PHYS102 - Electric Energy - Capacitors

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1 PHYS102 - lectric nerg - Cpcitors Dr. Suess Februr 14, 2007 Plcing Chrges on Conuctors Plcing Chrges on Conuctors II Plcing Chrges on Conuctors III Plcing Chrges on Conuctors IV fiel n nerg fiel n nerg II fiel n nerg III Clculting nerg from lectric Fiel Clculting nerg from lectric Fiel II mple - nerg n uniform -fiel Cpcitors 12 Cpcitors Cpcitors II Cpcitors II Clculting Cpcitnce Clculting Cpcitnce II

2 Plcing Chrges on Conuctors It tkes work to plce totl chrge on n isolte conuctor. This mount of work is energ store on the conuctor. It is often ver ifficult to clculte the store potentil energ. Let s look t sstem which we cn clculte the store energ. Consier two ver lrge, prllel, conucting pltes s shown on the right. w PHYS102 -NOT: w. lectric nerg slie 2 Plcing Chrges on Conuctors II First eposit q onto the left plte (this is ccomplishe b plcing btter cross both conucting pltes). The -fiel is uniform ner the center of the pltes with mgnitue: q = σ ε 0 σ = q A There is potentil ifference between the two conuctors: b PHYS102 - V b = lectric nerg slie 3 l Plcing Chrges on Conuctors III V b = b l So fr, there hs been no mention of the reference of zero potentil. q This is becuse we re concerne with the potentil ifference between the two pltes! V 0 = q Aε 0 PHYS102 - lectric nerg slie 4 2

3 Plcing Chrges on Conuctors IV We nee to buil on the plte, but moving n itionl mount of chrge q requires n mount W of work. Fin the totl mount of work b summing ll contributions (W). U = W = q W = q 0 Aε 0 W = Aε 0 W = 0 V q q PHYS102 - U 2 lectric nerg slie 5 -fiel n nerg W = 1 2 Aε 0 2 This eqution is specific to prllel plte ssembl. This energ store is relte to the - fiel between the pltes. We cn solve in terms of the electric fiel mgnitue. = (Finl fiel strength once eposite.) Aε 0 PHYS102-2 = A 2 ε lectric nerg slie 6 3

4 -fiel n nerg II 2 = A 2 ε Rewriting U in terms of electric fiel strength: U = 2Aε 0 ( A 2 ε 2 0 2) U = A 1 2 ε 0 2 This energ store is relte to the squre of the -fiel between the pltes. The term A is the volume where the electric fiel is present! PHYS102 - lectric nerg slie 7 -fiel n nerg III U = V 1 2 ε 0 2 (V is volume where fiel present.) Rewriting s U/V gives the energ ensit with units of J/m 3 : u = 1 2 ε 0 2 This result is Generl n onl requires knowlege of n electric fiel. This implies tht if n electric fiel is present in spce, then there eists store PHYS102 energ-in tht spce. lectric nerg slie 8 4

5 Clculting nerg from lectric Fiel If the electric fiel is known for region in spce (with volume V ), one cn clculte the store energ (IF the electric fiel is constnt!): U = V u If the electric fiel is not constnt (like tht for point prticle), then one nees to clculte U for ver smll region of spce, V where the electric fiel is uniform. U = u V U = U = 1 2 ε 0 2 V PHYS102 - lectric nerg slie 9 Clculting nerg from lectric Fiel II U = U = 1 2 ε 0 2 V Where the integrtion limits re over region of spce where the electric fiel eists. Let s look t n emple. PHYS102 - lectric nerg slie mple - Uniform -fiel mple - nerg n uniform -fiel Consier tpicl thunerclou tht rises to n ltitue of 10 km n hs imeter of 20 km. Assuming n verge electric fiel strength of 10 5 V/m, estimte the totl electrosttic energ store in the clou. Solution The energ ensit is given b: u = 1 2 ε 0 2 = 1 2 ( C 2 /Nm 2 )(10 5 V/m) 2 = J/m 3 We ssume tht the energ ensit is the sme throughout the storm, so: U = u V = u V = u V = u π r 2 h = J PHYS102 - lectric nerg slie 11 5

6 Cpcitors slie Prllel plte emple Cpcitors When using pir of conuctors to store energ, we term the pir of conuctors cpcitor. cpcitor. Cpcitors tpicll use to store shorttermshort-term electricl energ. Consier our previous emple: V = ε 0 A V ( ) V = ε 0 A PHYS102 - lectric nerg slie 12 Cpcitors II V = ( ) ε 0 A Rewriting to fin the mount of chrge. ( ) ε0 A = V V = ε 0 A Notice tht the rtio /V epens onl the geometr of the specific problem. The rtio C /V is terme the cpcitnce. PHYS102 - lectric nerg slie 13 6

7 0.3 Cpcitnce Cpcitors II C is mesure of the cpcit to store chrge for given potentil ifference cross two conuctors. The unit of cpcitnce is one Fr. [C] = [] = 1C / 1 V 1Fr [V ] 1 Fr is ver lrge vlue, n tpicl vlues of cpcitnce re:pf, nf, n µf.note: 1 milifr is lso lrge. PHYS102 - lectric nerg slie 14 Clculting Cpcitnce Clculte the cpcitnce for long coil cble of length L. Represent the cble s two concentric clinricl conuctors with rii n b (b > ) s shown on the right. Let the inner conuctor crr chrge + uniforml istribute over its length. V b = b l λ ( < r < b) = 2π ε 0 r ˆr PHYS102 - V b = 2π ε 0 L ln(b/) V b > 0. lectric nerg slie 15 Clculting Cpcitnce II b V b = 2π ε 0 L ln(b/) C = /V = 2π ε 0 L ln(b/) The cpcitnce of coil cble vries s the length of the cble vries.this is ver importnt result for mn eperiments. b It is convenient to tlk bout the cpcitnce per unit length for coil cble. C/L = 2π ε 0 PHYS102 - ln(b/) lectric nerg slie 16 7

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