Phys102 General Physics II
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1 Phys1 Generl Physics II pcitnce pcitnce pcitnce definition nd exmples. Dischrge cpcitor irculr prllel plte cpcitior ylindricl cpcitor oncentric sphericl cpcitor Dielectric Sls 1
2 pcitnce Definition of cpcitnce A cpcitor is useful device in electricl circuits tht llows us to store chrge nd electricl energy in controllle wy. The simplest to understnd consists of two prllel conducting pltes of re A seprted y nrrow ir gp d. If chrge is plced on one plte, nd - on the other, the potentil difference etween them is, nd then the cpcitnce is defined s /. The SI unit is /, which is clled the Frd, nmed fter the fmous nd cretive scientist Michel Frdy from the erly 18 s. Applictions Rdio tuner circuit uses vrile cpcitor Blocks D voltges in c circuits Act s switches in computer circuits Triggers the flsh ul in cmer onverts A to D in filter circuit Prllel Plte pcitor
3 hrging pcitor Electric Field of Prllel Plte pcitor Guss Lw E σ ε d Ed ε A σ A ε A d E ε A ε EA d oulom/olt Frd ε A 3
4 irculr prllel plte cpcitor r s r r 1 cm A πr π(.1) A.3 m S 1 mm.1 m ε A S (1 11 ).3.1 oulom olt }Frd F 3 pf p pico 1-1 Model of coxil cle for clcultion of cpcitnce Outer metl rid Signl wire 4
5 Find the cpcitnce of ordinry piece of coxil cle (T cle) For long wire we found tht Er kλ outer insultor r where r is rdil to the wire. r E. ds kλ dr kλlnr rdius r E. ds Edscos18 Eds Edr metl rid with - signl wire rdius with Insultor (dielectric ε) ds - dr ecuse pth of integrtion is rdilly inwrd kλln or kλln is higher thn λ L k 1 ir 4πε πεl ln πεl ln πεl ln L πε ln 6 1 L ln 4 43 PF m L 86 PF m L.5 mm. mm κ ε (ir) κ pcitnce of two concentric sphericl shells E. ds Edr E. ds Edscos18 Eds Edr ds - dr Edr k /r dr k k 1 r / k( 1 1 ) ) k( k( ) 4πε dr r 5
6 Sphericl cpcitor or sphere Recll our fvorite exmple for E nd is sphericl symmetry R The potentil of chrged sphere is (k)/r with t r. The cpcitnce is R 4πε R k R k Where is the other plte (conducting shell)? It s t infinity where it elongs, since tht s where the electric lines of flux terminte. k 1 1 nd R in meters we hve R R(m) 1 1 R(cm) R( cm) PF Erth: (6x1 8 cm)pf 6µF Mrle: 1 PF Bsketll: 15 PF You: 3 PF Prllel omintion of pcitors Typicl electric circuits hve severl cpcitors in them. How do they comine for simple rrngements? Let us consider two in prllel / We wish to find one euivlent cpcitor to replce 1 nd. Let s cll it. The importnt thing to note is tht the voltge cross ech is the sme nd euivlent to. Also note wht is the totl chrge stored y the cpcitors?. 1 1 ( 1 ) 1 1 6
7 Series omintion of pcitors Wht is the euivlent cpcitor? oltge cross ech cpcitor does not hve to e the sme. The chrges on ech plte hve to e eul nd opposite in sign y chrge conservtion. The totl voltge cross ech pir is: ( ) ( ) 1 1 So ; Therefore, Smple prolem µf 5. µf 3 4. µf ) Find the euivlent cpcitnce of the entire comintion. 1 nd re in series µF nd 3 re in prllel. e µ F 7
8 Smple prolem (continued) µf 5. µf 3 4. µf ) If 1 volts, wht is the chrge 3 on 3? / ouloms c) Wht is the totl energy stored in the circuit? U e U J J Electric Potentil Energy of pcitor As we egin chrging cpcitor, there is initilly no potentil l difference etween the pltes. As we remove chrge from one plte nd put it on the other, there is lmost no energy cost. As it chrges up, this chnges. - At some point during the chrging, we hve chrge on the positive plte. - The potentil difference etween the pltes is /. As we trnsfer n mount d of positive chrge from the negtive plte to the positive one, its potentil energy increses y n mount du. du d The totl potentil energy increse is d. U d Also U using / 8
9 /c U Grphicl interprettion of integrtion d d where / /c 1 U N i 1 i i Are under the tringle is the vlue of the integrl Are of the tringle is lso 1/ h d du d Are 1 ()(h) 1 ()( ) 1 Are under the tringle d Where is the energy stored in cpcitor? Find energy density for prllel plte cpcitor. When we chrge cpcitor we re creting n electric field. We cn think of the work done s the energy needed to crete tht electric field. For the prllel plte cpcitor the field is constnt throughout, so we cn evlute it in terms of electric field E esily. Use U (1/) σ σ E nd Ed κε ε εa Solve for εae, ES nd sustitute in We re now including dielectric effects: ε U ( εae)( Ed) εe ( Ad) U 1 volume occupied y E ε E u Ad Electrosttic energy density generl result for ll 1 geometries. u ε E To get totl energy you need to integrte over volume. 9
10 How much energy is stored in the Erth s tmospheric electric field? (Order of mgnitude estimte) tmosphere h Erth R km E 1 1 m 1 U ε E olume olume 4πR h olume π (6 1 ) ( 1 ) m R 6x1 6 m U (1 )(1 )(8.6 1 ) 11 U J This energy is renewed dily y the sun. Is this lot? The totl solr influx is Wtts/m Usun (6 1 ) 1 J s 1 U Usun 1 1 Only n infinitesiml frction gets converted to electricity. 1 J dy World consumes out 1 18 J/dy. This is 1/ of the solr flux. Dielectrics A dielectric is ny mteril tht is not conductor, ut polrizes well. Even though they don t conduct they re electriclly ctive. Exmples. Stressed plstic or piezo-electric electric crystl will produce sprk. When you put dielectric in uniform electric field (like in etween the pltes of cpcitor), dipole moment is induced on the molecules throughout the volume. This produces volume polriztion tht is just the sum of the effects of ll the dipole moments. If we put it in etween the pltes of cpcitor, the surfce chrge densities due to the dipoles ct to reduce the electric field in the cpcitor. 1
11 Dielectrics The mount tht the field is reduced defines the dielectric constnt κ from the formul E E / κ,, where E is the new field nd E is the old field without he dielectric. Since the electric field is reduced nd hence the voltge difference is reduced (since E d), the cpcitnce is incresed. / / (( / κ) ) κ κ is typiclly etween 6 with wter eul to 8 Permnent dipoles Induced dipoles E the pplied field E the field due to induced dipoles E E -E 11
12 d Ed σ E ε σ A E ε A S ε A ε A d E E κ Ed κ κ κ κ Wht is the electric field in sphere of uniform distriution of positive chrge. (nucleus of protons) R r E ρ 4 3 πr3 EdA E 4 π r E ρr 3ε enc ε ρ 4 π r 3 3 ε 4πε R 3 r 1
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