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.

48 8 hpte () Epess the potentil t due to the element of chge dq locted t u: u du k kdq d λ o, ecuse λ Q/, u du d Integte fom u / to u / nd simplify to otin: ( ) ( ) ( ) ( ) [ ] u u du ln ln ln ln () Divide the numeto nd denominto of the gument of the logithm y to otin: ln ln ln whee /. Divide y nd simplify to otin: ln ln ln ln povided >>.

49 Electic Potentil 9 Epnd otin: ln inomilly to ln povided >>. Sustitute to epess () fo >> /: ( ), the potentil due to point chge Q. 56 A cicle of dius is emoved fom the cente of unifomly chged thin cicul disk of dius nd chge pe unit e σ. () Find n epession fo the potentil on the is distnce fom the cente of the disk. () Show tht fo >> the electic potentil on the is of the unifomly chged disk with cutout ppoches /, whee Q σπ( ) is the totl chge on the disk. Pictue the Polem The potentil on the is of the unifomly chged disk is the sum of the potentil due to the disk of dius nd the potentil due to the disk of dius tht hs een emoved. We cn think of the chged disk tht hs een emoved s hving negtive chge density σ. Note tht if >>, then it is lso tue tht >>. ()The potentil on the is of the cicul disk is: ( ) ( ) ( ) whee π kσ ( ) [( ) ] nd ( ) π k( σ )[( ) ] Sustitute fo ( ) nd ( ) nd simplify to otin: ( ) πkσ [( ) ] ( )[( ) ] πk σ πkσ [( ) ( ) ] πkσ ( ) () Epnding yields: inomilly highe ode tems

50 hpte Epnding yields: inomilly Sustituting in the epession fo ( ) nd simplifying yields: πkσ ( ) ( ) π kσ πkσ The totl chge on the disk is: Q ( ) Q σπ σ π ( ) Sustituting fo σ yields: Q πk π ( ) ( ) ( ) 57 The epession fo the electic potentil inside unifomly chged solid sphee is given y ( ), whee is the dius of the sphee nd is the distnce fom the cente. This epession ws otined in Emple - y fist finding the electic field. In this polem, you deive the sme epession y modeling the sphee s nested collection of thin spheicl shells, nd then dding the potentils of these shells t field point inside the sphee. The potentil d tht is distnce fom the cente of unifomly chged thin spheicl shell tht hs dius nd chge dq is given y d kdq/ fo nd d kdq/ fo (Eqution -). onside sphee of dius contining chge Q tht is unifomly distiuted nd you wnt to find t some point inside the sphee (tht is fo < ). () Find n epession fo the chge dq on spheicl shell of dius nd thickness d. () Find n epession fo the potentil d t due to the chge in shell of dius nd thickness d, whee. (c) Integte you epession in Pt () fom to to find the potentil t due to ll the chge in the egion fthe fom thn the cente of the sphee. (d) Find n epession fo the potentil d t due to the chge in shell of dius nd thickness d, whee. (e) Integte you epession in Pt (d) fom to to find the potentil t due to ll the chge in the egion close thn to the cente of the sphee. (f) Find the totl potentil t y dding you Pt (c) nd Pt (e) esults. Pictue the Polem The digm shows unifomly chged sphee of dius nd the field point P t which we wish to find the totl potentil. We cn use the definition of chge density to find the chge inside sphee of dius nd the potentil t due to this chge. We cn epess the potentil t due to the

51 Electic Potentil chge in shell of dius nd thickness d t using then integte this epession fom to to find. d kdq' nd Q ' P () The chge dq in shell of dius nd thickness d t > is given y: dq ρ d ' ρa' d' Becuse A' π ' : dq π ' ρd' Becuse the sphee is unifomly chged: Sustituting fo ρ nd simplifying yields: Q Q ρ π Q π Q dq π ' d' π Q ' d' () Epess the potentil d in the kdq d intevl ' due to dq: ' Sustituting fo dq nd simplifying yields: k Q d ' d' ' d' ' (c) Integte d fom to 'd' ( ) to find : (d) The potentil d t due to the chge in shell of dius nd thickness d, whee, is given y: d kdq kρ kρd ' kρa' d' ( π ' ) d' π kρ ' d'

52 hpte Sustituting fo ρ nd simplifying yields: d πk Q π ' ' d' d' (e) Integte fom to to find the potentil t due to ll the chge in the egion close thn to the cente of the sphee: ' d' (f) The sum of ou esults fom Pt (c) nd Pt (e) is: ( ) 58 lculte the electic potentil t the point distnce / fom the cente of unifomly chged thin spheicl shell of dius nd chge Q. (Assume the potentil is zeo f fom the shell.) Pictue the Polem We cn find the potentil eltive to infinity t distnce / fom the cente of the spheicl shell y integting the electic field fo to. We cn pply Guss s lw to find the electic field oth inside nd outside the spheicl shell. The potentil eltive to infinity t distnce / fom the cente of the spheicl shell is given y: E d E> d E< d Becuse thee is no chge inside the spheicl shell, E < nd: Apply Guss s lw to spheicl sufce of dius > to otin: Solving fo E > yields: Sustitute fo E > in eqution () to otin: < S E d nd E d () E ( π ) Q inside nda E > Q E > π d > Q

53 Electic Potentil Evluting this integl yields: 59 [SSM] A cicle of dius is emoved fom the cente of unifomly chged thin cicul disk of dius. Show tht the potentil t point on the centl is of the disk distnce z fom its geometicl cente is given y ( ) ( ) z k z z π σ, whee σ is the chge density of the disk. Pictue the Polem We cn find the electosttic potentil of the conducting wshe y teting it s two disks with equl ut opposite chge densities. The electic potentil due to chged disk of dius is given y: ( ) k σ π Supeimpose the electosttic potentils of the two disks with opposite chge densities nd simplify to otin: ( ) ( ) k k k k k k k k σ π σ π σ π σ π σ π σ π σ π σ π The chge density σ is given y: ( ) Q π σ Sustituting fo σ yields: ( ) ( ) ( )

54 hpte Equipotentil Sufces 6 An infinite flt sheet of chge hs unifom sufce chge density equl to.5 μ/m. How f pt e the equipotentil sufces whose potentils diffe y? Pictue the Polem We cn equte the epession fo the electic field due to n infinite plne of chge nd Δ/Δ nd solve the esulting eqution fo the seption of the equipotentil sufces. Epess the electic field due to the σ E infinite plne of chge: elte the electic field to the potentil: Equte these epessions nd solve fo Δ to otin: Sustitute numeicl vlues nd evlute Δ : Δ E Δ Δ σ Δ 8.85 N m.5μ/m Δ.56 mm ( ) 6 [SSM] onside two pllel unifomly chged infinite plnes tht e equl ut oppositely chged. () Wht is (e) the shpe(s) of the equipotentils in the egion etween them? Eplin you nswe. () Wht is (e) the shpe(s) of the equipotentils in the egions not etween them? Eplin you nswe. Pictue the Polem The two pllel plnes, with thei opposite chges, e shown in the pictoil epesenttion. z Q Q y

