) 8.k () By definition, ΔU is given y: ΔU qδ (.μ)( 8. k). mj

13 Electic Potentil (c) Use consevtion of enegy to elte ΔU nd ΔK: ΔK ΔU o K K ΔU m Becuse K : K m ΔU.mJ Use the definition of finite potentil diffeence to otin: ( ) ( ) E ( ) (. k/m)( ) (d) Fo () : ( ) (. k/m)( ) o ( ) (.k/m) In potssium chloide molecule the distnce etween the potssium ion (K ) nd the chloine ion (l ) in potssium chloide molecule is.8 m. () lculte the enegy (in e) equied to septe the two ions to n infinite distnce pt. (Model the two ions s two point pticles initilly t est.) () If twice the enegy detemined in Pt () is ctully supplied, wht is the totl mount of kinetic enegy tht the two ions hve when they wee n infinite distnce pt? Pictue the Polem In genel, the wok done y n etenl gent in septing the two ions chnges oth thei kinetic nd potentil enegies. Hee we e ssuming tht they e t est initilly nd tht they will e t est when they e infinitely f pt. Becuse thei potentil enegy is lso zeo when they e infinitely f pt, the enegy W et equied to septe the ions to n infinite distnce pt is the negtive of thei potentil enegy when they e distnce pt. () Epess the enegy equied to septe the ions in tems of the wok equied y n etenl gent to ing out this seption: W ΔK ΔU et kqq k U ( e) i e ke Sustitute numeicl vlues nd evlute W et : W et 9 9 ( N m / )(.6 ) J.8 m

14 hpte onvet this enegy to e: W et 9 (.8 J) 5.e e J () Apply the wok-enegy theoem to the system of ions to otin: Solving fo K yields: f W ΔK ΔU K U et whee K f is the kinetic enegy of the ions when they e n infinite distnce pt. K f W et U i f i Fom Pt (), U i W et : K f W et Wet Wet 5.e [SSM] Potons e elesed fom est in n de Gff cceleto system. The potons initilly e locted whee the electicl potentil hs vlue of 5. M nd then they tvel though vcuum to egion whee the potentil is zeo. () Find the finl speed of these potons. () Find the cceleting electicfield stength if the potentil chnged unifomly ove distnce of. m. Pictue the Polem We cn find the finl speeds of the potons fom the potentil diffeence though which they e cceleted nd use E Δ/Δ to find the cceleting electic field. () Apply the wok-kinetic enegy theoem to the cceleted potons: Solve fo v to otin: Sustitute numeicl vlues nd evlute v: W ΔK e Δ mv eδ v m v K f 9 ( )( 5. M) m/s kg () Assuming the sme potentil chnge occued unifomly ove the distnce of. m, we cn use the eltionship etween E, Δ, nd Δ epess nd evlute E: Δ E Δ 5. M.m.5M/m

15 Electic Potentil 5 The pictue tue of television set ws, until ecently, invily cthode-y tue. In typicl cthode-y tue, n electon gun ngement is used to ccelete electons fom est to the sceen. The electons e cceleted though potentil diffeence of. k. () Which egion is t highe electic potentil, the sceen o the electon s stting loction? Eplin you nswe. () Wht is the kinetic enegy (in oth e nd joules) of n electon s it eches the sceen? Pictue the Polem The wok done on the electons y the electic field chnges thei kinetic enegy. Hence we cn use the wok-kinetic enegy theoem to find the kinetic enegy nd the speed of impct of the electons. () Becuse positively chged ojects e cceleted fom highe potentil to lowe potentil egions, the sceen must e t the highe electic potentil to ccelete electons towd it. () Use the wok-kinetic enegy theoem to elte the wok done y the electic field to the chnge in the kinetic enegy of the electons: Sustitute numeicl vlues nd evlute K f : W ΔK K f o K f eδ () K ( )(. k). e e f onvet this enegy to e: K f (. e).8 5 J.6 e 9 J 5 () A positively chged pticle is on tjectoy to collide hed-on with mssive positively chge nucleus tht is initilly t est The pticle initilly hs kinetic enegy K i. In ddition, the pticle is initilly f fom the nucleus. Deive n epession fo the distnce of closest ppoch. You epession should e in tems of the initil kinetic enegy K of the pticle, the chge ze on the pticle, nd the chge Ze on the nucleus, whee oth z nd Z e integes. () Find the numeicl vlue fo the distnce of closest ppoch etween 5. Me α-pticle nd etween 9. Me α-pticle nd sttiony gold nucleus. (The vlues 5. Me nd 9. Me e the initil kinetic enegies of the lph pticles. Neglect the motion of the gold nucleus following the collisions.) (c) The dius of the gold nucleus is out 7 5 m. If α-pticles ppoch the nucleus close thn 7 5 m, they epeience the stong nucle foce in ddition to the electic foce of epulsion. In the ely th centuy, efoe the stong nucle foce ws known, Enest uthefod omded gold nuclei with α-pticles tht hd kinetic enegies of out 5 Me. Would you

