Physics 11b Lecture #2. Electric Field Electric Flux Gauss s Law

Transcription

1 Physcs 11b Lectue # Electc Feld Electc Flux Gauss s Law

2 What We Dd Last Tme Electc chage = How object esponds to electc foce Comes n postve and negatve flavos Conseved Electc foce Coulomb s Law F Same nvese-squae law as gavty Sgn makes the dynamcs dffeent Supeposton Pncple e = 1 F Q qq 1 ˆ Q q = Q ˆ Q

3 Today s Goals Contnue on the supeposton pncple Wok out a couple of poblems Electc dpole Contnuous chage dstbuton Intoduce electc feld E Fst (and the easest) of the E&M felds Feld lnes Defne electc flux Φ E Gauss s Law

4 Supeposton Pncple Suppose we have many chages dstbuted n space What s total foce on chage Q fom all the othe chages? Pncple of Supeposton Just add them up Foce fom each one not affected by the othes F Q = = Q 1 Qq q Q Q ˆ ˆ Q Q Foce fom the -th chage q q q -q 3q -q -q Q

5 Electc Dpole Two chages +q and q sepaated by d Such pa s called an electc dpole Bng anothe chage q neaby We call t a test chage What s the foce on q? Ths s easy Calculate the foce fom +q What s the dstance between +q and q? Calculate the foce fom q Add the vectos q + q q d

6 Electc Dpole Foce F 1 fom +q s F 1 qq 1 qq = = ( ) d + ( ) 1 πε πε + d Same sze fo F fom q Vecto addton a hozontal F What s the length? F F d d : = + 4: 1 q + q q d F 1 F F F = 1 qq d 4 πε ( + d 4) 3

7 Contnuous Dstbuton Electc chage s often dstbuted ove objects, e.g. Along a length of we (1-d) On the suface of a metallc plate (-d) Insde the volume of a conductng ball (3-d) Supeposton pncple needs to be modfed Q q F ˆ Q = Q Sum ove ndvdual chages Q Q dq F ˆ Q = Integal ove lne/suface/volume

8 Unfom Lnea Chage Chage Q unfomly dstbuted on a thn od of length l Test chage q s nea the mddle of the od Foce on q? y q x dx x Smat set up s eveythng Defne x-y coodnates Pck a small pece of the od and call t dx Calculate the foce fom dx and ntegate

9 Unfom Lnea Chage The pece dx has chage Foce fom ths pece s 1 Q dx q df = + x Note that the foce fom the geen pece s same sze and flpped n x Only the y-component wll eman afte addng up Q 1 dx q dfy = + x + x Q dx Lnea chage densty (C/m) df q x dx y-component Integate ths!

10 Unfom Lnea Chage Integate df y F = = Qq πε 1 Qq 4 πε ( + x ) ( x ) dx dx A lttle (?) math wok hee df y q F = Qq x dx yˆ Ths wasn t easy Is t eally coect? Check the decton, -dependence, lage- lmt

11 Electc Feld Look at the supeposton pncple agan What ths chage s F Q Q = Most of the nteestng pat has nothng to do wth the value of Q tself But the poston of Q s mpotant Let s call t Q = ˆ = Q = Q q Q ˆ Q What all the othe chages ae and whee they ae F Q = Q q( ) 3 4 πε A functon of

12 Electc Feld How about wtng E() s a vecto functon of poston F Q 1 q( ) = Q E() = Q 3 4 πε Fo each poston n space, thee s a vecto E Also called a vecto feld It tells you what wll happen f you put a chage hee We call t the electc feld Unt of E s Newton/Coulomb (N/C) Smple case: only one chage q at = 1 q 1 E () = = q 3 ˆ

13 Electc Feld E feld ceated by a sngle chage 1 q E () = ˆ E ponts outwad Guess why t s called feld We ve ntoduced a mddle-man fo the electc foce Chage ceates E feld E feld ceates foce on chage End esult s unchanged (of couse)

14 Feld Lnes Tedous (and busy) to daw lttle aows eveywhee Took me 1 mnutes to daw Faaday came up wth an ease way = feld lnes Instead of lttle aows, daw long, contnuous aows ognatng fom the chage Feld lnes epesent decton of the E feld just as well as the lttle aows What about the sze of E?

15 Feld Lne Densty E feld weakens wth dstance Invese-squae law 1 q E = Feld lnes become less cowded wth dstance Consde a sphee at Suface aea S = 4π Numbe of feld lnes N s constant Densty N N = = = Nε S 4π q E Densty of feld lnes s popotonal to the E feld stength E

16 Feld Lne Rules You can often daw feld lnes wthout calculaton Just follow a few ules Feld lnes stat fom a postve chage and end n a negatve chage Excepton: t s OK to come fom/go to nfntely fa away Feld lnes cannot splt, mege, o coss each othe The numbe of feld lnes attached to a chage s popotonal to the amount of the chage The last ule ensues feld lne densty E feld

17 Examples wth Two Chages Examples fom the textbook

18 Numbe of Feld Lnes Imagne a small aea of a suface n an electc feld Let s call the aea A It could be any shape, any angle How many feld lnes un though ths aea? A E Remembe: densty of feld lnes E Numbe of feld lnes should be E A Rght?

19 Numbe of Feld Lnes The answe depends on the angle between E and the suface Defne a unt vecto n standng nomal to the suface A n θ E Pojected n the decton of E, the aea A A cosθ Numbe of feld lnes E A cosθ We can also wte ths as E A whee A A n We can ntegate ths to fnd the numbe of feld lnes passng though any suface

20 Electc Flux We now defne electc flux as S s a suface aea, da = da n Φ E = E d A Φ E s popotonal to the numbe of feld lnes gong thu S Unt of Φ E s Newton m /Coulomb (N m /C) S NB: sgn of flux depends on the decton of n That s, you must defne whch sde of S s postve S = 1m n Φ E = 1 Nm /C n E =1N C Φ E = 1 Nm /C

21 Smple Flux Example 1 Consde a sphee of adus aound a chage q Defne n to pont outwad We know E and n ae paallel 1 q We also know E = Φ E = E d A = 4π E = sphee q ε q n Ths s hadly supsng Φ E should be popotonal to the numbe of feld lnes comng out of chage q, whch should be popotonal to q We just ddn t know that the constant was 1/ε

22 Smple Flux Example Consde a sphee n a unfom E feld Take pola coodnates (θ, φ) elatve to the decton of E Φ = E E cosθ sn d d = = π E cos sn θ θ φ π θ θdθ θ n Agan hadly supsng Incomng E on one sde s balanced by outgong E We know feld lnes neve dsappea wthout a chage

23 Gauss s Law Net flux though a closed suface s gven by the net chage nsde the suface by Φ E = E d A = qn ε Ths seems natual, fom what we ve found so fa Fomal poof s found n textbook 4.5 Ths s moe useful than t looks We wll see n the next lectue

24 Summay Intoduced electc feld E by Feld lnes and the ules Fom a postve chage to a negatve chage No splttng, megng, o cossng Numbe of feld lnes amount of chage Densty E feld Defned electc flux numbe of feld lnes though the suface Gauss s Law qn Φ E = ε S F Q Φ E = E d A = Q E()