2 mv2 qv (0) = 0 v = 2qV (0)/m. Express q. . Substitute for V (0) and simplify to obtain: v = q

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1 Pof Anchodoui Polems set # Physics 69 Mch 3, 5 (i) Eight eul chges e locted t cones of cue of side s, s shown in Fig Find electic potentil t one cone, tking zeo potentil to e infinitely f wy (ii) Fou point chges e fixed t cones of sue centeed t oigin, s shown in Fig The length of ech side of sue is The chges e locted s follows: + is t (, +), + is t (+, +), 3 is t (+, ), nd +6 is t (, ) A fifth pticle tht hs mss m nd chge + is plced t oigin nd elesed fom est Find its speed when it is vey f fom oigin Solution: (i) To compute potentil ll you need to know is tht e e 3 chges distnce s wy, 3 distnce s wy, nd one chge distnce s 3 wy You ( find potentil ) due to ech chge septely, nd dd esults vi supeposition: V πɛ s πɛ s (ii) The digm shows fou point chges fixed t cones of sue nd fifth chged pticle tht is elesed fom est t oigin We cn use consevtion of enegy to elte initil potentil enegy of pticle to its kinetic enegy when it is t get distnce fom oigin nd electosttic potentil t oigin to expess U i Use consevtion of enegy to elte initil potentil enegy of pticle to its kinetic enegy whenit is t get distnce fom oigin: K + U, o ecuse K i U f, K f U i Expess initil potentil enegy of pticle to its chge nd electosttic potentil t oigin: U i V () Sustitute fo K f nd U i to otin: mv V () v V ()/m Expess electosttic potentil t oigin: V () 6 πɛ ( ) πɛ Sustitute fo V () nd simplify to otin: v 6 πɛ m Five identicl point chges + e nged in two diffeent mnnes s shown in Fig : in once cse s fce-centeed sue, in o s egul pentgon Find potentil enegy of ech system of chges, tking zeo of potentil enegy to e infinitely f wy Expess you nswe in tems of constnt times enegy of two chges + septed y distnce Solution: Using pinciple of supeposition, we know tht potentil enegy of system of chges is just sum of potentil enegies fo ll uniue pis of chges The polem is n educed to figuing out how mny diffeent possile piings of chges e e, nd wht enegy of ech piing is The potentil enegy fo single pi of chges, oth of mgnitude, septed y distnce d is just: P E pi πɛ d Since ll of chges e sme in oth configutions, ll we need to do is figue out how mny pis e e in ech sitution, nd fo ech pi, how f pt chges e How mny uniue pis of chges e e? Thee e not so mny tht we couldn t just list m y ute foce - which we will do s check - ut we cn lso clculte how mny e e In oth configutions, we hve chges, nd we wnt to choose ll possile goups of chges tht e not epetitions So f s potentil enegy is concened, pi (, ) is sme s (, ) Piings like this e known s comintions, s opposed to pemuttions

2 whee (, ) nd (, ) e not sme It is stightfowd to see tht wys of choosing pis fom five chges 5 5 (5 ) So e e uniue wys to choose chges out of 5 Fist, lets conside fce-centeed sue lttice In ode to enumete possile piings, we should lel chges to keep m stight Lel cone chges, nd cente chge 5 (it doesnt mtte which wy you nume cones, just so long s 5 is middle chge) Then ou possile piings e: (, ) (, 3) (, ) (, 5) (, 3) (, ) (, 5) (3, ) (3, 5) (, 5) Thee e ten, just s we expect In this configution, e e only thee diffeent distnces tht cn septe pi of chges: pis on djcent cones e distnce pt, centecone piing is distnce pt, nd f cone-f cone pi is pt We cn tke ou list ove, nd sot it into pis tht hve sme seption We hve fou pis of chges