Problem Set 3 SOLUTIONS

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1 Univesity of Albm Deptment of Physics nd Astonomy PH 10- / LeCli Sping 008 Poblem Set 3 SOLUTIONS points. Remembe #7 on lst week s homewok? Clculte the potentil enegy of tht system of thee chges, fo cicle of dius. Tke the zeo of potentil enegy to be infinitely f wy fom ll chges. Expess you nswe in tems of the enegy of chges q 1 nd q septed by e.g., constnt times k eq 1q /. Fom lst week, we hd n ngement of thee chges on cicul tck of dius. Two of the chges (q 1 nd q, q 1 q ) wee septed by 90 on the tck, nd hence distnce 1 pt. The thid chge (q 3) ws distnce d fom both of those. We found tht d could be expessed in tems of : q d + We lso found tht the mgnitude of q 3 ws fixed by the geomety given in the poblem, nd could expess the mgnitude of q 3 eltive to q 1 o q : q 3 1 h + i 3 q 1 1 q + h + i 3.15q 1 How do we find the potentil enegy of this system of chges? We just hve to dd togethe the potentil enegy of ech unique pi of chges. In this cse, thee e thee piings: P E system P E 1& + P E 1&3 + P E &3 The potentil enegy of pi of chges q 1 nd q septed by distnce 1 is stightfowd: P E pi keq1q 1 All we need to do is evlute the sum bove, plugging in the vlues we know: P E system keq1q 1 + keq1q keqq3 3 keq1q + keq1q3 p + + keq1q + keq1q keq1q keq1q k eq 1» 1 keq1q3 p + 3 ` + q 1 p ˆ + + p + +» 1 + 1» » 3 1 k eq 1 ` + q 1 + p ˆ + p » keq1q 4.1 keq1q So the enegy of this system of chges is bout 4.1 times highe thn if we hd chges lone septed by, o bout.9 times highe thn if we emoved q 3 nd just hd q 1 nd q septed by.

2 . 15 points. () Find the equivlent cpcitnce of the cpcitos in the figue below. (b) Find the chge on ech cpcito. (c) Find the potentil diffeence coss ech cpcito. 8μF 4μF 8μF + - 6μF 3μF 1V Befoe we stt, it is useful to emembe tht one fd times one volt gives one coulomb: 1 [F] 1 [V] 1 [C], nd tht cpcitnce times voltge gives stoed chge: QCV. Knowing this now will sve some confusion on units lte on. Fo tht mtte, it is lso good to emembe tht the pefix µ mens In ode to find single equivlent cpcito tht could eplce ll five in the digm bove, we need to look fo puely seies nd pllel combintions tht cn be eplced by single cpcito. The uppemost 8 µf nd 4 µf cpcitos e puely in seies, so they cn be eplced by single equivlent we will cll C 84, s shown below: Using ou ule fo combining seies cpcitos, we cn find the vlue of C 84 esily: 1 C µf µf C 84 8 µf.67 µf 3 Now we hve this equivlent cpcitnce puely in pllel with the second 8 µf cpcito. We cn eplce C 84 nd the second 8 µf cpcitos with single equivlent, which we will cll C 884: Using ou ddition ule fo pllel cpcitos, we cn find its vlue: C 884 C µf µf