55 Electic Potentil 5 () Becuse the electic field etween the chged pltes is unifom nd pependicul to the pltes, the equipotentil sufces e plnes pllel to the chged plnes. () The egions to eithe side of the two chged plnes e equipotentil egions, so ny sufce in eithe of these egions is n equipotentil sufce. 6 A Geige tue consists of two elements, long metl cylindicl shell nd long stight metl wie unning down its centl is. Model the tue s if oth the wie nd cylinde e infinitely long. The centl wie is positively chged nd the oute cylinde is negtively chge. The potentil diffeence etween the wie nd the cylinde is. k. () Wht is the diection of the electic field inside the tue? () Which element is t highe electic potentil? (c) Wht is (e) the shpe(s) of the equipotentils inside the tue? (d) onside two equipotentils descied in Pt (c). Suppose they diffe in electic potentil y. Do two such equipotentils ne the centl wie hve the sme spcing s they would ne the oute cylinde? If not, whee in the tue e the equipotentils tht e moe widely spced? Eplin you nswe. Detemine the oncept () The diection of the electic field inside the tue is the diection of the foce the electic field eets on positively chged oject. Becuse the wie is positively chged nd the tue is negtively chged, nd ecuse of the cylindicl geomety of the Geige tue, the electic field is diected dilly wy fom the centl wie. () Becuse it would equie moe wok to ing positively chged oject fom infinity to the sufce of the wie thn it would to ing this test oject to the sufce of the cylindicl tue, the centl wie is t the highe electic potentil. (c) Becuse of the cylindicl geomety of the Geige tue, the equipotentil sufces e cylindes concentic with the centl wie. (d) No. Becuse the mgnitude of the electic field, which is the te of chnge with distnce (lso known s the gdient) of the potentil deceses with distnce fom the wie, the spcing etween djcent equipotentil sufces hving the sme potentil diffeence etween them deceses s you get fthe fom the centl wie. 6 [SSM] Suppose the cylinde in the Geige tue in Polem 6 hs n inside dimete of. cm nd the wie hs dimete of.5 mm. The cylinde is gounded so its potentil is equl to zeo. () Wht is the dius of the equipotentil sufce tht hs potentil equl to 5? Is this sufce close to the wie o to the cylinde? () How f pt e the equipotentil sufces tht

56 6 hpte hve potentils of nd 5? (c) ompe you esult in Pt () to the distnce etween the two sufces tht hve potentils of 7 nd 75 espectively. Wht does this compison tell you out the electic field stength s function of the distnce fom the centl wie? Pictue the Polem If we let the electic potentil of the cylinde e zeo, then the sufce of the centl wie is t nd we cn use Eqution - to find the electic potentil t ny point etween the oute cylinde nd the centl wie. () Fom Eqution - we hve: ef () kλ ln () whee ef is the dius of the oute cylinde nd is the distnce fom the cente of the centl wie nd < ef. Solving fo kλ yields: ( ) kλ ln ef At the sufce of the wie, nd.5 mm. Hence: k λ 8.. cm ln.5 mm nd. cm () ( 8. ) ln Setting 5 yields: 5 ( 8. ) ln. cm o. cm. cm ln.9 e.9 Solve fo to otin:. cm e.9. cm, close to the wie. () The seption of the equipotentil sufces tht hve potentil vlues of nd 5 is: Solving eqution () fo yields: Δ () 5 kλ ef e 8. (. cm) e

57 Electic Potentil 7 Sustitute fo the dii in eqution (), simplify, nd evlute Δ to otin: Δ (. cm) e (. cm) e (. cm).86 mm e 5 8. e 8. (c) The distnce etween the 7 nd the 75 equipotentils is: Δ (. cm) e e mm This close spcing of these two equipotentil sufces ws to e epected. lose to the centl wie, two equipotentil sufces with the sme diffeence in potentil should e close togethe to eflect the fct tht the highe electic field stength is gete close to the wie. 6 A point pticle tht hs chge of. n is t the oigin. () Wht is (e) the shpes of the equipotentil sufces in the egion ound this chge? () Assuming the potentil to e zeo t, clculte the dii of the five sufces tht hve potentils equl to.,., 6., 8. nd., nd sketch them to scle centeed on the chge. (c) Ae these sufces eqully spced? Eplin you nswe. (d) Estimte the electic field stength etween the.- nd 6.- equipotentil sufces y dividing the diffeence etween the two potentils y the diffeence etween the two dii. ompe this estimte to the ect vlue t the loction midwy etween these two sufces. Pictue the Polem We cn integte the epession fo the electic field due to point chge to find n epession fo the electic potentil of the point pticle. () The equipotentil sufces e sphees centeed on the chge. () Fom the eltionship etween the electic potentil due to the point chge nd the electic field of the point chge we hve: Tking the potentil to e zeo t yields: o d E d d

58 8 hpte Becuse Q. 8 : N m 8 (. ) 9 N m () Use eqution () to complete the following tle: () (m) The equipotentil sufces e shown in coss-section to the ight: point chge (c) No. The equipotentil sufces e closest togethe whee the electic field stength is getest. (d) The vege vlue of the mgnitude of the electic field etween the.- nd 6.- equipotentil sufces is given y: Δ E Δ 6 Δ Dop pependiculs to the is fom. nd 6. to ppoimte the dii coesponding to ech of these potentil sufces: E 6. m.7 m est 9 m The ect vlue of the electic field t the loction midwy etween these two sufces is given y E, whee is the vege of the dii of the.- nd 6.- equipotentil sufces. Sustitute numeicl vlues nd evlute E ect.