distnce pt, fou tht e pt, nd two tht e pt Wite down enegy fo ech type of pi listed in Tle, multiply y nume of those pis, nd dd esults toge: ( P E sue (cente cone pi) + (f cone pi) + (djcent cone pi) πɛ + + / ) 783 πɛ Fo pentgon lttice, things e even esie This time, just pick one chge s, nd lel os fom -5 in clockwise o counte-clockwise fshion Since we still hve 5 chges, e e still piings, nd y e sme s list ove Fo pentgon, howeve, e e only two distinct distnces - ei chges cn e djcent, nd thus distnce pt, o y cn e next-neest neighos Wht is next-neest neigho distnce? In egul pentgon, ech of ngles is 8, nd in ou cse, ech of sides hs length, s shown in Fig We cn use lw of cosines to find distnce d etween nextneest neighos; d + cos 8 ( cos ) d cos 8 φ 68, whee nume φ is known s Golden Rtio The distnces nd d utomticlly stisfy golden tio in egul pentgon, d/ φ Given neest neigho distnce in tems of, we cn n cete tle of piings fo pentgon; se e listed in Tle Now once gin we wite down enegy fo ech type of pi, nd multiply y nume of pis: P E pentgon 5(enegy of djcent pi) + 5(enegy of next neest neigho pi) 5 πɛ ( + d ) 5 πɛ [ + ( cos 8 ) ] 89 πɛ So enegy of pentgonl lttice is highe, mening it is less fvole thn sue lttice Nei one is enegeticlly fvoed though - since enegy is positive, it mens tht ei configution of chges is less stle thn just hving ll five chges infinitely f fom ech o This mkes sense - if ll five chges hve sme sign, y dont wnt to nge next to one no, nd thus se ngements cost enegy to keep toge If we didnt foce chges toge in se pttens, positive enegy tells us tht y would fly pt given hlf chnce Fo this eson, nei one is vlid sot of cystl lttice, el cystls hve eul numes of positive nd negtive chges, nd e ovell electiclly neutl 3 Conside system of two chges shown in Fig 3 Find electic potentil t n ity point on x xis nd mke plot of electic potentil s function of x/ Solution The electic potentil cn e found y [ supeposition ] pinciple At point on x xis, we hve V (x) πɛ x + ( ) πɛ x+ πɛ x x+ The ove expession my e ewitten s V (x) V x/ x/+, whee V πɛ The plot of dimensionless electic

3 potentil s function of x/ is depicted in Fig 3 A point pticle tht hs chge of + nc is t oigin (i) Wht is (e) shpes of euipotentil sufces in egion ound this chge? (ii) Assuming potentil to e zeo t, clculte dii of five sufces tht hve potentils eul to V, V, 6 V, 8 V nd V, nd sketch m to scle centeed on chge (iii) Ae se sufces eully spced? Explin you nswe (iv) Estimte electic field stength etween -V nd 6-V euipotentil sufces y dividing diffeence etween two potentils y diffeence etween two dii Compe this estimte to exct vlue t loction midwy etween se two sufces Solution: (i) The euipotentil sufces e sphees centeed on chge (ii) Fom eltionship etween electic potentil due to point chge nd electic field of point chge we hve: dv E d ( πɛ d o V V πɛ ) Tking potentil to e zeo t yields: V πɛ V πɛ πɛ V Becuse 8 C, it follows tht N m /C 8 C V Now you cn use pevious eution to detemine vlues of : V [V] 6 8 [m] The euipotentil sufces e shown in coss-section in Fig 5 (iii) No The euipotentil sufces e closest toge whee electic field stength is getest (iv) The vege vlue of mgnitude of electic field etween -V nd 6-V euipotentil sufces is given y: E est V V 6 V 9 m 66 m 9 V m The exct vlue of electic field t loction midwy etween se two sufces is given y E πɛ, whee is vege of dii of -V nd 6-V euipotentil sufces Sustitute numeicl vlues nd evlute E exct