Gauss's Law. EAcos (for E = constant, surface flat ) 1 of 11

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1 1 of 11 Gauss's Law Gauss's Law is one of the 4 funmental laws of electicity and magnetism called Maxwell's quations. Gauss's law elates chages and electic fields in a subtle and poweful way, but befoe we can wite down Gauss's Law, we need to intoduce a new concept: the electic flux though a suface. suface with aea Conside an imaginay suface which cuts acoss some -field lines. We say that thee is some electic flux though this suface. To make the notion of flux pecise, we must fist define a suface vecto. Definition: suface vecto = nˆ, associated with a flat suface of aea. Magnitude of vecto = aea of suface. Diection of vecto = diection pependicula (nomal) to aea suface = diection of unit nomal ˆn. Notice that thee is an ambiguity in the diection ˆn. vey flat suface has two pependicula diections. smalle aea, shote The electic flux though a suface is defined as cos (fo = constant, suface flat ) The flux has the following geometical intepetation: flux the numbe of electic field lines cossing the suface. Think of the -field lines as ain flowing thew an open window of aea. The flux is a measue of the amount of ain flowing though the window. To get a big flux, you need a lage, a lage, and you need the aea pependicula to the -field vecto, which means the aea vecto is paallel to. (In the ain analogy, you need the window to be facing the ain diection.) =, cos =1, max = 9 o, cos =, = PHY112 Lectue Notes, Dubson, 9/9/214

2 2 of 11 ( cos is the pojection of the aea onto the plane pependicula to. The plane pependicula to is the aea which "faces the ain". Only the aea facing the ain contibutes to the flux. aea = cos Let s conside this flux business in a little moe detail. In the diagam below, we have a constant electic field, passing though suface 1, epesented by vecto, tilted at angle. [We use bold font fo vectos.] This tilted suface 1 has aea = = LW. The pojection of this suface 1 onto the plane pependicula to the -field is suface 2, which has an aea that we call (fo aea of suface pependicula to diection of ). The aea of this plane, this suface 2, is = L W = LcosW = cos. o we have = cos. uface 2: aea = = cos W uface 1: aea = = LW L = L cos L Now we ae going to show that the flux though a suface is popotional to the numbe of field lines passing though the suface. Recall that, fom the definition of a field line diagam, the magnitude of the -field is popotional to the density of the lines: (# field lines)/ N / (N is the numbe of field lines though the aea ). o, we have N. Now, we showed above that cos, so we have N cos. Done! The numbe N of field lines though a suface is popotional to the flux. PHY112 Lectue Notes, Dubson, 9/9/214

3 3 of 11 We can now see that, since the same numbe of -field lines pass though both sufaces 1 and 2, they must have the same magnitude flux. The math shows the same thing: Fo suface 1, 1 cos. Fo suface 2, 2 cos. Now, the fomula is a special case fomula: it only woks if the suface is flat and the -field is constant. If the -field vaies with position and/o the suface is not flat, we need a moe geneal definition of flux: "suface integal of " To undestand a suface integal, do this: in you imagination, beak the total suface up into many little segments, labeled with an index i. The suface vecto of segment i is i. If the segment is vey, vey tiny, it is effectively flat and the -field is constant ove that tiny suface, so we can use ou special case fomula. The flux though segment i is theefoe i i i. ( i is the field at the segment i) The total flux is the sum: ii i (In the limit that the segments become infinitesimal, thee ae an infinite numbe of segments and the sum becomes an integal.) In geneal, computing suface integal can be extemely messy. o why do we cae about this thing called the electic flux? The electic flux is elated to chage by Gauss's Law. PHY112 Lectue Notes, Dubson, 9/9/214

4 4 of 11 Gauss's Law (the 1 st of 4 Maxwell's quations) q enclosed In wods, the electic flux though any closed suface is a constant (1/ ) times the total chage inside. "closed" closed suface integal suface is closed if it has no edges, like a sphee. Fo a closed suface, the diection of is always the outwad nomal. The constant is elated to k by k F k q q q q coul (I units) Gauss' Law can be deived fom Coulomb's Law if the chages ae stationay, but Gauss's Law is moe geneal than Coulomb's Law. Coulomb's Law is only tue if the chages ae stationay. Gauss's Law is always tue, whethe o not the chages ae moving. It is easy to show that Gauss's Law is consistent with Coulomb's Law. Fom Coulomb's k Q 1 Q Law, the -field of a point chage is. We get the same esult by applying Gauss's Law: imaginay spheical suface, adius 1 Q olving fo, we have Done. Q (since is paallel to on ) (since is constant on ) (says M. Gauss) PHY112 Lectue Notes, Dubson, 9/9/214