16 6 hpte epect this epeiment to evel the eistence of this stong nucle foce? Eplin you nswe. Pictue the Polem We know tht enegy is conseved in the intection etween the α pticle nd the mssive nucleus. Unde the ssumption tht the ecoil of the mssive nucleus is negligile, we know tht the initil kinetic enegy of the α pticle will e tnsfomed into potentil enegy of the two-ody system when the pticles e t thei distnce of closest ppoch. () Apply consevtion of enegy to the system consisting of the positively chged pticle nd the mssive nucleus: ΔK ΔU o K K U U f i f i Becuse K f U i : K U i f etting e the seption of the pticles t closest ppoch, epess U : f U kq q k ( Ze)( ze) nucleus pticle f kzze Sustitute fo U f to otin: K kzze i kzze K i () Fo 5.-Me α pticle nd sttiony gold nucleus: 9 9 ( N m / )( )( 79)(.6 ) 9 ( 5. Me)(.6 J/e) 5 Fo 9.-Me α pticle nd sttiony gold nucleus: 9 9 ( N m / )( )( 79)(.6 ) 9 ( 9. Me)(.6 J/e) 9 6 fm 5 fm (c) No. The distnce of closest ppoch fo 5-Me lph pticle found ove (6 fm) is much lge thn the 7 fm dius of gold nucleus. Hence the sctteing ws solely the esult of the invese-sque oulom foce. Potentil Due to System of Point hges 6 Fou point chges, ech hving mgnitude of. μ, e fied t the cones of sque whose edges e.-mlong. Find the electic potentil t the cente of the sque if () ll the chges e positive, () thee of the chges

17 Electic Potentil 7 e positive nd one chge is negtive, nd (c) two chges e positive nd two chges e negtive. (Assume the potentil is zeo vey f fom ll chges.) Pictue the Polem et the numels,,, nd denote the chges t the fou cones of sque nd the distnce fom ech chge to the cente of the sque. The potentil t the cente of sque is the lgeic sum of the potentils due to the fou chges. Epess the potentil t the cente of the sque: kq k kq kq kq ( q q q q ) k q i i () If the chges e positive: N m / μ. m 5. k ( )(. ) () If thee of the chges e positive nd one is negtive: N m / μ. m.7 k ( )(. ) (c) If two e positive nd two e negtive: 7 [SSM] Thee point chges e fied t loctions on the -is: q is t. m, q is t. m, nd q is t 6. m. Find the electic potentil t the point on the y is t y. m if () q q q. μ, () q q. μ nd q. μ, nd (c) q q. μ nd q. μ. (Assume the potentil is zeo vey f fom ll chges.) Pictue the Polem The potentil t the point whose coodintes e (,. m) is the lgeic sum of the potentils due to the chges t the thee loctions given. Epess the potentil t the point whose coodintes e (,. m): q k q q k q i i i

18 8 hpte () Fo q q q. μ: 9 ( N m / )(. ) μ.9 k.m. m. 5 m () Fo q q. μ nd q. μ: 9 ( N m / )(. ) μ 7.55 k.m. m. 5 m (c) Fo q q. μ nd q. μ: 9 ( N m / )(. ) μ.k.m. m. 5 m 8 Points A, B, nd e fied t the vetices of n equiltel tingle whose edges e.-m long. A point pticle with chge of. μ is fied t ech of vetices A nd B. () Wht is the electic potentil t point? (Assume the potentil is zeo vey f fom ll chges.) () How much wok is equied to move point pticle hving chge of 5. μ fom distnce of infinity to point? (c) How much dditionl wok is equied to move the 5.-μ point pticle fom point to the midpoint of side AB? Pictue the Polem () The potentil t vete is the lgeic sum of the potentils due to the point chges t vetices A nd B. () The wok equied to ing chge fom infinity to vete equls the chnge in potentil enegy of the system duing this pocess. (c) The dditionl wok equied to move the 5.-μ point pticle fom point to the midpoint of side AB is the poduct of 5.-μ nd the diffeence in potentil etween point nd the midpoint of side AB. () Epess the potentil t vete s the sum of the potentils due to the point chges t vetices A nd B: q k A A q B B