59 E 9 N m m.9 m 8 (. ) ect Electic Potentil 9 m The estimted vlue fo E diffes y out % fom the ect vlue. Electosttic Potentil Enegy 65 Thee point chges e on the -is: q is t the oigin, q is t. m, nd q is t 6. m. Find the electosttic potentil enegy of this system of chges fo the following chge vlues: () q q q. μ; () q q. μ nd q. μ; nd (c) q q. μ nd q. μ. (Assume the potentil enegy is zeo when the chges e vey f fom ech othe.) The electosttic potentil enegy of this system of thee point chges is the wok needed to ing the chges fom n infinite seption to the finl positions shown in the digm. q q q 6, m Epess the wok equied to ssemle this system of chges: U kqq, qq k, kqq, qq, kqq, qq, Find the distnces,,,, nd, :, m,, m,nd, 6m () Evlute U fo q q q. μ: 9 N m U mj (. μ)(. μ) (. μ)(. μ).m (. μ)(. μ).m 6.m

60 5 hpte () Evlute U fo q q. μ nd q. μ: U 9 N m mJ (.μ)(.μ) (.μ)(.μ).m (.μ)(.μ).m 6.m (c) Evlute U fo q q. μ nd q. μ: U 9 N m mj (. μ)(. μ) (. μ)(. μ).m (. μ)(. μ).m 6.m 66 Point chges q, q, nd q e fied t the vetices of n equiltel tingle whose sides e.5 m-long. Find the electosttic potentil enegy of this system of chges fo the following chge vlues: () q q q. μ, () q q. μ nd q. μ; nd (c) q q. μ nd q. μ. (Assume the potentil enegy is zeo when the chges e vey f fom ech othe.) Pictue the Polem The electosttic potentil enegy of this system of thee point chges is the wok needed to ing the chges fom n infinite seption to the finl positions shown.5 m q.5 m in the digm. q q.5 m Epess the wok equied to ssemle this system of chges: U kqq, qq k, kqq, qq, kqq, qq, Find the distnces,,,, nd, :,,,.5m

61 Electic Potentil 5 () Evlute U fo q q q. μ: U 9 N m mJ (.μ)(.μ) (.μ)(.μ) (.μ)(.μ).5m.5m.5m () Evlute U fo q q. μ nd q. μ: U 9 N m mj (.μ)(.μ) (.μ)(.μ).5m (.μ)(.μ).5m.5m (c) Evlute U fo q q. μ nd q. μ: U 9 N m mj (. μ)(. μ) (. μ)(. μ).5m (. μ)(. μ).5m.5m 67 [SSM] () How much chge is on the sufce of n isolted spheicl conducto tht hs.-cm dius nd is chged to. k? ()Wht is the electosttic potentil enegy of this conducto? (Assume the potentil is zeo f fom the sphee.) Pictue the Polem The potentil of n isolted spheicl conducto is given y, whee Q is its chge nd its dius, nd its electosttic potentil enegy yu Q. We cn comine these eltionships to find the sphee s electosttic potentil enegy. () The potentil of the isolted spheicl conducto t its sufce is elted to its dius: Q k whee is the dius of the spheicl conducto.

62 5 hpte Sustitute numeicl vlues nd evlute Q: (. cm)(. k) Q N m.5 n. n () Epess the electosttic potentil enegy of the isolted spheicl conducto s function of its chge Q nd potentil : U Q Sustitute numeicl vlues nd evlute U: U (.5 n)(. k). J μ 68 Fou point chges, ech hving chge with mgnitude of. μ, e t the cones of sque whose sides e. m-long. Find the electosttic potentil enegy of this system unde the following conditions: () ll of the chges e negtive, () thee of the chges e positive nd one of the chges is negtive, nd (c) the chges t two djcent cones e positive nd the othe two chges e negtive. (d) the chges t two opposite cones e positive nd the othe two chges e negtive. (Assume the potentil enegy is zeo when the point chges e vey f fom ech othe.) Pictue the Polem The electosttic potentil enegy of this system of fou point chges is the wok needed to ing the chges fom n infinite seption to the finl positions shown in the digm. q. m q. m q q. m The wok equied to ssemle this system of chges equls the potentil enegy of the ssemled system: kqq U, qq k, kqq, qq, kqq, qq, kqq qq,, kqq qq,, qq kqq,,

63 Electic Potentil 5 Find the distnces,,,,,,,,,, nd,,:,,,, nd,,. m.m () Evlute U fo q q q q. μ: U 9 N m mj (. μ)(. μ) (. μ)(. μ). m (. μ)(. μ) (. μ)(. μ). m (. μ)(. μ) (. μ)(. μ). m. m. m. m () Evlute U fo q q q μ nd q. μ: U 9 N m (. μ)(. μ) (. μ)(. μ). m (. μ)(. μ) (. μ)(. μ). m (. μ)(. μ) (. μ)(. μ). m. m. m. m (c) et q q. μ nd q q. μ: 9 N m U mj (. μ)(. μ) (. μ)(. μ).m (. μ)(. μ) (. μ)(. μ) (. μ)(. μ) (. μ)(. μ)..m m. m.m.m

64 5 hpte (d) et q q. μ nd q q. μ: U 9 N m mj (. μ)(. μ) (. μ)(. μ). m (. μ)(. μ) (. μ)(. μ). m (. μ)(. μ) (. μ)(. μ). m. m. m. m 69 [SSM] Fou point chges e fied t the cones of sque centeed t the oigin. The length of ech side of the sque is. The chges e locted s follows: q is t (, ), q is t (, ), q is t (, ), nd 6q is t (, ). A fifth pticle tht hs mss m nd chge q is plced t the oigin nd elesed fom est. Find its speed when it is vey f fom the oigin. Pictue the Polem The digm shows the fou point chges fied t the cones of the sque nd the fifth chged pticle tht is elesed fom est t the oigin. We cn use consevtion of enegy to elte the initil potentil enegy of the pticle to its kinetic enegy when it is t get distnce fom the oigin nd the electosttic potentil t the oigin to epess U i. q 6q y m,q q q Use consevtion of enegy to elte the initil potentil enegy of the pticle to its kinetic enegy when it is t get distnce fom the oigin: Epess the initil potentil enegy of the pticle to its chge nd the electosttic potentil t the oigin: Sustitute fo K f nd U i to otin: ΔK ΔU o, ecuse K i U f, K U f i U q i ( ) ( ) mv q v q m ()

65 Electic Potentil 55 Epess the electosttic potentil t the oigin: () kq 6kq kq kq 6kq Sustitute fo () nd simplify to otin: v q 6kq m q 6 k m 7 onside two point pticles tht ech hve chge e, e t est, nd e septed y.5 5 m. () How much wok ws equied to ing them togethe fom vey lge seption distnce? () If they e elesed, how much kinetic enegy will ech hve when they e septed y twice thei seption t elese? (c) The mss of ech pticle is. u (. AMU). Wht speed will ech hve when they e vey f fom ech othe? Pictue the Polem () In the sence of othe chged odies, no wok is equied to ing the fist poton fom infinity to its initil position. We cn use the wok- enegy theoem to find the wok equied to ing the second poton to position.5 5 m wy fom the fist poton. () nd (c) We cn pply consevtion of mechnicl enegy to the two-poton system to find the kinetic enegy of ech poton when they e septed y twice thei seption t elese nd when they e septed y lge distnce. () Apply the wok-enegy theoem to the second poton to otin: W ke Δ K ΔU et ke Sustitute numeicl vlues nd evlute W et : W et 96 ke 9 (.6 ) 9 N m e.58 J 5.5 m.6 9 J () Apply consevtion of mechnicl enegy to the septing potons to otin: Δ K ΔU K K U U o, ecuse K i, K U U f f i f i f i Sustituting fo U i nd U f nd simplifying yields: K U f i ke U f ke i ke f ke ke