5 5 of 11 When viewed in tems of field lines, Gauss's Law is almost obvious (afte a while). Recall that flux is popotional to the numbe of field lines passing though the suface. Notice also that flux can be positive o negative depending on the angle between the - field vecto and the aea vecto. Whee the field lines exit a closed suface, the flux thee is positive; whee the field lines ente a closed suface, the flux thee is negative. o the total flux though a closed suface is popotional to [(# field lines exiting) (# field lines enteing)] (exiting ) If a closed suface encloses no chages, then the numbe of lines enteing must equal the numbe of lines exiting, since thee ae no chages inside fo the field lines to stop o stat on. (enteing ) imaginay suface o only chages inside the suface can contibute to the flux though the suface. Positive chages inside poduce positive flux; negative chages poduce negative flux. The net flux is due only to the net chage inside: q enclosed. Using Gauss's Law to solve fo the -field Gauss's Law is always tue (it's a LW). But it is not always useful. Only in situations with vey high symmety is it easy to compute the flux integal. In these few cases of high symmety, we can use Gauss's Law to compute the -field. PHY112 Lectue Notes, Dubson, 9/9/214

6 6 of 11 xample of pheical ymmety: Compute -field eveywhee inside a unifomly-chaged spheical shell. Q By symmety, must be adial (along a adius), so = (). We choose an imaginay suface concentic with and inside the chaged sphee. ince the -field is adial and the suface vecto on is also adial, we have. (The dot poduct of the paallel vectos is just the poduct of the magnitudes.) o we have whee is the aea of suface. We ae able to take outside the integal only because = () and so = constant on the suface. qenc Gauss's Law says, so we have =, =. Conclusion: = eveywhee inside a hollow unifom sphee of chage. If you daw the spheical gaussian suface outside the chaged shell, you can quickly 1 Q show that 2 ˆ eveywhee outside the shell. The -field outside a unifom 4 shell of chage, o outside any spheically symmetic chage distibution, is exactly the same as if all the chage was concentated at the cente. xample of Cylindical ymmety: Compute -field aound an infinite line of chage with chage pe length =. gaussian suface, adius, length L L By symmety, is in the cylindically adial diection and = ()., is in plane to line aea of cuved side = = 2 L L PHY112 Lectue Notes, Dubson, 9/9/214

7 7 of 11 ends side side (2 L) side The chage inside the gaussian suface is (chage/length)length = L, so Gauss gives gives q L 2 enc (2 L ) xample of Plana ymmety: Compute the -field nea an infinite plane of chage with Q chage pe aea. By symmety, the -field must be pependicula to the plane (eithe away o towads). end cap aea gaussian suface chage enclosed On end caps, On cuved side, q enc 2 2 = constant, egadless of position! PHY112 Lectue Notes, Dubson, 9/9/214

8 8 of 11 = constant = constant unifom plane of chage (seen edge-on) Conductos in lectostatic quilibium "lectostatic equilibium" means that all chages ae stationay; so the net foce on evey chage must be zeo (othewise the chage would be acceleating). Useful facts about metals (conductos) in electostatic equilibium: The electic field in the inteio of a metal must be zeo. The -field must be zeo in the inteio, othewise the conduction electons in the metal would feel a foce F = q = e and would move in esponse. lectons in motion would mean we ae not in electostatic equilibium. In a conducto, the chages aange themselves so the = eveywhee in the inteio (othewise, the chages ae not yet in equilibium and continue to move). The inteio of a conducto in equilibium can have no net chage (electons and potons must have equal density). Poof fom Gauss's Law: conside any closed suface within the conducto, (since = ) q enclosed conducto ny net chage on the conducto esides only on the suface of the conducto. PHY112 Lectue Notes, Dubson, 9/9/214

9 9 of 11 We just showed that no net chage can exist in the inteio, so it must be on the suface. The electic field must be pependicula to the suface of the conducto. The -field must be pependicula to the suface (in electostatic equilibium), othewise the component of the -field along the suface would push electons along the suface causing movement of chages (and we would not be in equilibium). x On the suface of metal, if x was not zeo, thee would be a foce F x = q x = e x on electons in the metal pushing them along the suface. metal net chage on suface only = inside -field suface metal in electostatic equilibium The -field nea the chaged suface of the conducto has magnitude. [Note: this is simila, but diffeent than the fomula fo the field nea a plane of chage: 2. ] Poof by Gauss's Law: ai metal gaussian "pillbox", = aea of one cap = chage pe aea q enc PHY112 Lectue Notes, Dubson, 9/9/214

10 1 of 11 Why not 2? Because = inside metal ai metal = inside metal But now, a vey puzzling situation has aisen! We poved befoe that an infinite plane of chage ceates a unifom field. But now we have also poved that a plane of 2 chage on the suface of a metal makes a field. 2 How can both be coect? The -field nea the suface of a metal is not only due to the chages on that neaby suface. The -field is always due to all chages, including chages on fa-away sufaces! PHY112 Lectue Notes, Dubson, 9/9/214

11 11 of 11 Metal slab with net positive chage. suface 1 suface 2 (ssume that the sufaces on the top and bottom ae small and fa away so that tot is due to sufaces 1 and 2 only.) tot The chages all aange themselves so that = inside the metal just outside metal PHY112 Lectue Notes, Dubson, 9/9/214