1 Physics 1402: Lecture 7 Tody s gend nnouncements: Lectures posted on: www.phys.uconn.edu/~rcote/ HW ssignments, solutions etc. Homework #2: On Msterphysics tody: due Fridy Go to msteringphysics.com Ls: Begin this week Tody s Topic : End Chpter 24: Definition of Cpcitnce Exmple Clcultions (1) Prllel Plte Cpcitor (2) Cylindricl Cpcitor (3) Isolted Sphere Energy stored in cpcitors Dielectrics Cpcitors in Circuits
2 Definitions & Exmples L d Cpcitnce cpcitor is device whose purpose is to store electricl energy which cn then e relesed in controlled mnner during short period of time. cpcitor consists of 2 sptilly seprted conductors which cn e chrged to Q nd Q respectively. The cpcitnce is defined s the rtio of the chrge on one conductor of the cpcitor to the potentil difference etween the conductors.
3 Exmple 1: Prllel Plte Cpcitor Clculte the cpcitnce. We ssume σ, σ chrge densities on ech plte with potentil difference V: d Need Q: Need V: from defn: Use Guss Lw to find E Recll: Two Infinite Sheets (into screen) Field outside the sheets is zero Gussin surfce encloses zero net chrge Field inside sheets is not zero: Gussin surfce encloses nonzero net chrge σ σ E=0 E=0 E
4 Exmple 1: Prllel Plte Cpcitor Clculte the cpcitnce: ssume Q,Q on pltes with potentil difference V. d s hoped for, the cpcitnce of this cpcitor depends only on its geometry (,d). Dimensions of cpcitnce C = Q/V => [C] = F(rd) = C/V = [Q/V] Exmple: Two pltes, = 10cm x 10cm d = 1cm prt => C = ε 0 /d = = 0.01m 2 /0.01m * 8.852e12 C 2 /Jm = 8.852e12 F
5 Lecture 7 CT 1 Suppose the cpcitor shown here is chrged to Q nd then the ttery disconnected. Now suppose I pull the pltes further prt so tht the finl seprtion is d 1. d 1 > d d 1 d If the initil cpcitnce is C 0 nd the finl cpcitnce is C 1, is ) C 1 > C 0 B) C 1 = C 0 C) C 1 < C 0 Exmple 2: Cylindricl Cpcitor Clculte the cpcitnce: ssume Q, Q on surfce of cylinders with potentil difference V. r L
6 Recll: Cylindricl Symmetry Gussin surfce is cylinder of rdius r nd length L Cylinder hs chrge Q E r E r pply Guss' Lw: Q L Exmple 2: Cylindricl Cpcitor Clculte the cpcitnce: ssume Q, Q on surfce of cylinders with potentil difference V. r σ σ L If we ssume tht inner cylinder hs Q, then the potentil V is positive if we tke the zero of potentil to e defined t r = : gin: depends only on system prmeters (i.e., geometry)
7 Lecture 7, CT 2 In ech cse elow, chrge of Q is plced on solid sphericl conductor nd chrge of Q is plced on concentric conducting sphericl shell. Let V 1 e the potentil difference etween the spheres with ( 1, ). Let V 2 e the potentil difference etween the spheres with ( 2, ). Wht is the reltionship etween V 1 nd V 2? Q Q 1 Q Q 2 () V 1 < V 2 () V 1 = V 2 (c) V 1 > V 2 Exmple 3: Isolted Sphere Cn we define the cpcitnce of single isolted sphere? The sphere hs the ility to store certin mount of chrge t given voltge (versus V=0 t infinity) Need ΔV: V = 0 V R = k e Q/R So, C = R/k e
8 V Cpcitors in Prllel Q 1 Q 2 V Q Find equivlent cpcitnce C in the sense tht no mesurement t, could distinguish the ove two situtions. Prllel Comintion: => Totl chrge: Q = Q 1 Q 2 Equivlent Cpcitor: C = C 1 C 2 Cpcitors in Series Q Q Q Q Find equivlent cpcitnce C in the sense tht no mesurement t, could distinguish the ove two situtions. The chrge on C 1 must e the sme s the chrge on C 2 since pplying potentil difference cross cnnot produce net chrge on the inner pltes of C 1 nd C 2. RHS: LHS:
9 Exmples: Comintions of Cpcitors How do we strt?? Recognize C 3 is in series with the prllel comintion on C 1 nd C 2. i.e. Energy of Cpcitor How much energy is stored in chrged cpcitor? Clculte the work provided (usully y ttery) to chrge cpcitor to / Q: Clculte incrementl work dw needed to dd chrge dq to cpcitor t voltge V: The totl work W to chrge to Q is then given y: In terms of the voltge V:
10 Lecture 7 CT 3 The sme cpcitor s lst time. The cpcitor is chrged to Q nd then the ttery disconnected. Then I pull the pltes further prt so tht the finl seprtion is d 1. d 1 > d If the initil energy is U 0 nd the finl cpcitnce is U 1, is d d 1 ) U 1 > U 0 B) U 1 = U 0 C) U 1 < U 0 Summry Suppose the cpcitor shown here is chrged to Q nd then the ttery disconnected. d Now suppose I pull the pltes further prt so tht the finl seprtion is d 1. How do the quntities Q, W, C, V, E chnge? Q: W: C: V: E: remins the sme.. no wy for chrge to leve. increses.. dd energy to system y seprting decreses.. since energy, ut Q remins sme increses.. since C, ut Q remins sme remins the sme.. depends only on chg density How much do these quntities chnge?.. exercise for student!! nswers:
11 Where is the Energy Stored? Clim: energy is stored in the Electric field itself. Think of the energy needed to chrge the cpcitor s eing the energy needed to crete the field. To clculte the energy density in the field, first consider the constnt field generted y prllel plte cpcitor: The Electric field is given y: The energy density u in the field is given y: Units: J/m 3 Dielectrics Empiricl oservtion: Inserting nonconducting mteril etween the pltes of cpcitor chnges the VLUE of the cpcitnce. Definition: The dielectric constnt of mteril is the rtio of the cpcitnce when filled with the dielectric to tht without it. i.e. κ vlues re lwys > 1 (e.g., glss = 5.6; wter = 78) They INCRESE the cpcitnce of cpcitor (generlly good, since it is hrd to mke ig cpcitors They permit more energy to e stored on given cpcitor thn otherwise with vcuum (i.e., ir)
12 Prllel Plte Exmple Chrge prllel plte cpcitor filled with vcuum (ir) to potentil difference V 0. n mount of chrge Q = C V 0 is deposited on ech plte. Now insert mteril with dielectric constnt κ. Chrge Q remins constnt Voltge decreses from V 0 to Electric field decreses lso: So, C = κ C 0 Lecture 7, CT 3 Two prllel plte cpcitors re identicl (sme, sme d) except tht C 1 hs hlf of the spce etween the pltes filled with mteril of dielectric constnt κ s shown. If oth cpcitors re given the sme mount of chrge Q, wht is the reltion etween E 1, the electric field in the ir of C 1, nd E 2, the electric field in the ir of C 2 κ Q E 1 =? Q Q C 1 E 2 =? C 2 Q () E 1 < E 2 () E 1 = E 2 (c) E 1 > E 2