19 Electic Potentil 9 Becuse q A q B q : kq A B Sustitute numeicl vlues nd evlute : 9 ( N m / )(. ) μ.98 k. k.m.m () Epess the equied wok in tems of the chnge in the potentil enegy of the system: Sustituting fo U yields: Sustitute numeicl vlues nd evlute W : W U Δ U U U W q5 W ( 5.μ)(.98k) 59.9mJ (c) Epess the equied wok in tems of the chnge in the potentil enegy of the system: W midpoint Δ U U midpoint of AB U Sustituting fo yields: U of midpoint nd U AB W midpoint q 5 q 5 midpoint of AB midpoint of AB q 5 () The potentil t the midpoint of AB is the sum of the potentils due to the point chges t vetices A nd B: midpoint of AB q k A A q B B Becuse q A q B q nd A B : midpoint of AB kq whee is the distnce fom vete A (nd vete B) to the midpoint of side AB of the tingle. Sustituting in eqution () nd simplifying yields: q kq W midpoint 5

20 hpte Sustitute numeicl vlues nd evlute W midpoint : W midpoint 9 ( ) ( N m / )(. μ) 5. μ 59.9 mj.5 m.98 k 9 Thee identicl point pticles with chge q e t the vetices of n equiltel tingle tht is cicumscied y cicle of dius tht lies in the z plne nd is centeed t the oigin. The vlues of q nd e. μ nd 6. cm, espectively. (Assume the potentil is zeo vey f fom ll chges.) () Wht is the electic potentil t the oigin? () Wht is the electic potentil t the point on the z is t z? (c) How would you nswes to Pts () nd () chnge if the chges wee still on the cicle ut one is no longe t vete of the tingle? Eplin you nswe. Pictue the Polem The electic potentil t the oigin nd t z is the lgeic sum of the potentils t those points due to the individul chges distiuted long the equto. () Epess the potentil t the oigin s the sum of the potentils due to the chges plced t intevls long the equto of the sphee: q k i oigin i i kq Sustitute numeicl vlues nd evlute oigin : oigin 9 ( N m / )(. μ) m 5k () Using geomety, find the distnce fom ech chge to z :.6 m Poceed s in () with.6 m : q k i i ' i kq ' 9 ( N m / )(. μ) m 95.k

21 Electic Potentil (c) Becuse the two field points e equidistnt fom ll points on the cicle, the nswes fo Pts () nd () would not chnge. Two point chges q nd q e septed y distnce. At point / fom q nd long the line joining the two chges the potentil is zeo. (Assume the potentil is zeo vey f fom ll chges.) () Which of the following sttements is tue? () The chges hve the sme sign. () The chges hve opposite signs. () The eltive signs of the chges cn not e detemined y using dt given. () Which of the following sttements is tue? () q > q'. () q < q'. () q q'. () The eltive mgnitudes of the chges cnnot e detemined y using the dt given. (c) Find the tio q/q. Pictue the Polem We cn use the fct tht the electic potentil t the point of inteest is the lgeic sum of the potentils t tht point due to the chges q nd q to find the tio q/q'. () The only wy tht, in the sence of othe point chges, the potentil cn e zeo t / is if q nd q hve opposite signs. is coect. () Becuse the point of inteest is close to q, the mgnitude of q must e less thn the mgnitude of q. is coect. (c) Epess the potentil t the point of inteest s the sum of the potentils due to the two chges: kq kq' Simplify to otin: q q' q q'

22 hpte [SSM] Two identicl positively chged point pticles e fied on the -is t nd. () Wite n epession fo the electic potentil () s function of fo ll points on the -is. () Sketch () vesus fo ll points on the is. Pictue the Polem Fo the two chges, nd espectively nd the electic potentil t is the lgeic sum of the potentils t tht point due to the chges t nd. () Epess () s the sum of the potentils due to the chges t nd : kq () The following gph of s function of / ws plotted using spedsheet pogm: / A point chge of e is t the oigin nd second point chge of e is on the -is t. () Sketch the potentil function () vesus fo ll points on the is. () At wht point o points, if ny, is zeo on the is? (c) Wht point o points, if ny, on the -is is the electic field zeo? Ae these loctions the sme loctions found in Pt ()? Eplin you nswe. (d) How much wok is needed to ing thid chge e to the point on the -is? Pictue the Polem Fo the two chges, nd espectively nd the electic potentil t is the lgeic sum of the potentils t tht point due to the chges t nd. We cn use the gph nd the function found in Pt () to identify the points t which (). We cn find the wok needed to ing thid chge e to the point on the is fom the chnge in the potentil enegy of this thid chge.