66 56 hpte ememeing tht K f is the kinetic enegy of oth potons, sustitute numeicl vlues nd evlute K f, ech poton : K f, ech poton ke N m 9 (.6 ) 5 ( m) e J.6 9 J (c) Apply consevtion of mechnicl enegy to the septing potons to otin: Solving fo v yields: Δ K ΔU K f Ki U f U i o, ecuse K i U f, K U m v U f i v U m p i p whee U i is hlf the initil potentil enegy of the two-poton system. i Sustitute numeicl vlues nd evlute v : v.6 8 ke e 7.67 kg m/s 9 J 7 onside n electon nd poton tht e initilly t est nd e septed y. nm. Neglecting ny motion of the much moe mssive poton, wht is the minimum () kinetic enegy nd () speed tht the electon must e pojected t so it eches point distnce of. nm fom the poton? Assume the electon s velocity is diected dilly wy fom the poton. (c) How f will the electon tvel wy fom the poton if it hs twice tht initil kinetic enegy? Pictue the Polem We cn pply the consevtion of mechnicl enegy to the electon-poton system to find the minimum initil kinetic enegy (tht is, the initil kinetic enegy tht coesponds to finl kinetic enegy of zeo) equied in Pt () nd, in Pt (c), to find how f the electon will tvel wy fom the poton if it hs twice the initil kinetic enegy found in Pt ().

67 Electic Potentil 57 () Apply consevtion of mechnicl enegy to the electonsttiony poton system with K f to otin: Solving fo fo U f nd Ki, min U i yields: nd sustituting Δ K ΔU K U U i, min f i ke ke K i, min U f U i f i o, simplifying futhe, K i, min ke i f Sustitute numeicl vlues nd evlute K i, min : K i, min N m J 9 (.6 ) 9.6 J. nm. nm () elte the minimum initil kinetic enegy of the electon to its initil speed: Sustitute numeicl vlues nd evlute v i : K v v i, min m evi ( J) i.59 5 i m/s kg K i, min m e (c) Apply consevtion of mechnicl enegy to the electon- poton system to otin: Assuming, s we did in Pt (), tht K f yields: Sustituting fou nd Solve fo f to otin: f U i yields: Δ K ΔU K K U U K U U i, min f i K ke f i f i ke i, min f i i f K ke i, min

68 58 hpte Sustituting numeicl vlues nd evluting f yields negtive vlue; esult tht is not physicl nd suggests tht, conty to ou ssumption, K f. To confim tht this is the cse, ssume tht the electon escpes fom the poton (it s finl electosttic potentil will then e equl to zeo) nd find the initil kinetic enegy equied fo this to occu. If the electon is to escpe the influence of the poton, its finl electosttic potentil enegy will e zeo nd: ke ke K i, escpe U i i i Sustitute numeicl vlues nd evlute K i, escpe : K i, escpe N m. nm 9 (.6 ).5 9 J Becuse K i, min > Ki, escpe, the electon escpes fom the poton with esidul kinetic enegy. Genel Polems 7 A positive point chge equl to.8 9 is septed fom negtive point chge of the sme mgnitude y 6. m. Wht is the electic potentil t point 9. m fom ech of the two chges? Pictue the Polem Becuse the chges e point chges, we cn use the epession fo the oulom potentil to find the field t ny distnce fom them. Using the epession fo the potentil due to system of point chges, epess the potentil t the point 9. m fom ech of the two chges: kq k kq ( q q ) Becuse q q : 7 [SSM] Two positive point chges, ech hve chge of q, nd e fied on the y-is t y nd y. () Find the electic potentil t ny point on the -is. () Use you esult in Pt () to find the electic field t ny point on the -is.

69 Electic Potentil 59 Pictue the Polem The potentil t ny point on the is is the sum of the oulom potentils due to the two point chges. Once we hve found, we cn use E ( )iˆ to find the electic field t ny point on the is. y q q () Epess the potentil due to system of point chges: i kqi i Sustitute to otin: ( ) chge t kq kq chge t - kq () The electic field t ny point on the is is given y: E( ) iˆ kq ( ) d d iˆ kq iˆ 7 If conducting sphee is to e chged to potentil of. k, wht is the smllest possile dius of the sphee so tht the electic field ne the sufce of the sphee will not eceed the dielectic stength of i? Pictue the Polem The dius of the sphee is elted to the electic field nd the potentil t its sufce. The dielectic stength of i is out M/m. elte the electic field t the sufce of conducting sphee to the potentil t the sufce of the sphee: When E is mimum, is minimum: E min E ( ) ( ) m ( ) E Sustitute numeicl vlues nd evlute min : min. k M/m mm

70 6 hpte 75 [SSM] Two infinitely long pllel wies hve unifom chge pe unit length λ nd λ espectively. The wies e pllel with the z-is. The positively chged wie intesects the -is t, nd the negtively chged wie intesects the -is t. () hoose the oigin s the efeence point whee the potentil is zeo, nd epess the potentil t n ity point (, y) in the y plne in tems of, y, λ, nd. Use this epession to solve fo the potentil eveywhee on the y is. () Using 5. cm nd λ 5. n/m, otin the eqution fo the equipotentil in the y plne tht psses though the point, y. (c) Use spedsheet pogm to plot the equipotentil found in pt (). Pictue the Polem The geomety of the wies is shown elow. The potentil t the point whose coodintes e (, y) is the sum of the potentils due to the chge distiutions on the wies. λ (,y) y λ () Epess the potentil t the point whose coodintes e (, y): (, y) wie t ef kλ ln λ ln π whee (). wie t k ( λ) ef kλ ln ln ef ln ef y Becuse ( ) nd ( ) : y (, y) λ ln π ( ) y ( ) y On the y-is, nd: (, y) λ ln π λ ln π y y ()