N m /C 8 C 3 V (66 m+9 m) / m The estimted vlue fo E diffes y out % fom exct vlue 5 Two coxil conducting cylindicl shells hve eul nd opposite chges The inne shell hs chge + nd n oute dius, nd oute shell hs chge nd n inne dius The length of ech cylindicl shell is L, nd L is vey long comped with Find potentil diffeence, V V etween shells Solution: The digm shown in Fig 5 is coss-sectionl view showing chges on inne nd oute conducting shells A potion of Gussin sufce ove which well integte E in ode to find V in egion < < is lso shown Once weve detemined how E vies with, fom V V E d we cn find V V E d Apply Guss lw to cylindicl Gussin sufce of dius nd length L, E ˆndA E πl ɛ Solving fo E yields: E Sustitute fo E nd integte fom to : V V πɛ L d πɛ L ln πɛ L

4 6 An electic potentil V (z) is descied y function V m z + V, z > m, m < z < m V (z) 3 V 3 V m 3 z 3, m < z < m 3 V + 3 V m 3 z 3, m < z < m, m < z < m V m z + V, z < m The gph in Fig 6 shows vition of n electic potentil V (z) s function of z (i) Give electic field vecto E fo ech of six egions (ii) Mke plot of z-component of electic field, E z, s function of z Mke sue you lel xes to indicte numeic mgnitude of field (iii) ulittively descie distiution of chges tht gives ise to this potentil lndscpe nd hence electic fields you clculted Tht is, whee e chges, wht sign e y, wht shpe e y (plne, sl)? Solution (i) Using E V, nd noting tht electic potentil only depends on vile z, we hve tht z-component of electic field is given y E z dv dz The electic field vecto is n given y E dv ˆk dz Fo z > m, E dv ˆk dz d dz ( V m z + V)ˆk V m ˆk Fo m < z < m, E dv ˆk dx Fo m < z < m, E dv ˆk dz ( 3 V 3 V m 3 z 3) ˆk V m 3 z ˆk Note tht z hs units of [m ], so vlue of d dz electic field t point just inside, z m ɛ m (whee ɛ > is vey smll nume), is given y E V m 3 ( m) ˆk V ˆk m Note tht z-component of electic field, V m, hs coect units Fo m < z < m, E dv ˆk ( dz d dz 3 V + 3 V m 3 z 3) ˆk V m 3 z ˆk The vlue of electic field t point just inside, z + m + ɛ m, is given y E + V m 3 ( m) ˆk V ˆk m Fo m < z < m, E dv ˆk dz Fo z < m, E dv ˆk dz d dz ( V m z + V)ˆk V m ˆk (ii) The z-component of electic field s function of z is shown in Fig 6 (iii) In egion m < z < m e is non-unifom (in z-diection) sl of positive chge Note tht z-component of electic field is zeo t z m, negtive fo egion m < z < m, nd positive fo m < z < m s it should fo positive sl tht hs zeo field t cente In egion m < z < m e is conducto whee field is zeo On plne z m, e is positive unifom chge density σ tht poduces constnt field pointing to ight in egion z > m (hence positive component of electic field) On plne z m, e is negtive unifom chge density σ In egion m < z < m e is conducto whee field is zeo On plne z m, e is positive unifom chge density σ tht poduces constnt field pointing to left in egion z < m (hence positive component of electic field) On plne z m, e is negtive unifom chge density σ 7 Two conducting, concentic sphees hve dii nd The oute sphee is given chge Wht is chge on inne sphee if it is ed? Solution: The system of conducting concentic sphees is shown in Fig 7 When oject is ed, it mens its potentil is zeo, ut note tht chge on it my not e zeo