3 This leves us with thee cpcitos in seies, s shown below: Adding togethe these thee in seies, we hve the ovell equivlent cpcitnce, C eq: 1 C eq 1 C µf µf C eq 1.68 µf Since we now hve one single cpcito connected to single voltge souce, we cn find the totl chge stoed in the equivlent cpcito, Q eq: Q eq C eqv (1.68 µf) (1 V) 0.16 µc Now in ode to get the chge nd voltge on ech single cpcito, we hve to wok bckwds nd ebuild ou oiginl cicuit. We know tht the C eq cpcito is elly thee cpcitos in seies - the 6 µf, the 3 µf, nd C 884. Seies cpcitos lwys hve the sme chge, nd one must hve the sme chge s the equivlent cpcito: Q 6µF Q 3µF Q 884 Since we know the chge nd cpcitnce fo ll thee of these cpcitos, we cn now find the voltge on ech, since V Q/C: V 6µF Q6µF 6µF V 3µF Q3µF 3µF V 884 Q884 C µc 6µF 0.16 µc 3µF 3.4 V 6.7 V 0.16 µc µf 1.9 V Notice tht the voltge on ll thee of these seies cpcitos dds up to the totl bttey voltge - it must be so, bsed on consevtion of enegy. Next, we know tht C 884 is elly two cpcitos in pllel - the lowe 8 µf cpcito nd C 84. Pllel cpcitos hve the sme voltge, so we know tht both of these hve to hve V lowe 8µF V 84 V V coss them. We know the voltge nd the cpcitnce fo C 84 nd the lowe 8 µf cpcitos now, so we cn find the stoed chge, QCV : Q lowe 8µF 8 µf V µf 1.9 V 15. µc Q 84 C 84 V µf 1.9 V 5.1 µc Finlly, the cpcito C 84 is elly two cpcitos in seies, which must both hve the sme chge: Q 4µF Q uppe 8µF Q 84. Given the chge on both of the emining cpcitos nd thei cpcitnces, we cn find the voltges: V 4µF Q4µF 4µF V uppe 8µF Quppe 8µF 8µF 5.1 µc 4µF 1.6 V 5.1 µc 8µF 0.63 V

4 Now we know the chge nd voltge on evey single cpcitnce, s well s the ovell chge (Q eq) nd effective cpcitnce (C eq). You numbes my be vey slightly diffeent thn those bove due to diffeent choices in ounding, this is noml. The esults e summized in the tble below: Tble 1: Equivlent cpcitnces, chges, nd voltges Cpcito [µf] Chge [µc] Voltge [V] top 8 µf µf lowe 8 µf µf µf C eq points. A pllel-plte cpcito hs 4.00 cm pltes septed by 6.00 mm of i. If 1.0 V bttey is connected to this cpcito, how much enegy does it stoe in Joules? In electon volts? If we wnt to find the enegy stoed in the cpcito, we need to know two of thee things, minimlly: the mount of chge stoed, the voltge pplied, nd the cpcitnce. Any two of these thee e sufficient, bsed on ou fomul fo the potentil enegy stoed in cpcito: P E 1 Q V 1 C ( V ) Q C We ledy know the pplied voltge, V 1.0 Volts. Since this is pllel plte cpcito nd we know its e A nd plte spcing d we cn esily clculte the cpcitnce... if we e vey ceful with units. Recll tht the dielectic constnt of i is essentilly one (κ 1). C κɛ0a d 1 `4.00 ɛ0 cm ` 1 m 100 cm m ` F/m 1 ` m m F F pf Now we know the cpcitnce nd the voltge, we cn find the enegy edily: P E 1 ` F (1.0 V) F V J ` J «1 ev ev 66 MeV J This poblem bings to mind few hnd SI unit convesions, which you should be ble to veify: 1 J1 F 1 V 1 C 1 V, 1 C1 F 1 V points. A cpcito with i between its pltes is chged to 150 V nd then disconnected fom the bttey. When piece of glss is plced between the pltes, the voltge coss the cpcito dops to 5 V. Wht is the dielectic constnt of the glss? (Assume the glss completely fills the spce between the pltes.)

