ocepts: pctce, delectrc costt, resstce, seres/prllel comtos () coxl cle cossts of sultor of er rdus wth chrge/legth +λ d outer sultg cylder of rdus wth chrge/legth -λ. () Fd the electrc feld everywhere spce. () Fd the potetl dfferece etwee the er d outer edges. (c) Fd the cpctce of the system f the cle hs totl legth L. (2) delectrc mterl s chrcterzed y delectrc costt κ d, f ler, whe serted to cpctor, wll produce cpctce κ geo where geo s clled the geometrc cpctce d s the cpctce of the cpctor wthout mterl serted etwee the pltes. s rge, vcuum hs delectrc costt of.000000 whle wter hs delectrc costt of 80. () Suppose you mesured cpctce of 00 pf cpctor whch ws empty d you mesured cpctce of 300 pf whe the cpctor ws mmersed flud. Wht s the delectrc costt of the flud? () Suppose sl of ths mterl of thckess d d re ws serted sde of prllel plte cpctor wth pltes of re d thckess s (s>d). Fd the cpctce of ths system. (3) Ot the ddto formuls for cpctors seres d prllel. (4) Ot the equvlet resstce for resstors seres d prllel. (5) Suppose hd plces two mterls seres. Both of the mterls hve the sme legth (L) d the sme cross sectol re. The frst mterl hs resstvty ρ whle the secod mterl hs resstvty ρ 2. Fd the resstce of the seres comto
() coxl cle cossts of sultg cylder of er rdus wth chrge/legth +λ (o ts surfce) d outer cylder of rdus wth chrge/legth -λ. () Fd the electrc feld everywhere spce. () Fd the potetl dfferece etwee the er d outer edges. (c) Fd the cpctce of the system f the cle hs totl legth h. We kow how to clculte the electrc feld from log wre: hoose the Guss cylder s show of rdus r d heght h. I the rego etwee the coxl cles, we the hve: Φ E E 2πrh The eclosed chrge s gve y: qec λh (we ssume the sgle wre hs zero rdus here). We c equte the two expressos v Guss s lw: λh λ E 2π rh E r ε0 π ε0 Now let me expd ths to clude the cse of sgle cylder of rdus. I ths cse, the chrge (f t s coductg wre) wll e ll locted o the surfce of the wre. 2 r ˆ If r<, the o et chrge s eclosed. Ths mes: E r< 0 If you re outsde the cylder of rdus, the you re eclosg et chrge. s efore, we the hve the electrc feld s gve y: λ E r ˆ r> 2πε0 We c ow esly expd ths to clude the secod cylder. If you re t rdus r>, the o et chrge s eclosed. Thus, outsde the coxl cle, the et electrc feld s gve y: E r> 0 So we c ow come the solutos: 0 ( r < ) λ E ˆ 2πε r ( < r < ) 0 ( r > 0 ) Now we eed to clculte the potetl dfferece etwee the two cylders.
No-lculus: It s t more dffcult for o-clculus studets to fd the potetl dfferece etwee the er d outer coductors. However, here t s: V E r 0 0 λ r λ 2πε r 2πε You my rememer our ofte-used pproxmto tht sde-steps lots of clculus whch ws: x x Oce you hve the potetl dfferece, you ext eed to clculte the totl chrge seprto. Ths s gve y: Q λ h We c fd the cpctce from: λ πε πε πε Q h 2 0h 2 0 λ V h 2 0 If the cle hs totl legth h, the 2πε ( 0 h ) Note: I hve purposely used h to represet the legth rther th L. We wll use L lter to represet ductce.