5 To detemine chge, tke potentil on inne sphee s zeo nd ssume tht chge on it is Since V () V ( ) E( )d, potentil diffeence t is n V () V ( ) πɛ + +, whee we hve tken zeo of potentil t infinity Theefoe, + +, yielding 8 Conside two nested, spheicl conducting shells The fist hs inne dius nd oute dius The second hs inne dius c nd oute dius d The system is shown in Fig 8 In following fou situtions, detemine totl chge on ech of fces of conducting sphees (inne nd oute fo ech), s well s electic field nd potentil eveywhee in spce (s function of distnce fom cente of spheicl shells) In ll cses shells egin unchged, nd chge is n instntly intoduced somewhee (i) Both shells e not connected to ny o conductos (floting) tht is, i net chge will emin fixed A positive chge + is intoduced into cente of inne spheicl shell Tke zeo of potentil to e t infinity (ii) The inne shell is not connected to gound (floting) ut oute shell is gounded tht is, it is fixed t V nd hs whteve chge is necessy on it to mintin this potentil A negtive chge is intoduced into cente of inne spheicl shell (iii) The inne shell is gounded ut oute shell is floting A positive chge + is intoduced into cente of inne spheicl shell (iv) Finlly, oute shell is gounded nd inne shell is floting This time positive chge + is intoduced into egion in etween two shells In this cse uestion Wht e E() nd V ()? cnnot e nsweed nlyticlly in some egions of spce In egions whee se uestions cn e nsweed nlyticlly, give nswes In egions whee y cnnot e nsweed nlyticlly, explin why, ut ty to dw wht you think electic field should look like nd give s much infomtion out potentil s possile Solution: (i) Thee is no electic field inside conducto In ddition, net chge on n isolted conducto is zeo (ie + c + d ), yielding, +, c, d + Using Guss lw, πɛ ˆ, > d, c < < d E() πɛ ˆ, < < c, < < πɛ ˆ, < The field lines e shown in Fig 9 Since V () V ( ) E( )d, potentil diffeence is n: πɛ, > d πɛ d (, ) c < < d V () V ( ) πɛ c + d, < < c πɛ c + d, < < πɛ + c + d, < (ii) Since oute shell is now gounded, d to mintin E() outside oute shell

6 We hve,,, c +, d Agin using Guss lw yields:, > c E() πɛ ˆ, < < c, < < πɛ ˆ, < The field lines e shown in Fig 9 The potentil diffeence is n V () V ( ), > c πɛ c, < < c πɛ c, < < πɛ + c, < (iii) The inne shell is gounded nd to mintin E() outside inne shell Becuse e is no electic field on oute shell,, c d Guss lw n yields {, > E() πɛ ˆ, < The field lines e shown in Fig 9 The potentil diffeence is n V () V ( ) {, > πɛ, < (iv) The electic field within cvity is zeo If e is ny field line tht egn nd ended on inne wll, integl E d s ove closed loop tht includes field line would not e zeo This is impossile since electosttic field is consevtive, nd efoe electic field must e zeo inside cvity The chge etween two conductos pulls minus chges to ne side on inne conducting shell nd epels plus chges to f side of tht shell Howeve, net chge on oute sufce of inne shell ( ) must e zeo since it ws initilly unchged (floting) Since oute shell is gounded, d to mintin E() outside oute shell Thus, d, c nd E(), fo < o > c Fo < < c, E() is in fct well defined ut functionl fom is vey complicted The field lines e shown in Fig 9 Wht cn we sy out electic potentil? V () fo > c, nd V () constnt fo <, ut etween two shells functionl fom of potentil is vey complicted 9 The hydogen tom in its gound stte cn e modeled s positive point chge of mgnitude +e ( poton) suounded y negtive chge distiution tht hs chge density ( electon) tht vies with distnce fom cente of poton s: ρ() ρ e / ( esult otined fom untum mechnics), whee 53 nm is most pole distnce of electon fom poton (i) Clculte vlue of ρ needed fo hydogen tom to e neutl (ii) Clculte electosttic potentil (eltive to infinity) of this system s function of distnce fom poton

