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devices - Devices Battegx Generates E —field using...

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Unformatted text preview: Devices Battegx : Generates E —field using chemical energy. .4» _. 6‘ = any” is actually potential difference between + and — plates. +O' — O' -——> 0 0 0' A E: —x —> go x=0 x=d air or vacuum between plates AV: —EoAS= ~3d SOV= 3d —> —> 80 80 A C = %= 807 x=0 l x= o— 0' . . = —(d _ l) — — Put conductor of thickness 1 1n 80 ‘90 80 80A , . C = d l (1) 0 Inlddle. Now E =0 inside conductor + HOLOHGNOO We have a — a £1": a, d — l 2 . . " . . . . 280A , That 1s, as if two capamtors were 1n senes, Each havmg C1 = C2 = d l (2) Ad 'Es(2)&(1) s 1 1+1 . ou e —= — — in comparing q _ y e C C1 C2 So when capacitors are connected in series, total equivalent capacitance is given by 1 1 ' —C—= 2 F - SERIES CONNECTION s i ‘——-— W5 Q o» ‘ Dielectric between plates. Dielectric consists of Dipoles. b 71% L6: Ct “(102": ,, 3: mm a W k 5 swag 2=5X§ 3 i302!) 6" —q LIE" ”f. CLMCK, H Q/fl +0' Dipole in E -field experiences torque, which causes each Dipole to line up along E . On surfaces of dielectric charge sheets + a band — ab appear. 42?? (’53 Gui/3.134; E -field inside dielecric 54; 66:; Jam. af (f’vkigd flaw by A v ”-‘ - LJ‘W : J 5.3,”, 7)/45,&7: [ «kg; {2‘ £3;ch (lam/57 (42:42.43) “fmémagr/ipfi Um ' E) [1.1172 21? 1»wa 0—0;” fin k: x a —> 0 E k 0—0,, 1 E a k [F Dielectric Const. [N .B. k is always > 1] . N V—Ed-Ed—id OW k~ k _ k _50k _2_ aAabk_k80A SO Ck~Vk_ 0d _ d Capacitance is increased by factor k. Note: in both cases conductor of thickness 1 between plates or Dielectric of thickness d, potential difference is reduced but physics is totally different! ”it/M; Next, put the two devices in a circuit: C Note: the lines connecting the devices are perfect conductors and so under stationary conditions they must become equipotentials. That is if we close the switch S and wait for a while the Potential difference across C will become 8. V: 6' Q=8C So the capacitor plates now have +Q (left) and —Q (right). How did this happen? Clearly, the +ive plate of the battery pulled electrons from the left sheet of capacitor while the ~ive plate pushed electrons on the right sheet. Effectively, you start with q=0 on either sheet, transfer charge from one sheet to the other, so one becomes —q & the other q f: {,4 4;; meme/a, J a? {Lian—U a"; (Li/1 W33 beiifiujem @3163 6,, It I‘M VP ’a/Qg; g] V=fi§q {Ia Now take A q from rt. To Left. Work done will be , 1 _ DW = Aq% area of rectanglet mix)” [70“? 6 {PI chum) q "I And we get CONSWWTI'L/z; FaACE: - WflRK hm”; 31::3‘ :Nbéffifi‘Ubb-rVT (9,, ”Tm? PATH» +q+dq ~q-dq 5/12. 1 2 To build 11p charge from 0 to Q you need area of A. This work is now UE— - — 2-Q(Q / C): 2Q C 1% V Q/C .c_ Aq if. Q Where does this work go? Notice, space between plates is not empty, there is an E -field in it. This energy is stored as potential energy in that E -f1eld. Apply it to ||-plate [air bet plates] all) 5 OF”! C: 1,; gin/j at U _ 021420] E _ 250A 1 E = 3 = —.90E2Ad 80 U U 1 Energy density— _ v—OEZ— ‘ A—g— — 380132 this IS like a “Pressure” ”75... :1. «2’0 €30 E2 ll-Plate [Dielectric] U — Q2 — U2A2d E ” 20k “ 2kgoA 1 =Ek80Ek2Ad 1 77E(/c) : ElcsoEkZ Next, consider the Expt: two identical capacitors. 5 //2m,. 