5 When we inset dielectic into pllel plte cpcito, fo the sme mount of chge stoed the voltge is educed by fcto κ, the dielectic constnt: V empty κ V filled If the voltge is 150 V while the cpcito is empty, nd 5 V when filled, then we must hve κ6. Anothe wy to see this is to think bout the chge stoed. If ou (idel) cpcito is fully chged t 150 V nd disconnected, it keeps totl mount of chge Q. We cn elte this chge to the voltge pplied when the cpcito is empty: Q C empty V empty 150C empty When we inset the dielectic between the pltes, the cpcito emins disconnected but is still fully chged. The chges cn t go nywhee while the cpcito is disconnected, so we hve the sme Q. With the dielectic, howeve, we know tht the cpcitnce inceses by fcto κ: C full κc empty. We cn elte the stoed chge Q to the new cpcitnce nd voltge, nd combine tht with the expession bove to find κ: Q C full V full κc emptyv full 5κC empty Q 150C empty 5κC empty 150C empty κ 6 Remembe, dielectics incese the totl chge stoed nd the cpcitnce, but decese the voltge equied to stoe the sme mount of chge points. A potentil diffeence of 100 mv exists between the oute nd inne sufces of cell membne. The inne sufce is negtive eltive to the oute. How much wok is equied to move sodium ion N + outside the cell fom the inteio? Answe in electon volts nd Joules. A singly-chged ion hs chge of 1e. The wok done in moving chge q coss potentil diffeence V is edily clculted: W P E q V. In this cse the chge is e C, nd V 0.1 V. Wtch how esy it is to find the nswe in electon volts: W q V ` C (0.1 V) ` C (0.1 V) 0.1 [ C V] ev J 0.1 ev 1 ev J «1 ev J «emembe: 1 J1 C 1 V In fct, we didn t even need to go though ll tht. An electon volt is defined s the enegy equied to move one electon s equivlent of chge - 1e - though potentil diffeence of 1 Volt. Ou ion hs the sme mgnitude of chge s n electon, nd we move it though 0.1 V. Following the definition of n electon volt, we must hve P E 0.1 ev. This is one eson why electon volts e such hndy unit fo mny es of physics. Anywy: how bout the nswe in Joules? Well, if 1 ev J, then 0.1 ev J points. A poton nd n electon e cceleted fom est though potentil diffeence of 10 V. Clculte the speed nd kinetic enegy of ech. This one is just consevtion of enegy: electic potentil enegy is conveted into kinetic enegy. Let s wok it though ll the wy though, but fo geneic chge q of mss m. This solves the genel poblem, nd t the end we cn just plug

6 in the ppopite numbes fo poton nd electon. This sves us the wok of solving the sme poblem twice. Initilly, the chge is t est, nd thus hs no kinetic enegy. Ou potentil diffeence elly only tells us the diffeence in potentil enegy fom the initil to finl stte, we still need to define zeo point fo potentil enegy. We will define the finl potentil enegy to be zeo, fo convenience - this mens the chge stts out t potentil enegy of e V nd ends t 0, moving fom high potentil to low. With ll of this, we cn wite down consevtion of enegy, nd solve fo kinetic enegy nd velocity. P E i + KE i P E f + KE f q V KE f Aledy, we hve the kinetic enegy. Since the poton nd electon hve the sme mgnitude of chge e, nd the voltge is the sme in both cses, the kinetic enegy is the sme: KE f e V ` C (10 V) 1 ev 10 ev J We could hve just witten this down without clculting nything t ll. Recll tht n electon volt is defined s the enegy equied to move one electon s equivlent of chge - 1e - though potentil diffeence of 1 Volt. Both the poton nd electon hve chge of e, nd both move though potentil diffeence of 10 V. Theefoe, the potentil enegy chnge is 10 ev in both cses, so the kinetic enegy is lso 10 ev in both cses (o J). Well. We still need the speed. Just plug in the non-eltivistic kinetic enegy fomul, nd solve fo velocity: KE f 1 mv f q V vf q V m q V v f m Now we hve something of poblem. Fo electons, the chge is q e, doesn t tht led to n imginy numbe bsed on the fomul bove? Yes, it would... if we don t think bit fist. The poblem stted tht the electon ws cceleted though potentil diffeence of 10 V, it did not stte whethe the potentil ws highe t the stt o the end! Fom the phsing of the poblem, we hve to ssume tht the electon stts out t lowe potentil nd moves to highe potentil, since negtive chge will only be cceleted towd highe potentil. If this is the cse, the finl potentil is highe, then V must be negtive. In the fomul bove, both q nd V will be negtive, nd thee is no poblem. The poton, on the othe hnd, will be ttcted to egions of lowe potentil, so we hve to ssume tht it moves fom high to low potentil, mening V must be positive. The key is to emembe tht the chnge in kinetic enegy must be equl nd opposite the chnge in potentil enegy. Both chges gin kinetic enegy nd lose potentil enegy, so in both cses the initil potentil enegy must be positive (since the finl potentil enegy is set to zeo). Fo potons, this mens stting t high potentil nd moving to lowe one. Fo electons, this mens stting t low potentil nd moving to highe one, owing to its negtive chge. Anothe wy to go bout it is to use the now known vlue of kinetic enegy, which sneks ound this poblem ll togethe: KE f 1 mv f vf KE m KE v f m