7 Solution: (i) Expess chge d in spheicl shell of volume dv π d t distnce fom poton: d ρdv ρ e / π d Expess condition fo chge neutlity: e πρ e / d Fom tle of integls we hve x e x dx ex ( x x+) Using 3 this esult yields e / d 3 / Sustitute in expession fo e to otin: e πρ 3 ρ e Sustitute numeicl vlues to otin ρ π C 356 π(53 nm) 8 C/m 3 3 (ii) The electosttic potentil of this poton-electon system is sum of electosttic ( potentils due to poton nd electon s chge density: V V + V, whee V e πɛ + ), V ρ( ) πɛ π d, nd ρ( )π d Sustituting fo ρ( ) in expession fo yields: πρ e / d Using gin x e x dx ex ( x x + ) we evlute 3 x e x/ dx 3 e x/ 8 x + x + 3 e / nd [ ] πρ 3 e / Sustituting fo in expession fo V yields: V { ]} e πɛ + πρ [ 3 e / Sustitute fo ρ fom (i) nd simplify to otin: { [ ]} ( V e πɛ e 3 e / e πɛ e / + ) + Sustituting fo ρ( ) nd simplifying yields: V πɛ ρ e / π d ρ ɛ e / d Fom tle of integls we hve: xe x dx ex (x ) Using this esult we evlute ( e x/ xdx ) e x/ x + e x/ + Sustitute fo e / d nd ρ in expession fo V to otin V ɛ ) ( [ ] ( e e π / 3 + πɛ e e / ) + Sustituting fo V nd V in V V + V nd simplifying yields: V ( e πɛ e / + e πɛ + ) e / e πɛ e / A pticle tht hs mss m nd positive chge is constined to move long x-xis At x L nd x L e two ing chges of dius L Ech ing is centeed on x-xis nd lies in plne pependicul to it Ech ing hs totl positive chge unifomly distiuted on it (i) Otin n expession fo potentil V (x) on x xis due to chge on ings (ii) Show tht V (x) hs minimum t x (iii) Show tht fo x << L, potentil ppoches fom V (x) V () + αx (iv) Use esult of Pt (iii) to deive n expession fo ngul feuency of oscilltion of mss m if it is displced slightly fom oigin nd elesed (Assume potentil euls zeo t points f fom ings) Solution: (i) Expess potentil due to ing chges s sum of potentils due to ech of i chges: V (x) Vingto The potentil fo ing of chge is V (x) πɛ x + hve: Vingto left left + Vingto ight whee is dius of ing nd is its chge Fo ing to left we (x+l) +L πɛ Fo ing to ight we hve: Vingto ( πɛ ight πɛ (x L) +L Sustitute fo Vingto nd Vingto to otin V (x) + ) (ii) To left ight (x+l) +L (x L) +L show tht V (x) is minimum t x, we must show tht fist deivtive of V (x) t x nd tht second deivtive is positive Evlute fist deivtive to otin dv dx { } L x L+x πɛ [(L x) +L ] 3/ [(L+x) +L ] { 3/ 3(L x) πɛ + [(L x) +L ] 5/ [(L x) +L ] 3/ fo extem Solving fo x yields x Evl- } Evluting this ute d V dx 3(L+x) [(L+x) +L ] 5/ [(L+x) +L ] 3/

8 ving L A peugh p i k h k cones of cue of edge s, s shown in Figue P369 m () Detemine x, y, nd z components of esultnt m o chges foce exeted y on chge locted 3 mj t point A ()ewht e mgnitude nd diection of this esultnt foce? 69 [SSM] Fou point chges e fixed t cones of sue centeed t oigin The length of ech side of sue is The chges e locted s follows: + is t (, +), + is t (+, +), 3 is t (+, ), nd +6 z is t (, ) A fifth pticle tht hs mss m nd chge + is plced t oigin nd elesed fom est Find its speed when it is vey f fom oigin he pentgon lttice, things e even esie This time, just pick on s fom -5 in clockwise o counte-clockwise fshion Since we Pictue Polem y The digm neg piings, nd y e sme Fo pen shows fou point chges fixed ts list ove ed cones of sue nd fifth )zk istinct distnces - ei chges cn e djcent, nd thus dis chged pticle Pointtht is elesed fom is m, est t Aoigin We cn use s x distnce? n if neest neighos