23 Electic Potentil () The potentil t is the sum of the potentils due to the point chges e nd e: ( ) k ( e) k( e) The following gph of () fo ke nd ws plotted using spedsheet pogm. 5 5 () (m) () Fom the gph we cn see tht () when: ± Emining the function, we see tht () is lso zeo povided: Fo >, () when: Fo < <, () when: (c) The electic field t is the sum of the electic fields due to the point chges e nd e:. 6 E ( ) k ( ) ( ) e k e ( ) Setting E() nd simplifying yields: 6 Solve this eqution to find the points on the -is whee the electic field is zeo: 5. nd. 55

24 hpte Note tht the zeos of the electic field e diffeent fom the zeos of the electic potentil. This is genelly the cse lthough, in specil cses, they cn e the sme. (d) Epess the wok tht must e done in tems of the chnge in potentil enegy of the chge: W ΔU q ( ) Evlute the potentil t : ( ) k ( e) k( e) 6ke ke ke Sustitute to otin: W ke e ke [SSM] A dipole consists of equl ut opposite point chges q nd q. It is locted so tht its cente is t the oigin, nd its is is ligned with the z- is (Figue -) The distnce etween the chges is. et e the vecto fom the oigin to n ity field point nd θ e the ngle tht mkes with the z diection. () Show tht t lge distnces fom the dipole (tht is fo >> ), the dipole s electic potentil is given y (, θ ) kp ˆ kpcos θ, whee p is the dipole moment of the dipole nd θ is the ngle etween nd p. () At wht points in the egion >>, othe thn t infinity, is the electic potentil zeo? Pictue the Polem The potentil t the ity field point is the sum of the potentils due to the equl ut opposite point chges. () Epess the potentil t the ity field point t lge distnce fom the dipole: kq kq k ( q) kq efeing to the figue, note tht, fo the f field ( >> ): Sustituting nd simplifying yields: cosθ nd cosθ o, ecuse p q, kqcosθ (, θ ) kq

25 Finlly, ecuse p ˆ pcosθ : kpcosθ (, θ ) kp ˆ (, θ ) Electic Potentil 5 () (, θ ) whee cos θ : θ cos 9 t points on the z is. Note tht these loctions e equidistnt fom the two oppositelychged ends of the dipole. A chge configution consists of thee point chges locted on the z is (Figue -). One hs chge equl to q, nd is locted t the oigin. The othe two ech hve chge equl to q, one is locted t z nd the othe is locted t z. This chge configution cn e modeled s two dipoles: one centeed t z / nd with dipole moment in the z diection, the othe centeed t z / nd with dipole moment in the z diection. Ech of these dipoles hs dipole moment tht hs mgnitude equl to q. Two dipoles nged in this fshion fom line electic qudupole. (Thee e othe geometicl ngements of dipoles tht cete qudupoles ut they e not line.) () Using the esult fom Polem, show tht t lge distnces fom the qudupole (tht is fo >> ), the electic potentil is given y (, qud θ ) kbcos θ, whee B q. (B is the mgnitude of the qudupole moment of the chge configution.) () Show tht on the positive z is, this potentil gives n electic field (fo z >> ) of E ( 6kB z ) k. ˆ (c) Show tht you get the esult of Pt () y dding the electic fields fom the thee point chges. Pictue the Polem () The electic potentil due to the line electic qudupole is the sum of the potentils of the two dipoles. () The electic field on the y is cn e otined fom the electic potentil on the y is using z is Ez kˆ is. z () The electic potentil due to the line electic qudupole is given y: Fom Polem -: qud kq cosθ kqcosθ nd

26 6 hpte Sustituting fo nd yields: qud kqcosθ kqcosθ Simplify to otin: qud kqcosθ ( )( ) kqcosθ efeing to the figue, note tht, fo the f field ( >> ): Sustitute nd simplify to otin: Becuse B q : cosθ,, nd qud qud (, θ ) cosθ kqcosθ kq cos θ kbcos θ (, θ ) () The electic field on the z is is elted to the electic potentil on the z is: E z is z z is kˆ On the y is, θ nd cos θ. Hence: kb y is y Sustituting fo z is yields: (c) The E field on the z is is given kb kˆ z z E z is kq 6kB kˆ z y: is ( ) z z ( z ) E z kˆ etting w z yields: E z is kq z ( w) ( w) kˆ

27 Electic Potentil 7 Epnd ( w) nd ( ) w inomilly to otin: nd ( w ) w w highe - ode tems ( w ) w w highe - ode tems Fo w << : ( ) Sustituting in the epession fo kq z w w w nd w w w ( ) E z is nd simplifying yields: kq ( w w w w ) kˆ ( 6w )kˆ E z is z Finlly, sustitute fo w to otin: E z is kq 6 z z 6kB kˆ z 6kq kˆ z kˆ omputing the Electic Field fom the Potentil 5 A unifom electic field is in the diection. Points nd e on the -is, with t. m nd t 6. m. () Is the potentil diffeence positive o negtive? () If is k, wht is the mgnitude of the electic field? Pictue the Polem We cn use the eltionship E (d/d) to decide the sign of nd E Δ/Δ to find E. () Becuse E (d/d), is gete fo lge vlues of. So: is positive. () Epess E in tems of nd the seption of points nd : E Δ Δ Δ Sustitute numeicl vlues nd evlute E : E k 6. m.m 5. k/m