7 This fomul is pehps esie to use, since we ledy know KE. Plug in the known msses nd chge, nd you should find: v f m/s v f m/s poton electon 7. 5 points A pllel plte cpcito is held t constnt voltge. Initilly thee is only i between the pltes. If dielectic with dielectic constnt of is inseted into the cpcito, wht hppens to the enegy stoed in the cpcito? This is lot like question 4. If we inset dielectic into the cpcito with fixed voltge, the cpcitnce goes up by fcto κ, s does the totl mount of stoed chge. Without dielectic, we cn wite the stoed enegy in thee diffeent wys: P E empty 1 Cempty ( V ) P E empty Q empty C empty P E empty 1 Qempty V We will veify tht ll thee fomuls give the sme esult, just fo fun. Once we fill the cpcito with dielectic, the voltge is held constnt t V, but the cpcitnce nd chge stoed incese: C full κc empty nd Q full κq empty. Using this nd the eltions bove, we cn find the enegy stoed with dielectic pesent in tems of the oiginl stoed enegy without: P E filled 1 C full ( V ) 1 κcempty ( V ) κ P E empty P E filled Q full C filled κ Q empty κc empty κ Q empty C empty κ P E empty P E filled 1 Q full V 1 κqempty V κ P Eempty No mtte which fomul we choose to solve the poblem, the nswe is the sme: the stoed enegy lso inceses by fcto κ, o times. Ntully you didn t need to solve it ll thee wys, one ws enough - I just wnted to show you tht it mde no diffeence which fomul you stted with points. Two chges, +q nd q, e septed by distnce d. Show tht the electic potentil f fom both kqd cos θ chges is ppoximtely V. The following ppoximtions my be useful (efeing to the figue below, with the oigin between the two chges): +, + xd d cos θ. + - θ d +q -q Fist, we should pick coodinte system nd n n oigin. The esiest choice seems to be putting the oigin t the midpoint between the two chges. We will pick n x y system with the x xis unning long line connecting the two chges, nd y xis unning pependicul. Finlly, define the +x nd +y diections s the diection fom the oigin towd the distnt point, which we will cll P. This puts point P t coodintes (x, y, ), nd the positive nd negtive

8 chge t (x d, y) nd (x + d, y) espectively. With these choices, we cn esily wite down the distnce fom P to the oigin (), to the positive chge ( +), nd to the negtive chge ( ) just using the distnce fomul: p x + y s x d + «+ y s x + d «+ y Aledy, we know enough to wite down the potentil t point P due to the two chges. The pinciple of supeposition sys tht the totl electic potentil t point P cn be found by dding togethe the electic potentils fom individul chges. Since potentil is scl, we don t even need to woy bout vectos, we just find the potentil due to the positive chge, the potentil due to the negtive chge, nd dd them togethe s numbes. We will choose the zeo fo electic potentil to be infinitely f wy, so the potentil t point P is just: V P keq + + ke ( q) keq ( +) + At this point, it is cle why you e given the ppoximtions bove. Plug them in, nd we e done: V P keqd cos θ Whee did these mgicl ppoximtions come fom? They wee given in the poblem, so one could just plug them in, but we will quickly deive them just to be sfe. Fist wite down + nd expnd: + s x d «+ y xd + d 4 1 xd + d 4 One of the conditions of the poblem is tht the point P is distnt, which mens d. If this is the cse, then the d / tem unde the sque oot is negligible: + 1 xd Agin, since d, the fction unde the dicl should be smll comped to 1, nd we cn use the ppoximtion (1 + ) n 1+x. Comping this with wht we hve bove, n 1 xd nd : + 1 xd «Now, do the sme thing fo - you get the sme esult, except the minus sign becomes plus. Now we cn ppoximte the diffeence between the two esily: xd «1 xd «xd d cos θ Fo the lst step, we used the fct tht cos θ x/, which you should be ble to veify. You cn lso find this ppoximtion geometiclly, noting tht + d cos θ nd + d cos θ. Ty dwing line fom one chge tht meets pependicully to see how this woks.