The followg s dvced pplcto: you wo t see ths o test. Here s terestg vrto o ths prolem. Suppose ths coxl cle s serted to delectrc flud wth delectrc costt κ. Depedg upo the ture of the flud, t my ether rse or lower wth the spce of the cle. You mght woder why d lso to wht heght t wll rse. Let s suppose the potetl dfferece s strctly mted to e V. hrges ove the flud wll exert force o the flud. If the flud s polr (such s wter) the the force exerted y the chrges s tlly s show the sketch. Ths force wll ele the flud to rse up to the re etwee the pltes. How hgh the flud wll rse s possle to clculte. Whe colum of flud rses through dstce h, the chge grvttol potetl eergy s gve y the verge heght to whch dsk of heght h s rsed: 2 2 2 2 2 U Mgh ρ π h gh ρ π gh { } verge 2 verge flud flud The work requred to rse ths flud cme from the electrosttc potetl dfferece, whch ws mted to e costt. Ths mes tht ddto chrge hd to e seprted. We c fd out how much ddtol chrge eeded to e seprted pretty esly. We hve essetlly two cpctors prllel whe the system hs flly cheved equlrum. Lter ths worksheet you wll fd tht cpctces prllel dd to gve equvlet cpctce. Ths the tells up wht ddtol chrge ws requred to e seprted: 2 πε 0 h h h 2πε0 + + κ h + h κ { } { } prllel eq 2 The mout of ddtol chrge whch eeded to e seprted s the gve y: Q Q Q V 2πε0 { h h 0 ( 2 ) h } V πε + κ h ( κ ) The ddtol work requred to produce ths chrge s the gve y: 2 2 0 W QV V πε h κ Ths work wet to rsg the delectrc flud to the system. Thus: 2 2 0 2 2 2 V πε h κ 2 ρfludπ gh Now we c determe how hgh the flud rses: κ 2 2πε V 0 ( ) h ρ 2 flud π g 2 2
(2) delectrc mterl s chrcterzed y delectrc costt κ d, f ler, whe serted to cpctor, wll produce cpctce κ geo where geo s clled the geometrc cpctce d s the cpctce of the cpctor wthout mterl serted etwee the pltes. s rge, vcuum hs delectrc costt of.000000 whle wter hs delectrc costt of 80. () Suppose you mesured cpctce of 00 pf cpctor whch ws empty d you mesured cpctce of 300 pf whe the cpctor ws mmersed flud. Wht s the delectrc costt of the flud? () Suppose sl of ths mterl of thckess d d re ws serted sde of prllel plte cpctor wth pltes of re d thckess s (s>d). Fd the cpctce of ths system. () We fd the delectrc costt from: mesuremets of cpctce to fd: κ κ geo 300pf 00pf 3. Now, we compre the two () To swer the secod prt, we wll eed to use result whch wll e comg, mely tht for cpctors seres, the totl cpctce s gve y: equvlet Sce we re sertg ths mterl to prllel plte cpctor, d the mterl s the form of sl, we c fd the cpctce of ech dvdul prt of the system: We hd, for prllel plte cpctor: ε 0 d The cpctce of ech dvdul prt s ths: I(): sl rego wth out the sl: ε 0 d geo I(): the sl rego wth the sl: κ κε geo 0 d II: The rego wth out the sl. ε ( s d) 0 2 Now we c fd the equvlet cpctce: d ( s d) s d κ( s d) s d+κs κd ε0 eq + 2 κε0 + ε0 ε0 κs + κs ε0 κ s ceq κ d+κ( s d) You c verfy tht f ds the correct cpctce s oted. You c lso verfy tht f d0, the correct geometrcl cpctce results.
(3) Ot the ddto formuls for cpctors seres d prllel. Soluto: PLLEL If two cpctors re prllel, they hve commo potetl dfferece cross ther pltes ut the chrge stored o ech cpctor my e dfferet. Thus, we hve: Q Q2 V 2 We hve tht the totl chrge stored s gve y: Q Q + Q + V V 2 2 eq Thus, for cpctors prllel, equvlet cpctce s gve y: equvlet SEIES If two cpctors re seres, the wll geerl hve dfferet potetl drop cross ther pltes. However, the totl chrge seprto o ech cpctor wll e the sme. Ths s ecuse chrge etwee the two outer pltes s the sme... you c thk of ctully removg the ceter pece. The equvlet cpctce s the: V V + V + Q + Q Q Q 2 2 2 equvlet Ths shows the tht for cpctors seres, the equvlet cpctce s gve y: equvlet Not ll crcuts c e expressed s smple comto of seres or prllel cpctors. Wth those crcuts, you wll requre more powerful techques order to do proper lyss.