28 8 hpte 6 An electic field is given y the epession E iˆ, whee. k/m. Find the potentil diffeence etween the point t. m nd the point t. m. Which of these points is t the highe potentil? Pictue the Polem Becuse () nd E e elted though E d/d, we cn find fom E y integtion. Septe viles in E d/d nd sustitute fo to otin: k d Ed. d m Integte fom to nd fom. m to. m: d k. m k. m. m. m. m [ ] d. m Simplify to otin: 7.5k Becuse 7.5k, the point t. m is t the highe potentil. 7 The electic field on the is due to point chge fied t the oigin is given y E ( )iˆ, whee 6. k m nd. () Find the mgnitude nd sign of the point chge. () Find the potentil diffeence etween the points on the -is t. m nd. m. Which of these points is t the highe potentil. Pictue the Polem We cn integte E d d to otin () nd then use this function to find the electic potentil diffeence etween the given points. () We know tht this field is due to point chge ecuse it vies invesely with the sque of the distnce fom the point chge. Becuse E is positive, the sign of the chge must e positive.

29 Electic Potentil 9 Becuse the given electic field is tht due to point chge, it follows tht: k kq 6. m q k 6. m k Sustitute the numeicl vlue of k nd evlute q: k 6. q m 9 N m n () The potentil diffeence etween the points on the -is t. m nd. m is given y: Δ () Fom E d we hve: d o d ( ) ( ) E d kq kq d etting ( ) yields: ( ) kq 6. k m Sustituting in eqution () yields: 6. k m 6. k m. m. m. k Becuse. k, the point t. m is t the highe potentil. 8 The electic potentil due to pticul chge distiution is mesued t mny points long the -is. A plot of this dt is shown in Figue -. At wht loction (o loctions) is the component of the electic field equl to zeo? At this loction (o these loctions) is the potentil lso equl to zeo? Eplin you nswe. Pictue the Polem Becuse E d d, we cn find the point(s) t which E y identifying the vlues fo fo which d/d. Emintion of the gph indictes tht d/d t ppoimtely.5 m. Thus E t.5m. At this loction, the potentil is not zeo. The electic field is zeo when the slope of the potentil function is zeo.

30 hpte Use the gph of () to estimte the negtive of the slope t the given points: d E ( m), d m m d E ( m)., d m nd E ( 7 m) d d m 7 m.5 m 9 Thee identicl point chges, ech with chge equl to q, lie in the y plne. Two of the chges e on the y-is t y nd y, nd the thid chge is on the -is t. () Find the potentil s function of position long the is. () Use the Pt () esult to otin n epession fo E (), the component of the electic field s function of, on. heck you nswes to Pts () nd () t the oigin nd s ppoches to see if they yield the epected esults. Pictue the Polem et e the distnce fom (, ) to (, ), the distnce fom (, ), nd the distnce fom (, ) to (, ). We cn epess () s the sum of the potentils due to the chges t (, ), (, ), nd (, ) nd then find E fom d/d. () Epess () s the sum of the potentils due to the chges t (, ), (, ), nd (, ): ( ) kq kq kq whee q q q q At, the fields due to q nd q cncel, so E () kq/ ; this is lso otined fom () if. As, i.e., fo >>, the thee chges ppe s point chge q, so E kq/ ; this is lso the esult one otins fom () fo >>. Sustitute fo the i to otin: ( ) kq kq () Fo >, > nd:

31 Electic Potentil Use E d/d to find E : d kq kq E ( ) kq > d ( ) ( ) Fo <, < nd: ( ) Use E d/d to find E : d kq kq E ( ) kq < d ( ) ( ) lcultions of fo ontinuous hge Distiutions A chge of. μ is unifomly distiuted on thin spheicl shell of dius. cm. (Assume the potentil is zeo vey f fom ll chges.) () Wht is the mgnitude of the electic field just outside nd just inside the shell? () Wht is the mgnitude of the electic potentil just outside nd just inside the shell? (c) Wht is the electic potentil t the cente of the shell? (d) Wht is the mgnitude of the electic field t the cente of the shell? Pictue the Polem We cn constuct Gussin sufces just inside nd just outside the spheicl shell nd pply Guss s lw to find the electic field t these loctions. We cn use the epessions fo the electic potentil inside nd outside spheicl shell to find the potentil t these loctions. () Apply Guss s lw to spheicl Gussin sufce of dius <. cm: Apply Guss s lw to spheicl Gussin sufce of dius >. cm: S Q E da enclosed ecuse the chge esides on the oute sufce of the spheicl sufce. Hence E <.cm ( ) ( π ) q E nd q kq E ( >.cm) π