9 The second ppoximtion we need is moe stightfowd to find. We just wite down +, using the ppoximte foms bove, nd once gin dop tems tht hve d / in them: + 1 xd «1 + xd ««1 x d points. Five identicl point chges +q e nged in two diffeent mnnes s shown below - in once cse s fce-centeed sque, in the othe s egul pentgon. Find the potentil enegy of ech system of chges, tking the zeo of potentil enegy to be infinitely f wy. Expess you nswe in tems of constnt times the enegy of two chges +q septed by distnce. Bonus (3 points): could one mke two-dimensionl epeting cystl with eithe of these ngements? Justify you nswe. +q +q Using the pinciple of supeposition, we know tht the potentil enegy of system of chges is just the sum of the potentil enegies fo ll the unique pis of chges. The poblem is then educed to figuing out how mny diffeent possible piings of chges thee e, nd wht the enegy of ech piing is. The potentil enegy fo single pi of chges, both of mgnitude q, septed by distnce d is just: P E pi keq d Since ll of the chges e the sme in both configutions, ll we need to do is figue out how mny pis thee e in ech sitution, nd fo ech pi, how f pt the chges e. How mny unique pis of chges e thee? Thee e not so mny tht we couldn t just list them by bute foce - which we will do s check - but we cn lso clculte how mny thee e. In both configutions, we hve 10 chges, nd we wnt to choose ll possible goups of chges tht e not epetitions. So f s potentil enegy is concened, the pi (, 1) is the sme s (1, ). Piings like this e known s combintions, s opposed to pemuttions whee (1, ) nd (, 1) e not the sme. Clculting the numbe of possible combintions is done like this: i wys of choosing pis fom five chges 5! 5 C 5!! (5 )! So thee e 10 unique wys to choose chges out of 5. Fist, let s conside the fce-centeed sque lttice. In ode to enumete the possible piings, we should lbel the chges to keep them stight. Lbel the cone chges 1 4, nd the cente chge 5 (it doesn t mtte which wy you numbe the cones, just so long s 5 is the middle chge). Then ou possible piings e: i A nice discussion of combintions nd pemuttions is hee:

10 (1, ) (1, 3) (1, 4) (1, 5) (, 3) (, 3) (, 5) (3, 4) (3, 5) (4, 5) And thee e ten, just s we expect. In this configution, thee e only thee diffeent distnces tht cn septe pi of chges: pis on djcent cones e distnce pt, cente-cone piing is distnce pt, nd f cone-f cone pi is pt. We cn tke ou list bove, nd sot it into pis tht hve the sme seption: Tble : Chge piings in the sque lttice #, piing type seption pis 4, cente-cone (1, 5) (, 5) (3, 5) (4, 5) 4, djcent cones (1, 4) (3, 4) (, 3) (1, ), f cone (1, 3) (, 4) And we e nely done ledy. We hve fou pis of chges distnce pt, fou tht e pt, nd two tht e pt. Wite down the enegy fo ech type of pi, multiply by the numbe of those pis, nd dd the esults togethe: P E sque 4 (enegy of cente-cone pi) + (enegy of f cone pi) + 4 (enegy of djcent cone pi)»»» keq keq keq »4 keq keq h 5 + i 7.83 kq Fo the pentgon lttice, things e even esie. This time, just pick one chge s 1, nd lbel the othes fom -5 in clockwise o counte-clockwise fshion. Since we still hve 5 chges, thee e still 10 piings, nd they e the sme s the list bove. Fo the pentgon, howeve, thee e only two distinct distnces - eithe chges cn be djcent, nd thus distnce pt, o they cn be next-neest neighbos. Wht is the next-neest neighbo distnce? In egul pentgon, ii ech of the ngles is 108, nd in ou cse, ech of the sides hs length, s shown below. We cn use the lw of cosines to find the distnce d between next-neest neighbos. 108 o d d + cos 108 (1 cos 108 ) d cos 108 φ ii See