71 Electic Potentil 6 () Evlute the potentil t,.5cm, ( ) ( ): (,) λ ln π λ ln π 5 ( ) ( ) Equte (,y) nd (,) : ( 5) y 5 ( 5) y Solve fo y to otin: y ±.5 5 (c) A spedsheet pogm to plot y ±.5 5 is shown elow. The fomuls used to clculte the quntities in the columns e s follows: ell ontent/fomul Algeic Fom A.5 A A.5 Δ B SQT(.5*A A^ 5) y.5 5 B B y.5 5 A B y pos y neg

72 6 hpte The following gph shows the equipotentil cuve in the y plne fo λ (,) ln. π y (cm) (cm) 76 The equipotentil cuve gphed in Polem 75 should e cicle. () Show mthemticlly tht it is cicle. () The equipotentil cicle in the y plne is the intesection of thee-dimensionl equipotentil sufce nd the y plne. Descie the thee-dimensionl sufce using one o two sentences. Pictue the Polem We cn use the epession fo the potentil t ny point in the y plne to show tht the equipotentil cuve is cicle. () Equipotentil sufces must stisfy the condition: λ ln π Solving fo / yields: π λ e whee is constnt. o [ y ] Sustitute fo nd to otin: ( ) y ( ) Epnd this epession, comine like tems, nd simplify to otin: y

73 Electic Potentil 6 omplete the sque y dding to oth sides of the eqution: y ( ) etting α nd β yields: ( α ) y β, the eqution of cicle in the y plne with its cente t (α,). () The thee-dimensionl sufces e cylindes whose es e pllel to the wies nd e in the y plne. 77 The hydogen tom in its gound stte cn e modeled s positive point chge of mgnitude e (the poton) suounded y negtive chge distiution tht hs chge density (the electon) tht vies with the distnce fom the cente of the poton s: ρ ( ) ρe ( esult otined fom quntum mechnics), whee.5 nm is the most pole distnce of the electon fom the poton. () lculte the vlue of ρ needed fo the hydogen tom to e neutl. () lculte the electosttic potentil (eltive to infinity) of this system s function of the distnce fom the poton. Pictue the Polem () Epessing the chge dq in spheicl shell of volume π d within distnce of the poton nd setting the integl of this epession equl to the chge of the electon will llow us to solve fo the vlue of ρ needed fo chge neutlity. () The electosttic potentil of this system is the sum of the electosttic potentils due to the poton nd electon s chge density. The potentil due to the poton is ke/. We cn use the given chge density to epess the potentil function due to the electon s chge distiution nd then integte this function to find the potentil due to the electon. () Epess the chge dq in spheicl shell of volume d π d t distnce fom the poton: dq ρd πρ ( ρ e )( π d) e d Epess the condition fo chge neutlity: e πρ e d Fom tle of integls we hve: e e d ( )

74 6 hpte Using this esult yields: Sustitute in the epession fo e to otin: e e d e π πρ πρ ρ Sustitute numeicl vlues nd.6 ρ evlute ρ : π (.5 nm) /m () The electosttic potentil of this poton-electon system is the sum of the electosttic potentils due to the poton nd the electon s chge density: Sustituting fo ( ' ) ρ in the epession fo Q yields: Fom tle of integls we hve: () whee ke, kρ( ' ) π ' d' () ' nd ( ' ) ' ' Q ρ π d ' Q πρ ' e d' e e d ( ) Using this esult to evlute ' ' e d' yields: nd e d' e 8 e 8 e 8 8 ( )

75 Electic Potentil 65 8 e Q πρ Sustituting fo Q in the epession fo yields: 8 e k ke ρ π Sustitute fo ρ fom Pt () nd simplify to otin: 8 e ke e e k ke π π Sustituting fo ( ) ' ρ in eqution () nd simplifying yields: d e k d e k ' ' ' ' ' ρ π π ρ Fom tle of integls we hve: ( ) e d e Using this esult to evlute d e ' ' yields: e e d e Sustitute fo d e ' ' nd ρ in the epession fo to otin: e ke e e k π π Sustituting fo nd in eqution () nd simplifying yields: e ke e ke e ke

76 66 hpte 78 hge is supplied to the metl dome of n de Gff geneto y the elt t the te of μ/s when the potentil diffeence etween the elt nd the dome is.5 M. The dome tnsfes chge to the tmosphee t the sme te, so the.5 M potentil diffeence is mintined. Wht minimum powe is needed to dive the moving elt nd mintin the.5 M potentil diffeence? Pictue the Polem We cn use the definition of powe nd the epession fo the wok done in moving chge though potentil diffeence to find the minimum powe needed to dive the moving elt. elte the powe needed to dive the moving elt to the te t which the geneto is doing wok: dw P dt Epess the wok done in moving chge q though potentil diffeence Δ: W qδ Sustitute fo W to otin: P [ qδ ] d dt Δ dq dt Sustitute numeicl vlues nd evlute P: (.5M)( /s) 5 W P μ 79 A positive point chge Q is locted on the is t. () How much wok is equied to ing n identicl point chge fom infinity to the point on the is t? () With the two identicl point chges in plce t nd, how much wok is equied to ing thid point chge Q fom infinity to the oigin? (c) How much wok is equied to move the chge Q fom the oigin to the point on the is t long the semicicul pth shown (Figue -5)? Pictue the Polem We cn use Wq finl position QΔ i f to find the wok equied to move these chges etween the given points. () Epess the equied wok in tems of the chge eing moved nd the potentil due to the chge t nd simplify to otin: W Q QΔ Q ( ) Q Q [ ( ) ( ) ]

77 () Epess the equied wok in tems of the chge eing moved nd the potentils due to the chges t nd nd simplify to otin: (c) Epess the equied wok in tems of the chge eing moved nd the potentils due to the chges t nd nd simplify to otin: W W Q Q QΔ Electic Potentil 67 Q Q ( ) Q Q [ ( ) ( ) ] chge t - QΔ Q[ ( ) ( ) ] Q chge chge t - t Q chge t ( ) 8 A chge of. n is unifomly distiuted on ing of dius. cm tht lies in the plne nd is centeed t the oigin. A point chge of. n is initilly locted on the is t 5. cm. Find the wok equied to move the point chge to the oigin. Pictue the Polem et q epesent the chge eing moved fom 5. cm to the oigin, Q the ing chge, nd the dius of the ing. We cn use Wq finl position qδ i f, whee is the epession fo the il field due to ing chge, to find the wok equied to move q fom 5. cm to the oigin. Epess the equied wok in tems of the chge eing moved nd the potentil due to the ing chge t 5. cm nd : W qδ q [ ( ) (.5 m) ] The potentil on the is of unifomly chged ing is given y: ( ) At 5. cm: (.5 m) (.5 m) At : ( )