Whe chrges strt flowg, oe mesures curret. For the preset tme, we re gog to cosder D curret whch mes tht the chrges re flowg t costt rte costt drecto. The SI ut whch mesures curret s the mpere d the curret s defed s: Q I t It s esy to see from ths tht mp / s the SI system of uts. Lter, fter we cover mgetc felds, we wll hve much etter wy to defe curret sce, relty, the electros re ot ctully flowg very much ut t s the electrosttc force whch gets trsmtted. Whle I m tlkg out curret, I eed to meto tht we regrd our curret s the flow of postve chrges (the covetol curret) due to d guess some tme go. I relty, however, wht does flow re the electros (ut the opposte drecto). There re ettes whch oppose the flow of curret clled resstors. It s oserved expermetlly tht the vst mjorty of resstces, Ohm s lw s oeyed. Ohm s lw sttes: V Now, fct the resstce s lmost lwys mesured s eg the rto of V to I ut ot ll mterls show ler reltoshp etwee V d I. These mterls re clled o-ohmtc resstces. I. The SI system of resstce s the Ohm Ω d t s esy to see tht the SI system, the uts for the Ohm re V/ or (J/)/(/s)Js/ 2. I wt to show you the ext prolem how resstce seres d prllel ehve. Before tht, however, I wt to show you how to clculte the resstce of mterl. To do so, we defe the resstvty of mterl d gve t the symol ρ. It hs the sme symol s volume chrge desty ut the meg s qute dfferet. I short, mge you hve cylder of mterl wth cross sectol re d legth L. The resstce of ths mterl s the gve y: ρ I the SI system, resstvty hs uts of Ωm. Metls hve resstvty of out 0-8 Ωm whle glss hs resstvty o the order of 0 0 Ωm. However, to e truthful, resstvty eeds to e specfed wth regrd to the temperture rge of use. Glss t room temperture s sultor ut f you het t up, t wll ecome coductor shortly efore t melts. Metls, o the other hd, ecome less coductg s the temperture s cresed. Ths s prmry dfferece etwee glssy mterls d metls. L
(4) Ot the equvlet resstce for resstors seres d prllel. Soluto: SEIES For resstors seres, the curret through ech resstor s the sme ut the potetl drop cross ech resstor s geerl dfferet. Thus, V V + V I + I I + I 2 2 2 equvlet Thus, for resstors seres: equvlet PLLEL For resstors prllel, the curret through ech resstor s geerl ot the sme. Isted, the potetl drop cross ech resstor s the sme. The reso ths s true s ech ed s coected to the sme wre whch we regrd theoretclly s equpotetl surfce. The totl curret put (or output) s the gve y: I I + I + V V V 2 2 equvlet It s esy to see from ths formulto, the tht the equvlet prllel resstce of resstors s gve y: equvlet Here s terestg oservto tht expls why you should ot plug too my thgs to sgle crcut: Suppose you hve detcl resstors prllel, ech of resstce. The eq eq Now suppose your crcut supples voltge V to the resstces. The curret s gve y: V V I eq s you dd more d more devces, ecomes lrge. The crcut comes closer d closer to short crcut, whch results fte mout of curret flowg. Sce the power compy frows upo your house tkg ll ther curret, the fuse wll low or your house wll ur dow ecuse the wres hve fte resstce tht cuses them to ct lke terl heters.
(5) Suppose hd plces two mterls seres. Both of the mterls hve the sme legth (L) d the sme cross sectol re. The frst mterl hs resstvty ρ whle the secod mterl hs resstvty ρ 2. Fd the resstce of the seres comto. Soluto: Ech of the resstors hs resstce gve y: L ρ Sce the two resstors re plced seres, the equvlet resstce s gve y: ρ + ρ L eq 2 If, however, the resstors were plced prllel, we would hve: ρρ2 L eq L ρ + ρ2 eq ρ +ρ2 Oe fl prtg ote: very my studets hve dffculty performg ths operto: + c o clcultors d otg the correct results. Ths dffculty hs ee see ll studets t ll levels. I recommed tht t wll e to your gret eeft f you mke sure tht you re le to put umers for d c d solve correctly for.