32 hpte Sustitute numeicl vlues nd evlute ( >.cm) E : E 9 ( ) ( N m / )(. μ) >.cm (. m) 6. M/m () Epess nd evlute the potentil just inside the spheicl shell: 9 kq ( ) ( N m / )(.μ).m 79k Epess nd evlute the potentil just outside the spheicl shell: 9 kq ( ) ( N m / )(.μ).m 79k (c) The electic potentil inside unifomly chged spheicl shell is constnt nd given y: 9 kq ( ) ( N m / )(.μ).m 79k (d) In Pt () we showed tht: ( <.cm) E [SSM] An infinite line chge of line chge density.5 μ/m lies on the z-is. Find the electic potentil t distnces fom the line chge of (). m, (). m, nd (c). m. Assume tht we choose t distnce of.5 m fom the line of chge. Pictue the Polem We cn use the epession fo the potentil due to line chge () kλ ln, whee t some distnce, to find the potentil t these distnces fom the line. Epess the potentil due to line chge s function of the distnce fom the line: () kλ ln

33 Electic Potentil Becuse t.5 m:.5m kλ ln.5m ln nd.5m ln ( ).5 m Thus we hve.5 m nd: () N m μ.5 ln m.5m ( ).696 N m/ ln.5m () Evlute (. m): (. m) k N m.m ln.5m () Evlute (. m): (. m) k N m.m ln.5m (c) Evlute (. m): (. m).696.k N m.m ln.5m () Find the mimum net chge tht cn e plced on spheicl conducto of dius 6 cm efoe dielectic ekdown of the i occus. () Wht is the electic potentil of the sphee when it hs this mimum chge? (Assume the potentil is zeo vey f fom ll chges.)

34 hpte Pictue the Polem We cn elte the dielectic stength of i (out M/m) to the mimum net chge tht cn e plced on spheicl conducto using the epession fo the electic field t its sufce. We cn find the potentil of the sphee when it cies its mimum chge using. m () Epess the dielectic stength of spheicl conducto in tems of the chge on the sphee: m E ekdown Q m E ekdown k Sustitute numeicl vlues nd evlute Q : m Q m ( M/m)(.6m) μ 9 N m 8.55μ () Becuse the chge cied y the sphee could e eithe positive o negtive: m m ± 9 N m ±.6m 5 ± 5 ( 8.55μ) Find the mimum sufce chge density σ m tht cn eist on the sufce of ny conducto efoe dielectic ekdown of the i occus. Pictue the Polem We cn solve the eqution giving the electic field t the sufce of conducto fo the getest sufce chge density tht cn eist efoe dielectic ekdown of the i occus. elte the electic field t the sufce of conducto to the sufce chge density: Solve fo σ unde dielectic ekdown of the i conditions: σ E σ E m eddown Sustitute numeicl vlues nd evlute σ m : σ m ( 8.85 /N m )( M/m) 5 /m

35 Electic Potentil 5 A conducting spheicl shell of inne dius nd oute dius c is concentic with smll metl sphee of dius <. The metl sphee hs positive chge Q. The totl chge on the conducting spheicl shell is Q. (Assume the potentil is zeo vey f fom ll chges.) () Wht is the electic potentil of the spheicl shell? () Wht is the electic potentil of the metl sphee? Pictue the Polem The digm is coss-sectionl view showing the chges on the sphee nd the spheicl conducting shell. A potion of the Gussin sufce ove which we ll integte E in ode to find in the egion > is lso shown. Fo < <, the sphee cts like point chge Q nd the potentil of the metl sphee is the sum of the potentil due to point chge t its cente nd the potentil t its sufce due to the chge on the inne sufce of the spheicl shell. () Epess > : > E> d Apply Guss s lw fo > : S Q E ˆ nda enclosed nd E > ecuse Q enclosed fo >. Sustitute fo E > to otin: ( ) > d () Epess the potentil of the metl sphee: Q t its cente sufce Epess the potentil t the sufce of the metl sphee: sufce k ( Q)

36 6 hpte Sustitute nd simplify to otin: 5 [SSM] Two coil conducting cylindicl shells hve equl nd opposite chges. The inne shell hs chge q nd n oute dius, nd the oute shell hs chge q nd n inne dius. The length of ech cylindicl shell is, nd is vey long comped with. Find the potentil diffeence, etween the shells. Pictue the Polem The digm is coss-sectionl view showing the chges on the inne nd oute conducting shells. A potion of the Gussin sufce ove which we ll integte E in ode to find in the egion < < is lso shown. Once we ve detemined how E vies with, we cn find fom E d. Epess the potentil diffeence : Apply Guss s lw to cylindicl Gussin sufce of dius nd length : Solving fo E yields: Ed q E nˆ da E ( π) S q π E E d Sustitute fo E nd integte fom to : q π d kq ln 6 Positive chge is plced on two conducting sphees tht e vey f pt nd connected y long vey-thin conducting wie. The dius of the smlle sphee is 5. cm nd the dius of the lge sphee is. cm. The electic field stength t the sufce of the lge sphee is k/m. Estimte the sufce chge density on ech sphee.