78 68 hpte Sustituting fo () nd (.5 m) yields: W q q ( ) ( ).5 m.5 m Sustitute numeicl vlues nd evlute W: W J e N m (. n)(. n).. m 7 J (.5 m) (. m).5 7 e J.6 9 J 8 Two metl sphees ech hve dius of. cm. The centes of the two sphees e 5. cm pt. The sphees e initilly neutl, ut chge Q is tnsfeed fom one sphee to the othe, ceting potentil diffeence etween the sphees of. A poton is elesed fom est t the sufce of the positively chged sphee nd tvels to the negtively chged sphee. () Wht is the poton s kinetic enegy just s it stikes the negtively chged sphee? () At wht speed does it stike the sphee? Pictue the Polem The poton s kinetic enegy just s it stikes the negtively chged sphee is the poduct of its chge nd the potentil diffeence though which it hs een cceleted. We cn find the speed of the poton s it stikes the negtively chged sphee fom its kinetic enegy nd, in tun, its kinetic enegy fom the potentil diffeence though which it is cceleted. () Apply the wok-kinetic enegy theoem to the poton to otin: The net wok done on the poton is given y: W net ΔK Kf Ki o, ecuse K i, W K K W net net f q p p Equting W net nd p K yields: K q e( ) e p p

79 Electic Potentil 69 () Use the definition of kinetic enegy to epess the speed of the poton when it stikes the negtively chged sphee: K p K p mpv v m p Δ m p Sustitute numeicl vlues nd evlute v: v 9 ( )( ) m/s kg 8 () Using spedsheet pogm, gph (z) vesus z fo unifomly chged ing in the z plne nd centeed t the oigin. The potentil on the z is is given y ( z) z (Eqution -9). () Use you gph to estimte the points on the z is whee the electic field stength is getest. Pictue the Polem () The electic field stength is getest whee the mgnitude of the slope of the gph of electic potentil is getest. () A spedsheet solution is shown elow fo. The fomuls used to clculte the quntities in the columns e s follows: ell ontent/fomul Algeic Fom A A. z Δz B /(A^)^(/) A B z/ (z/) z

80 7 hpte The following gph, plotted using spedsheet pogm, shows s function of z/:..8 () z/ Emining the gph we see tht the mgnitude of the slope is mimum t z.7 nd t z A spheicl conducto of dius is chged to k. When it is connected y long vey-thin conducting wie to second conducting sphee f wy, its potentil dops to k. Wht is the dius of the second sphee? Pictue the Polem et e the dius of the second sphee nd Q nd Q the chges on the sphees when they hve een connected y the wie. When the sphees e connected, the chge initilly on the sphee of dius will edistiute until the sphees e t the sme potentil. Epess the common potentil of the k () sphees when they e connected: nd k () Epess the potentil of the fist sphee efoe it is connected to the second sphee: k ( Q Q ) k () Solve eqution () fo Q : ( k) k Q

81 Solve eqution () fo Q : ( k) k Q Electic Potentil 7 Sustitute fo Q nd Q in eqution () nd simplify to otin: k k ( k) ( k) k k k o 8 k 8 A metl sphee centeed t the oigin hs sufce chge density tht hs mgnitude of.6 n/m nd dius less thn. m. A distnce of. m fom the oigin, the electic potentil is 5 nd the electic field stength is 5 /m. (Assume the potentil is zeo vey f fom the sphee.) () Wht is the dius of the metl sphee? () Wht is the sign of the chge on the sphee? Eplin you nswe. Pictue the Polem We cn use the definition of sufce chge density to elte the dius of the sphee to its chge Q nd the potentil function () to elte Q to the potentil t. m. () Use its definition, elte the sufce chge density σ to the chge Q on the sphee nd the dius of the sphee: σ Q π Q πσ elte the potentil to the chge on the sphee: ( ) () Q k Sustitute fo Q in the epession fo nd simplify to otin: ( ) π () πkσ σ () πσ

82 7 hpte Sustitute numeicl vlues nd evlute : ( 8.85 /N m )(.m)( 5).6 n/m 6.cm () The chge on the sphee is positive. The fomul fo the electic potentil outside unifomly chged spheicl shell is. If is positive, then so is Q. 85 Along the centl is of unifomly chged disk, t point.6 m fom the cente of the disk, the potentil is 8 nd the mgnitude of the electic field is 8 /m. At distnce of.5 m, the potentil is nd the mgnitude of the electic field is.5 /m. (Assume the potentil is zeo vey f fom the sphee.) Find the totl chge on the disk. Pictue the Polem We cn use the definition of sufce chge density to elte the dius of the disk to its chge Q nd the potentil function () to elte Q to the potentil t.5 m. Use its definition, elte the sufce chge density σ to the chge Q on the disk nd the dius of the disk: elte the potentil t to the chge on the disk: Q σ π Q πσ () () π kσ ( ) Sustitute (.6 m) 8 : ( ) 8 πkσ.6m.6m Sustitute (.5 m) : ( ) π kσ.5m.5m Divide the fist of these equtions y the second to otin: (.6m) (.5m).5m.6m Solving fo yields:.8m Epess the electic field on the is of disk chge: E π kσ

83 Electic Potentil 7 Solving fo σ yields: Sustitute fo σ in eqution () to otin: E σ πk E π E Q Sustitute numeicl vlues nd evlute Q: π 8.85 Q N m.5m (.5m) (.8m) (.8 m) (.5/m) 7.n 86 A dioctive Po nucleus emits n α-pticle tht hs chge e. When the α-pticle is lge distnce fom the nucleus it hs kinetic enegy of 5. Me. Assume tht the α-pticle hd negligile kinetic enegy s it left t the sufce of the nucleus. The dughte (o esidul) nucleus 6 P hs chge 8e. Detemine the dius of the 6 P nucleus. (Neglect the dius of the α pticle nd ssume the 6 P nucleus emins t est.) Pictue the Polem We cn use U kq q / to elte the electosttic potentil enegy of the pticles to thei seption. Epess the electosttic potentil enegy of the two pticles in tems of thei chge nd seption: kqq U kqq U Sustitute numeicl vlues nd evlute : 9 9 ( N m / )( )( 8)(.6 ).6fm Me e 87 [SSM] onfigution A consists of two point pticles, one pticle hs chge of q nd is on the is t d nd the othe pticle hs chge of q nd is t d (Figue -6). () Assuming the potentil is zeo t lge