Wht is next-neest neigho consevtion of enegy to elte s nticl point chges + e nged inoftwo diffeent mnnes s shown elow - in initil potentil enegy pticle to y its kinetic enegy when it is t get eed sue, xin s egul pentgon Find potentil enegy of ech o s distnce fom oiginngles nd is 68, nd in egul pentgon, ech of 3 ou cse, ec potentil t oigin f to wy Expess you nswe in ng zeo of potentil electosttic enegy to e infinitely neselow WeFigue cnp369 use expess Ui 69lw Polems nd 7 of cosines to find distnce d etwe enegy of two chges + septed distnce Figue y : Polem K U o, ecuse Ki Uf, Kf U i 8 o Use consevtion of enegy to elte 7 Conside chge distiution shownenegy in Figue initil potentil of P369 () Show tht mgnitude electic pticle toof its kinetic enegyfield whentit cente of ny fce of is cue hs vlue of 8k e /s t get distnce fom oigin: () Wht is diection of electic field t cente of top fce of cue? Expess initil potentil enegy U i V chged pticle # is 7 Review polem A negtively of pticle to its chge nd plced t cente of unifomly chged whee electosttic potentil t ing, oigin: ing hs totl positive chge s shown in Exmple +pticle, confined to + 38 The move long x xis, is Ui to otin: Sustitute fo Kf nd displced smll distnce x long xis (whee x '' ) mv V nd elesed Show tht pticle oscilltes in simple hmonic motion with feuency given yfigue : Polem d v V m C is supeposition, we know tht d/ V () enegy of system k e potentil expession fof x$ yields: π > ofthus, V (x) isof chges minimum is t just x (iii) Use dx L3 d of 3 ( m lced enegies fo ll uniue pis of chges The polem is n educed to figuing Tylo expnsion to show tht, fo x L, potentil ppoches fom V (x) V () + αx expnsion of V e of ech piing is ent possile piings ofwith chges nd wht +enegy The Tylo (x)density is: e, V (x) nc/m V () +lies V ()x 7 A line of chge unifom 35 ndu V ()x + highe ode tems Fo long L,line ychges, $ etween points withsustitute cox Vof(x) # V 5 (x) cm, V ()+V ()x+ y esults fom (i) nd to otin: ion y fo single pi oth distnce d is(ii)just: V ()x, septed of mgnitude odintes x $ nd x+$ xcm, Find, whee V () pt V (x) o V (x) electic V () field + αxit 3 π L π L nd α π L3 L cetes t oigin (iv) we cn otin potentil enegy function fom potentil function nd, noting tht it is 73 Review polem electic dipole in unifom e udtic in x,an find sping constnt nd electic ngul feuency of oscilltion of pticle pofield is displced slightly fom its euiliium position, s ficvided its displcement fom its euiliium position is smll Expess ngul feuency of oscilshown in Figue P373, whee is smll The seption d t k ltion of simple hmonic oscillto: ω whee m,of of chges is, nd moment of ineti k is estoing constnt Fom esult De- " # d + cos 8 ( p ) d cos 8 68 cos 8 ) nume is known s Golden Rtio The distnces how mny ges e sme in oth configutions, ll we need to dogiven is+figue out dipole is Iegul Assuming pentgon, dipoleof iselectic elesed fom Uthis nch tio in d/ in pt (iii) nd definition potentil (x) V () x neest V () + kx neigho, sitution, nd fo ech pi, how f pt chges e k Sustituting fo k in expession fo ω yields: ω cete whee tle of piings fo pentgon (Tle ) 8π L3 8π L3 8π m L3 pis of chges e e? Thee e not so mny tht we couldn t just list m ich we will do s check - ut we cn lso clculte how mny e e In oth hve chges, nd we wnt to choose ll possile goups of chges tht e not s potentil enegy is concened, pi (, ) is sme s (, ) Piings like comintions, s opposed to pemuttions whee (, ) nd (, ) e not sme ii Chge piings in pentgonl me of possile comintions is donetle like this:: lttic