37 Electic Potentil 7 Pictue the Polem et nd S efe to the lge nd smlle sphees, espectively. We cn use the fct tht oth sphees e t the sme potentil to estimte the electic fields ne thei sufces. Knowing the electic fields, we cn use σ E to estimte the sufce chge density of ech sphee. Epess the electic fields t the sufces of the two sphees: S E S nd E S Divide the fist of these equtions y the second to otin: E E S S S QS Q S Becuse the potentils e equl t the sufces of the sphees: Q S nd S S S Q Sustitute fo QS to otin: Q E E S S E S E S S S Sustitute numeicl vlues nd evlute E S : E.cm 5.cm ( k/m) 8k/m S Use σ E to estimte the sufce chge density of ech sphee: nd ( 8.85 /N m )( k/m).77 μ σ cm E cm /m ( 8.85 /N m )( 8 k/m).5μ σ 5 cm E5 cm /m 7 Two concentic conducting spheicl shells hve equl nd opposite chges. The inne shell hs oute dius nd chge q; the oute shell hs inne dius nd chge q. Find the potentil diffeence, etween the shells. Pictue the Polem The digm is coss-sectionl view showing the chges on the concentic spheicl shells. The Gussin sufce ove which we ll integte E in ode to find in the egion is lso shown. We ll lso find E in the egion fo which < <. We cn then use the eltionship Ed to find nd nd thei diffeence.

38 8 hpte q Gussin Sufce q Epess : E d Apply Guss s lw fo : S Q E ˆ nda enclosed nd E ecuse Q fo. enclosed Sustitute fo Epess : E to otin: E () d d Apply Guss s lw fo : ( π ) E nd q q E π kq Sustitute fo E to otin: d kq kq kq The potentil diffeence etween the shells is given y: kq 8 The electic potentil t the sufce of unifomly chged sphee is 5. At point outside the sphee t (dil) distnce of. cm fom its

39 Electic Potentil 9 sufce, the electic potentil is 5. (The potentil is zeo vey f fom the sphee.) Wht is the dius of the sphee, nd wht is the chge of the sphee? Pictue the Polem et e the dius of the sphee nd Q its chge. We cn epess the potentil t the two loctions given nd solve the esulting equtions simultneously fo nd Q. elte the potentil of the sphee t its sufce to its dius: Epess the potentil t distnce of. cm fom its sufce: Divide eqution () y eqution () to otin: 5 5.m.m 5 5 o.m () ().cm Solving eqution () fo Q yields: ( ) k Q 5 Sustitute numeicl vlues nd evlute Q: Q ( ) 5.n (. m) 5 9 ( N m / ) 9 onside two infinite pllel thin sheets of chge, one in the plne nd the othe in the plne. The potentil is zeo t the oigin. () Find the electic potentil eveywhee in spce if the plnes hve equl positive chge densities σ. () Find the electic potentil eveywhee in spce if the sheet in the plne hs chge density σ nd the sheet in the plne hs chge density σ. Pictue the Polem et the chge density on the infinite plne t e σ nd tht on the infinite plne t e σ. ll tht egion in spce fo which <, egion I, the egion fo which < <, egion II, nd the egion fo which <, egion III. We cn integte E due to the plnes of chge to find the electic potentil in ech of these egions.

40 hpte () Epess the potentil in egion I in tems of the electic field in tht egion: Epess the electic field in egion I s the sum of the fields due to the chge densities σ nd σ : I E I E d I σ iˆ σ iˆ σ iˆ σ iˆ σ iˆ Sustitute fo E I nd evlute I : I σ d σ σ σ ( ) Epess the potentil in egion II in tems of the electic field in tht egion: E d II II ( ) Epess the electic field in egion II s the sum of the fields due to the chge densities σ nd σ : E II σ ˆ σ i iˆ σ ˆ σ i iˆ Sustitute fo E II nd evlute II : Epess the potentil in egion III in tems of the electic field in tht egion: II III ( ) d ( ) E d III

41 Epess the electic field in egion III s the sum of the fields due to the chge densities σ nd σ : E III σ iˆ σ iˆ Electic Potentil σ iˆ σ iˆ σ iˆ Sustitute fo E III nd evlute III : III σ d σ ( ) σ σ () Poceed s in () with σ σ nd σ σ to otin: I, σ II, nd III These esults e summized in the following tle: egion Pt () σ σ ( ) Pt () σ 5 The epession fo the potentil long the is of thin unifomly chge disk is given y π kσ z (Eqution -), whee nd z σ e the dius nd the chge pe unit e of the disk, espectively. Show tht this epession educes to z fo z >>, whee Q σπ is the totl chge on the disk. Eplin why this esult is epected. Hint: Use the inomil theoem to epnd the dicl.