84 7 hpte distnces fom these chged pticles, show tht the potentil is lso zeo eveywhee on the plne. () onfigution B consists of flt metl plte of infinite etent nd point pticle locted distnce d fom the plte (Figue -6) The point pticle hs chge equl to q nd the plte is gounded. (Gounding the plte foces its potentil to equl zeo.) hoose the line pependicul to the plte nd though the point chge s the is, nd choose the oigin t the sufce of the plte neest the pticle. (These choices put the pticle on the is t d.) Fo configution B, the electic potentil is zeo oth t ll points in the hlf-spce tht e vey f fom the pticle nd t ll points on the plne just s ws the cse fo configution A. (c) A theoem, clled the uniqueness theoem, implies tht thoughout the hlf-spce the potentil function nd thus the electic field E fo the two configutions e identicl. Using this esult, otin the electic field E t evey point in the plne in the configution B. (The uniqueness theoem tells us tht in configution B the electic field t ech point in the plne is the sme s it is in configution A.) Use this esult to find the sufce chge density σ t ech point in the conducting plne (in configution B). Pictue the Polem We cn use the eltionship etween the potentil nd the electic field to show tht this ngement is equivlent to eplcing the plne y point chge of mgnitude q locted distnce d eneth the plne. In () we cn fist find the field t the plne sufce nd then use σ E to find the sufce chge density. In (c) the wok needed to move the chge to point d wy fom the plne is the poduct of the potentil diffeence etween the points t distnces d nd d fom q multiplied y the seption Δ of these points. () The potentil nywhee on the plne is nd the electic field is pependicul to the plne in oth configutions, so they must give the sme potentil eveywhee in the y plne. Also, ecuse the net chge is zeo, the potentil t infinity is zeo. () The sufce chge density is given y: At ny point on the plne, the electic field points in the negtive diection nd hs mgnitude: σ E () kq E cosθ d whee θ is the ngle etween the hoizontl nd vecto pointing fom the positive chge to the point of inteest on the z plne nd is the distnce long the plne fom the oigin (tht is, diectly to the left of the chge).

85 Electic Potentil 75 d Becuse cosθ : d kq E d qd π d d ( d ) kqd ( d ) Sustitute fo E in eqution () nd σ simplify to otin: ( ) / qd π d 88 A pticle tht hs mss m nd positive chge q is constined to move long the -is. At nd e two ing chges of dius (Figue -8). Ech ing is centeed on the -is nd lies in plne pependicul to it. Ech ing hs totl positive chge Q unifomly distiuted on it. () Otin n epession fo the potentil () on the is due to the chge on the ings. () Show tht () hs minimum t. (c) Show tht fo <<, the potentil ppoches the fom () () α. (d) Use the esult of Pt (c) to deive n epession fo the ngul fequency of oscilltion of the mss m if it is displced slightly fom the oigin nd elesed. (Assume the potentil equls zeo t points f fom the ings.) Pictue the Polem We cn epess the potentil due to the ing chges s the sum of the potentils due to ech of the ing chges. To show tht () is minimum t, we must show tht the fist deivtive of () t nd tht the second deivtive is positive. In Pt (c) we cn use Tylo epnsion to show tht, fo <<, the potentil ppoches the fom () () α. In Pt (d) we cn otin the potentil enegy function fom the potentil function nd, noting tht it is qudtic in, find the sping constnt nd the ngul fequency of oscilltion of the pticle povided its displcement fom its equiliium position is smll. () Epess the potentil due to the ing chges s the sum of the potentils due to ech of thei chges: ( ) ing to the left ing to the ight The potentil fo ing of chge is: ( ) whee is the dius of the ing nd Q is its chge.

86 76 hpte Fo the ing to the left we hve: ing the left to ( ) Fo the ing to the ight we hve: ing the ight to ( ) Sustitute fo ing the left to nd ing the ight to to otin: ( ) ( ) ( ) () Evlute d/d to otin: d d [( ) ] ( ) [ ] Solving fo yields: Evlute d /d to otin: d d ( ) ( ) Evluting this epession fo yields: fo etem ( ) ( ) 5 [ ] [( ) ] [ ] [( ) ] d d ( ) > 5 ( ) is mimum t. (c) The Tylo epnsion of () is: ( ) ( ) '( ) ''( ) highe ode tems Fo << : ( ) ( ) ( ) ( ) ' ''

87 Electic Potentil 77 Sustitute ou esults fom Pts () nd () to otin: (d) Epess the ngul fequency of oscilltion of simple hmonic oscillto: Fom ou esult fo Pt (c) nd the definition of electic potentil: ( ) ( ) o ( ) ( ) α whee ( ) nd α k' ω m whee k is the estoing constnt. U ( ) q ( ) q kqq ( ) k' kqq whee k ' Sustituting fo k in the epession kqq fo ω yields: ω m 89 Thee concentic conducting thin spheicl shells hve dii,, nd c so tht < < c. Initilly, the inne shell is unchged, the middle shell hs positive chge Q, nd the oute shell hs chge Q. (Assume the potentil equls zeo t points f fom the shells.) () Find the electic potentil of ech of the thee shells. () If the inne nd oute shells e now connected y conducting wie tht is insulted s it psses though smll hole in the middle shell, wht is the electic potentil of ech of the thee shells, nd wht is the finl chge on ech shell? Pictue the Polem The digm shows pt of the shells in coss-sectionl view unde the conditions of Pt () of the polem. We cn use Guss s lw to find the electic field in the egions defined y the thee sufces nd then find the electic potentils fom the electic fields. In Pt () we cn use the edistiuted chges to find the chge on nd potentils of the thee sufces.

88 78 hpte Q Q c () Apply Guss s lw to spheicl Gussin sufce of dius c to otin: E Qenclosed ( π ) nd E ecuse the net chge enclosed y the Gussin sufce is zeo. Becuse E (c) : ( c) Apply Guss s lw to spheicl Gussin sufce of dius < < c to otin: Use E ( < c) < to find the potentil diffeence etween c nd : Becuse (c) : The inne shell cies no chge, so the field etween nd is zeo nd: ( π ) Q E nd E ( < < c) ( ) ( c) d c c ( ) ( ) ( ) c c

89 Electic Potentil 79 () When the inne nd oute shells e connected thei potentils ecome equl s consequence of the edistiution of chge. Q Q c The chges on sufces nd c e elted ccoding to: Q Q Q () c Q does not chnge with the connection of the inne nd oute shells: Q Q Epess the potentils of shells nd c: In the egion etween the nd, the field is / nd the potentil t is then: ( ) ( c) ( ) () The enclosed chge fo < < c is Q Q, nd y Guss s lw the field in this egion is: E < < c k ( Q Q) Epess the potentil diffeence etween nd c: () c () k( Q Q) ecuse (c). ( ) c Solve fo () to otin: c ( ) k( Q Q) () Equte equtions () nd () nd solve fo Q to otin: Q ( c ) ( c ) Q () Sustitute eqution () in eqution () nd solve fo Q c to otin: Q c ( ) ( c ) c Q (5)