9 still piings, nd y e sme s list ove Fo pentgon, howeve, e e only two distinct distnces - ei chges cn e djcent, nd thus distnce pt, o y cn e next-neest neighos Wht is next-neest neigho distnce? 38 Solved Polems In egul pentgon, ech of ngles is 8, nd in ou cse, ech of sides hs length, s 38 Electic Due d toetween Systemnext-neest of Two Chges shown elow We cn use lw of cosines to findpotentil distnce neighos 8 o Conside system of two chges shown in Figue 38 Tle : Chge piings in sue lttice d #, piing type ) seption, cente-cone, djcent cones d + cos 8, f pcone d p ( pis (, 5) (, 5) (, ) (3, ) cos 8 ) cos 8 68 (3, 5) (, 5) (, 3) (, ) (, 3) (, ) Figue 38 Electic dipole Hee nume is known s Golden Rtio The distnces nd d utomticlly stisfy Find electic potentil t n ity point on x xis nd mke plot golden tio in egul pentgon, d/ Given neest neigho distnce in tems of, we cn e nely done ledy We hve fou pis of chges distnce pt, fou tht e n cete tle of piings fo Solution: pentgon (Tle ) d we t, nd two tht e pt Wite down enegy fo ech type of pi, multiply y num hose pis, nd dd esults toge: The electic potentil cn e found y supeposition pinciple At poin xis, we hve Tle : Chge piings in pentgonl lttice ) pi) + (djcent ( cone P Esue #, (cente-cone (f cone pi) piing type pi) + seption pis V ( x) x x x x ke ke ke d 5, next-neest neighos (, 3) (, ) (, ) (, 5) (3, 5) 5, djcent+ + p (,my ) e(, 3) (3, ) (, 5) (5, ) expession The ove ewitten s olems ke V ( x) ++ p V x / x / c Potentil Due to System of Two Chges p i ke h k whee V / The plot of dimensionless electic potentil s funct shown in Figue 38 ystem of two chges is depicted in Figue 38 pentgon lttice, things e even esie This time, just pick one chge s, nd lel es fom -5 in clockwise o counte-clockwise fshion Since we still hve 5 chges, e piings, nd y e sme s list ove Fo pentgon, howeve, e e o distinct distnces - ei chges cn e djcent, nd thus distnce pt, o y cn t-neest neighos Wht is next-neest neigho distnce? egul pentgon, ech of ngles is 8, nd in ou cse, ech of sides hs length wn elow We cn use lw of cosines to find distnce d etween next-neest neighos Figue 38 8 o Figue 3:dipole The lectic dipole of polem 3 Figue 38 Electic tic potentil t n ity point on x xis ndd mke plot

10 V (V) 6 8 (m) ufces e tion to V to otin: k k V 6 V point chge V k 8 V V o coxil conducting cylindicl shells hve eul nd nne shell hs chge + nd n oute dius, nd nd n inne dius The length of ech cylindicl long comped with Find potentil diffeence, tentil lls sufces e closest toge Figue whee : Polem electic field he digm is lue showing of V V 6 V lectic field etween E e nd oute -V euipotentil y: potion of which we ll s find to V in xis fom V 6 V V to ppoximte dii E est 9 o shown Once m 7 m m ch E vies of se with potentil V fom f electic field t loction midwy etween se two y E k, whee is vege of dii of -V Figue 5: Polem 5 otentil ffeence sufces Sustitute V V numeicl vlues nd evlute Ed V V Ed