42 hpte Pictue the Polem Epnd the dicl epession inomilly to otin: z z ()( )! z highe ode tems Fo z >> : nd z z z z Sustituting in Eqution - yields: π kσ z z πkσ z z The totl chge on the disk is given y: Q σπ Q π σ Sustitute fo σ nd simplify to Q k z Q π k z otin: z π z If z, then z z nd: kzq z z z If z <, then z z nd: k( z) Q z z z Thus, fo z >>, Eqution - educes to: z 5 [SSM] A od of length hs totl chge Q unifomly distiuted long its length. The od lies long the y-is with its cente t the oigin. () Find n epession fo the electic potentil s function of position long the -is. () Show tht the esult otined in Pt () educes to fo >>. Eplin why this esult is epected.

43 Electic Potentil Pictue the Polem et the chge pe unit length e λ Q/ nd dy e line element with chge λdy. We cn epess the potentil d t ny point on the is due to the chge element λdy nd integte to find (, ). () Epess the element of potentil d due to the line element dy: dy k d λ whee y nd Q λ. Sustituting fo nd λ yields: y dy d Use tle of integls to integte d fom y / to y /: ( ) ln, y dy () Fcto fom the numeto nd denominto within the pentheses to otin: ( ) ln, Use ln ln ln to otin: ( ) ln ln, et ε nd use ( )... 8 ε ε ε to epnd :... 8 fo >>.

44 hpte Sustitute fo to otin: et (, ) δ nd use ln( )... ln ln δ δ δ to epnd ln ± : ln fo >>. nd ln Sustitute fo ln nd ln nd simplify to otin: (, ) Becuse, fo >>, the chge cied y the od is f enough wy fom the point of inteest to look like point chge, this esult is wht we would epect. 5 A od of length hs chge Q unifomly distiuted long its length. The od lies long the y-is with one end t the oigin. () Find n epession fo the electic potentil s function of position long the -is. () Show tht the esult otined in Pt () educes to fo >>. Eplin why this esult is epected. Pictue the Polem et the chge pe unit length e λ Q/ nd dy e line element with chge λdy. We cn epess the potentil d t ny point on the is due to λdy nd integte to find (, ). () Epess the element of potentil d due to the line element dy: kλ d dy whee y nd Q λ.

45 Electic Potentil 5 Sustituting fo nd λ yields: y dy d Use tle of integls to integte d fom y to y : ( ) ( ) ( ) ( ) [ ] y y y dy ln ln ln, Becuse ln ln ln : ( ) ln, () () Fcto unde the dicl to otin: ln ln ln Becuse >> : ln ln ln Epnding ln inomilly yields: s highe ode tem ln Agin, ecuse >> : ln Sustitute in eqution () to otin: ln

46 6 hpte Finlly, sustituting in eqution () yields: (, ) Becuse, fo >>, the chge cied y the od is f enough wy fom the point of inteest to look like point chge, this esult is wht we would epect. 5 [SSM] A disk of dius hs sufce chge distiution given y σ σ / whee σ is constnt nd is the distnce fom the cente of the disk. () Find the totl chge on the disk. () Find n epession fo the electic potentil t distnce z fom the cente of the disk on the is tht psses though the disk s cente nd is pependicul to its plne. Pictue the Polem We cn find Q y integting the chge on ing of dius nd thickness d fom to nd the potentil on the is of the disk y integting the epession fo the potentil on the is of ing of chge etween the sme limits. σ d () Epess the chge dq on ing of dius nd thickness d: dq πσd π σ πσ d d Integte fom to to otin: Q πσ d πσ ()Epess the potentil on the is of the disk due to cicul element πσ of chge dq d : d kdq πkσ ' d Integte fom to to otin: πkσ d πkσ

47 Electic Potentil 7 5 A disk of dius hs sufce chge distiution given y σ σ / whee σ is constnt nd is the distnce fom the cente of the disk. () Find the totl chge on the disk. () Find n epession fo the electic potentil t distnce fom the cente of the disk on the is tht psses though the disk s cente nd is pependicul to its plne. Pictue the Polem We cn find Q y integting the chge on ing of dius nd thickness d fom to nd the potentil on the is of the disk y integting the epession fo the potentil on the is of ing of chge etween the sme limits. σ d () Epess the chge dq on ing of dius nd thickness d: dq πσd πσ πσ d d Integte fom to to otin: Q πσ d πσ () Epess the potentil on the is of the disk due to cicul element of chge dq πσd : d kdq ' πkσ d Integte fom to to otin: πkσ d πkσ ln 55 A od of length hs totl chge Q unifomly distiuted long its length. The od lies long the -is with its cente t the oigin. () Wht is the electic potentil s function of position long the -is fo > /? () Show tht fo >> /, you esult educes to tht due to point chge Q. Pictue the Polem We cn epess the electic potentil d t due to n elementl chge dq on the od nd then integte ove the length of the od to find (). In the second pt of the polem we use inomil epnsion to show tht, fo >> /, ou esult educes to tht due to point chge Q.