90 8 hpte Sustitute equtions () nd (5) in eqution () to otin: ( ) ( c )( ) ( c ) 9 onside two concentic spheicl thin metl shells of dii nd, whee >. The oute shell hs chge Q, ut the inne shell is gounded. This mens tht the potentil on the inne shell is the sme s the potentil t points f fom the shells. Find the chge on the inne shell. Pictue the Polem The digm shows coss-sectionl view of potion of the concentic spheicl shells. et the chge on the inne shell e q. The dshed line epesents spheicl Gussin sufce ove which we cn integte E nˆ da in ode to find E fo. We cn find () fom the integl of E etween nd. We cn otin second epession fo () y consideing the potentil diffeence etween nd nd solving the two equtions simultneously fo the chge q on the inne shell. q Q Apply Guss s lw to spheicl sufce of dius : ( π ) Q q k Q q E E ( ) Use E to find (): d ( ) k( Q q) k ( Q q) We cn lso detemine () y consideing the potentil diffeence etween, i.e., nd : ( ) kq Equte these epessions fo () to otin: ( Q q) k k q Q 9 [SSM] Show tht the totl wok needed to ssemle unifomly chged sphee tht hs totl chge of Q nd dius is given y Q π. Enegy consevtion tells us tht this esult is the sme s the ( )

91 Electic Potentil 8 esulting electosttic potentil enegy of the sphee. Hint: et ρ e the chge density of the sphee tht hs chge Q nd dius. lculte the wok dw to ing in chge dq fom infinity to the sufce of unifomly chged sphee tht hs dius ( < ) nd chge density ρ. (No dditionl wok is equied to sme dq thoughout spheicl shell of dius, thickness d, nd chge density ρ. Why?) Pictue the Polem We cn use the hint to deive n epession fo the electosttic potentil enegy du equied to ing in lye of chge of thickness d nd then integte this epession fom to to otin n epession fo the equied wok. If we uild up the sphee in lyes, then t given dius the net chge on the sphee will e given y: Q() Q When the dius of the sphee is, the potentil eltive to infinity is: () ( ) Q Q π π Epess the wok dw equied to ing in chge dq fom infinity to the sufce of unifomly chged sphee of dius : dw du Q π Q π ( ) 6 dq π d Q π d Integte dw fom to to otin: W Q U 6 π Q π Q π d Q π 9 () Use the esult of Polem 9 to clculte the clssicl electon dius, the dius of unifom sphee tht hs chge e hs nd n electosttic potentil enegy equl to the est enegy of the electon (5. 5 e). omment on the shotcomings of this model fo the electon. () epet the clcultion in Pt () fo poton using its est enegy of 98 Me. Epeiments indicte the poton hs n ppoimte dius of out. -5 m. Is you esult close to this vlue?

92 8 hpte Pictue the Polem We cn equte the est enegy of n electon nd the esult of Polem 9 in ode to otin n epession tht we cn solve fo the clssicl electon dius. () Fom Polem 9 we hve: U e π The est enegy of the electon is given y: E mc Equte these enegies to otin: e mc π Solving fo yields: e () π m c Sustitute numeicl vlues nd evlute : π (.6 ).69 m 5 N m e e 9 J This model does not eplin how the electon holds togethe ginst its own mutul epulsion. () Fo poton, eqution () yields: π (.6 ) 9. m 9 9 N m.6 98 Me e This esult is wy too smll to gee with the epeimentl vlue of. 5 m. 9 [SSM] () onside unifomly chged sphee tht hs dius nd chge Q nd is composed of n incompessile fluid, such s wte. If the sphee fissions (splits) into two hlves of equl volume nd equl chge, nd if these hlves stilize into unifomly chged sphees, wht is the dius of ech? () Using the epession fo potentil enegy shown in Polem 9, clculte the chnge in the totl electosttic potentil enegy of the chged fluid. Assume tht the sphees e septed y lge distnce. J

93 Electic Potentil 8 Pictue the Polem Becuse the post-fission volumes of the fission poducts e equl, we cn epess the post-fission dii in tems of the dius of the pefission sphee. () elte the initil volume of the unifomly chged sphee to the volumes of the fission poducts: ' Sustitute fo nd : ( ) π π ' Solving fo yields: () Epess the diffeence ΔE in the totl electosttic enegy s esult of fissioning: '. 79 Δ E E' E Fom Polem 9 we hve: E Q π Afte fissioning: Sustitute fo E nd E to otin: Q' E' ' π π Q.6E π ( Q) Δ E.6E E. 7E 9 Polem 9 cn e modified to e used s vey simple model fo nucle fission. When 5 U nucleus sos neuton, it cn fission into the fgments Xe, 9 S, nd neutons. The 5 U hs 9 potons, while Xe hs 5 potons nd 9 S hs 8 potons. Estimte the enegy elesed duing this fission pocess (in Me), ssuming tht the mss density of the nucleus is constnt nd hs vlue of 7 kg/m. Pictue the Polem We cn use the definition of density to epess the dius of nucleus s function of its tomic mss N. We cn then use the esult deived in Polem 9 to epess the electosttic enegies of the 5 U nucleus nd the nuclei of the fission fgments Xe nd 9 S. The enegy elesed y this fission pocess is: ( U 9 ) Δ E U () 5 U U Xe S

94 8 hpte Epess the mss of nucleus in tems of its density nd volume: Sustitute numeicl vlues nd evlute s function of N: Nm Nm ρπ πρ whee N is the nucle nume. 7 (.66 kg) 7 π ( kg/m ) 6 ( 9.97 m) N N The 'dius' of the 5 U nucleus is theefoe: U 6 ( 9.97 m)( 5) m Fom Polem 9 we hve: U Q π Sustitute numeicl vlues nd evlute the electosttic enegy of the 5 U nucleus: U 5 U 9 (.6 ) 5 ( 6.5 m) 9 π 8.85 N m e.9 J Me J/e The dii of Xe nd 9 S e: 6 9 ( 9.97 m)( ) nd Xe 9 S m 6 ( 9.97 m)( 9).5 5 m

95 Electic Potentil 85 Poceed s ove to find the electosttic enegy of the fission fgments Xe nd 9 S: nd U U Xe 9 S π 7.79 π 9 ( 5.6 ) 5 ( 8.85 / N m )( 5.8 m) e J Me J/e 9 ( 8.6 ) 5 ( 8.85 / N m )(.5 m). e J Me J/e Sustitute fo U 5, U, nd U Xe U 9 in eqution () nd evlute S ΔE: ΔE 89Me 8 Me ( 86Me 75Me)

96 86 hpte