11 Exm Pctice Polems Pt Solutions Polem Electic Field nd Chge Distiutions fom Electic Potentil An electic potentil V ( z ) is descied y function )(V "m - )z + V ; z > m + + ; m < z < m + 3 V # $ % 3 V & + "m-3 ' ( z3 ; m < z < m + V (z) * + 3 V+ # $ % 3 V & "m-3 ' ( z3 ; m < z < m ; m < z < m +,(V "m - )z + V ; z < m The gph elow shows vition of n electic potentil V ( z ) s function of z ) Mke plot of z-component of electic field, E z, s function of z Mke sue you lel xes to indicte numeic mgnitude of field ) Give electic field vecto E fo ech of six egions in (i) to (vi) elow? c) ulittively descie distiution of chges tht gives ise to this potentil Solution: We shll fct tht E " Figue 6: lndscpe Polem nd hence 6 electic fields you clculted Tht is, whee e chges, V Since electic potentil only depends wht on sign e y, wht shpe e y (plne, sl )? vile z, we hve tht z -component of electic field is given y In egion m < z < m e is non-unifom (in z-diection) sl of dv Ez positive chge Note tht z-component of electic field is zeo t z m, dz negtive fo egion m < z < m, nd positive fo m < z < m s it should fo positive sl tht hs zeo field t cente The electic field vecto is n given y In egion m < z < m e is conducto whee field is zeo On plne z m, e is positive unifom chge density tht poduces constnt field pointing to ight in egion z > m (hence positive component of electic field) On plne z m, e is negtive unifom chge density " In egion m < z < m e is conducto whee field is zeo On plne z m, e is positive unifom chge density tht poduces constnt field pointing to left in egion z < m (hence positive component of electic field) On plne z m, e is negtive unifom chge density " Figue 7: Polem 7

12 ,, c +, d Agin using Guss s Lw yields $, & &, && "# E() % &, &, & &' "# >c < <c << The potentil diffeence is n < > +, V () V ("), % ( - #$ '& *), < The potentil diffeence is n +, >c oute shell is gounded nd inne shell is floting - % (, d) Finlly, < < c + is intoduced into egion in etween t positive chge - #$ '& c *) this cse uestions Wht e E() nd V ()? cnnot V () V ("), % ( nlyticlly in some egions of spce In egions whee se u << - #$ '& c *), nsweed nlyticlly, give nswes In egions whee y cn nlyticlly, explin why, ut ty to dw wht you think elect - % look ( like nd give s much infomtion out potentil s + positive - ' *), <onegtive, fo exmple) #$ c & The cge electic within If e is ny field line tht eg Figue 8: The Fdy of field polem 8 cvity is zeo E d s ovea closed loop tht includes inne wll, integlto"gound c) The inne shell is gounded ut outeonshell is not connected positive chge + is intoduced into cente inne This spheicl shell since electosttic field is con would of not e zeo is impossile efoe electic field must e zeo inside cvity The chge e pulls minus chges to ne side on inne conducting sh w gounded, d to mintin E() outside oute conductos The inne shell is gounded nd to mintinplus outside E() to chges f side inne of thtshell shell Howeve, net chge on ou unchged (floti Becuse e is no electic field on oute shell,cinne ) must e zeo since it ws dshell ( initilly oute shell is gounded, d to mintin E() outside oute shell nside chge, Also,, netc conducto +on, n disolted conducto + d ), c d ields, +, c, d + Guss s Lw n yields $, >c & #%, > d & % ", < < c && %%,"# c < < d % %$, < < c << &, % " & %, << %, < & % &' % ""#, < #, > % E() $ % ", < & & d, c nd E(), < o > c Fo, E() is in fct well defined ut it is vey complicted The field lines figue elow hen n ()d, potentil diffeence is Figue 9: The electic field lines of polem 8; fom left to ight (i), (ii), (iii), (iv) +, >c - % (, < <c + -, #$ '& >cd*) - -#$ V (") -, % ( -, c < *<,d << - -#$ d#$ ' c & ) -- - % ( V ("), ' + *, < < c ( - -#$ & c % d) +, < - - % ' ( + & *), < < c *) - '& #$ - #$ - % c d (

U>, and is negative. Electric Potential Energy

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