2 First, put charges i Q on one UL, = E it it Next, connect left—to- left, right to right to make Equipotentials now charge Will be + — on each. "‘43; 2 TtlEnrgyU 2'1(Q)21 Q2 Q! /2 oa e E=——- j:— 2 2 C 4C (ii/Z «62/2 :CMP - What happened to half of the energy? In the second half of the experiment charge was transported from one set of plates to the other. This experiment tells us that it costs ener to ‘ transport charge through a conductor. This leads us to our third device. —RESISTOR First we define current A Q I: will? At E = E32 (36L? "'- gwbéu’hf'fi 0/; We 654mm? to . (“f/11.64;, céfiesétgjzbafl’ ' ”Tb CCLWQICJE/I MDléCfi‘ «.mtfi g quantity of charge flowing per second swif- Cl-OSS —sectional area is A and since only electrons are mobile one can write I = na (— e) A (— VD) where nu = # of mobile electrons/ m3 —e = l.6x10"19C VD = drift Speed “Hi 2 Notice: Direction of I is opposite to that of electron drift. Recall: VD E 10'4m/ 5. While Vrm, E 105m/ 3 at 300K. This is becaUSe ions are stationary and act as scattering centers. Electron has a very tortuous path so although the speed between collisions is high the entire electron “cloud” drifts rather slowly. V Since transport costs energy, potential must drop. Hence, definition of resistance V R = — I For a particular piece of conductor R = p; 4:0 I Z=length, A= cross—section p = resistivity (material property) DQ A T f. t' f l — = ~ M— [c conduc ion 0 lea At Ax ] V A VA V I = — = g = — 2 GA“ 0: electrical conductivity R t pl I It is instructive to write I = J oA J = current density vector —> —> —> and we know that V I SO .]:0'E _, __> That is, if you apply an E -field to a conductor it responds by setting up a current density. —> proportional to E, the proportionality factor being the conductivity (electrical). 8/)2, KIRCHHOFF’S RULES: PHYSICAL BASIS Loop Rule: Change of potential between two points is independent of the path because potential is derived from potential energy and the latter is defined for a CONSERVATIVE force so net change of potential on a closed loop must be zero. Pai"rem~h {QUE ad? flwéi’ P D M f,“ Vs U Mfg-g a El 9 [Recall that in a Thermodynamic cyclic process dU = 0 , Thermodynamic potential (Internal energy)] Junction Rule: Flow of charge is continuous, i.e., apart from what is involved in setting up the original field to drive a current, there can be no continuous accumulation (depletion) of charge at junction consequently, ' / Ci; (ff/lb ,, ‘ {'5‘ (let 1716577" 15 fl/{ux 652$ fi)CZ/VL<‘7€,/ 534:? a (15) VVS exam/at. ”’fi’ 6’. m 4,6 Mai/re“ e, W) fl‘g’é‘ “3‘3“: (gm/Ea- 5X flaawigw 11 \ 16 01044515" /' 15 z: 12 QSZM'fl/ 17 (oi/(5601,15 e W19 18 Z 1011!: Z [in Next, put all 3-devices togetherywa (1, C4) racer ‘3 '+ g - SWITCH l—-12 C At i=0, connect 1 to 2: charge will flow from battery to capacitor plates. CHARGING NOTICE NO CURRENT IN CAPACITOR. (M2, 10/12. At t=0, no charge 011 C, VC = 0, Potential at various points looks like _ 6‘ 8010:} That is, R limits the maximum current that can flow so it will take time to build up charge on C. + _ a little later C has ql ‘q , VC = %, and the Potential becomes A B A Eventually q —) Cg laié/Z; Mathematicallyzil" Cow; game. So now VC =8 , i=0 Switch from 2 9 1 to 2 9 3. circuit is S‘I’M/Yf C/VOC/k (1,3qu W {:6 “Etch €52, CHARACTERISTIC TIME 1': RC!!! g 1/ v 7*“ I k L? c: a 9,4, V 50 ”£6“??ng firs->9 7’ V Q I [EC-74 % a? 8' éjfl‘m efiJSybm (a; 'ZLTWJQ‘, 1» I/{G’U‘NG HQL‘DHF ‘9 65/: c ”79% GE [Seam NE TERM/NHL 0pc; )2 QNTAOLs" IQ/QTE To BE 'TTQWSFERISEQ WHMCF E RE—N <34: D ILS’C H7612 6’7 [N6], Wou N 'T (9/: (9 ...
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