11 Hee the numbe φ is known s the Golden Rtio. iii The distnces nd d utomticlly stisfy the golden tio in egul pentgon, d/ φ. Given the neest neighbo distnce in tems of, we cn complete tble of piings fo the pentgon: Tble 3: Chge piings in the pentgonl lttice #, piing type seption pis 5, next-neest neighbos d (1, 3) (1, 4) (, 4) (, 5) (3, 5) 5, djcent (1, ) (, 3) (3, 4) (4, 5) (5, 1) Now once gin we wite down the enegy fo ech type of pi, nd multiply by the numbe of pis: P E pentgon 5 (enegy of djcent pi) + 5 (enegy of next-neest neighbo pi)»» keq keq d»» keq k eq cos 108» keq cos 108 keq» kq So the enegy of the pentgonl lttice is highe, mening it is less fvoble thn the sque lttice. Neithe one is enegeticlly fvoed though - since the enegy is positive, it mens tht eithe configution of chges is less stble thn just hving ll five chges infinitely f fom ech othe. This mkes sense - if ll five chges hve the sme sign, they don t wnt to nge next to one nothe, nd thus these ngements cost enegy to keep togethe. If we didn t foce the chges togethe in these pttens, the positive enegy tells us tht they would fly pt given hlf chnce. Fo this eson, neithe one is vlid sot of cystl lttice, el cystls hve equl numbes of positive nd negtive chges, nd e ovell electiclly neutl. And the bonus? One cn tile floo with sque tiles, but neve with egul pentgons whee ll five sides e the sme length. Ty it - it won t wok unless you mke some of the pentgon s sides diffeent lengths, o squish it in some wy. In fct, only thee egul polygons cn tile floo without gps - tingles, sques, nd hexgons. A good explntion why cn be found hee: Also cceptble: since ll chges e positive, these ngements e inheently unstble nywy, nd mutul epulsion would pevent one fom mking cystl points. If ech of the chges in the pentgon ngement bove e 1 µc nd 1 m, wht is the electic potentil t the cente of the pentgon? Agin tke the zeo of potentil enegy infinitely f wy. This one is esie thn it sounds. The electic potentil t the cente due to one chge q distnce d wy is: V keq d Since evey chge is the sme, nd the sme distnce fom the cente of the pentgon, the pinciple of supeposition sys tht we just need to find the potentil due to one of the chges nd multiply it by five. How f is ech chge fom the cente? Hve look t the figue below. The inteio ngles of the pentgon, defined by dwing lines fom ech vetex to the cente, e 360 /57. Once gin we cn use the lw of cosines to elte the distnce d to : iii See nd the pio link

12 7 o d d d + d + d d cos 7 d p (1 cos 7 ) 0.85 m The potentil t the cente of the pentgon is just five times the potentil due to single chge of 1 µc t distnce of d 0.5 m: V cente 5V single 5 keq d 5 ` N m/c `1 10 C N m/c 5.9 kv 0.85 m Fo the vey lst pt, we note tht one newton pe coulomb is one volt pe mete, [N/C][V/m], so volts must be newtons times metes pe coulomb.

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