PHY2061 Enriched Physics 2 Lecture Notes. Gauss Law
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1 PHY61 Eniched Physics Lectue Notes Law Disclaime: These lectue notes ae not meant to eplace the couse textbook. The content may be incomplete. ome topics may be unclea. These notes ae only meant to be a study aid and a supplement to you own notes. Please epot any inaccuacies to the pofesso. Peamble In this lectue we lean a simple and poweful technique fo calculating the electic field fo situations involving a high degee of symmety. Flux Flux, Φ, is a measue of the amount of a vecto field, F, passing though a suface with aea A (diection defined pependicula to the plane of the suface): Φ= F A =FAcosθ whee θ is the angle between the field diection and the nomal to the aea suface. As one example, conside the flow of ai o wate though a suface. The voulume of a passing though the suface pe unit time is given by: Φ= v A =vacosθ Obviously this flow ate depends on the oientation of the suface. As anothe example, conside the light fom the un impinging onto the suface of the Eath. Obviously land nea the equato eceives a lage light flux than that at the poles. In this couse, we will conside the flux of electic and magnetic fields (essentially light!) passing though a suface. D. Acosta Page 1 9//5
2 PHY61 Eniched Physics Lectue Notes Flux Though a Closed uface Fig. fom HRW 7/e Conside the flux of electic field though a closed suface. Let s beak the suface into a lage numbe (N) of small elements of aea A that ae nealy flat. By convention, let s take the diection of the aea vecto to be pependicula to the suface and pointing out. The flux of electic field passing though this closed suface is then: N E i= 1 Φ= A If we let N and A, then the small elements become infinitesimal and we can wite: Φ= E d A This is a closed suface integal. It is a -dimensional integal ove a -dimensional suface. The cicle in the integal sign denotes an integal ove an entie closed suface (vesus just a single face). D. Acosta Page 9//5
3 PHY61 Eniched Physics Lectue Notes Example E y A bottom A top A ight x z Conside a squae cube immesed in a constant electic field: E= E x. Calculate the flux though each face. A i is the aea vecto pointing out fom cube face i, which has magnitude A i = s, whee s is the side length of the cube. Note that E Atop = since E Atop E Abottom = Also: E Afont = E A = back What is left is: E Aight = E Aight = Es since E Aight E A = E A = Es since E opposite A left left left ˆ Φ= E A = = sum ove all 6 faces 6 1 i Es i Es = Thus, the total flux into the cube is balanced by the flux going out. This will hold fo any abitay shape of the suface, povided the electic field is constant D. Acosta Page 9//5
4 PHY61 Eniched Physics Lectue Notes Flux of a Point Chage Conside a spheical shell suface of adius enclosing a point chage at the cente. da E q Note that da points out fom the closed suface, and that since E points adially as well, E da so that E da= E da q By Coulomb s Law, E = K o the flux though the spheical suface is: Φ= E da= E da = E da since E is constant at fixed adius d A = 4 π suface aea of sphee To see this explicitly, let s do the suface integal: π π da = sinθdθ dφ = 4π o o, the flux is: q 1 q 4π Φ= E 4π = K 4π = 4 1 q Φ= π Note that this is a constant, independent of the size of the sphee, and only dependent on the amount of chage enclosed. In fact, it tuns out that it does not even depend on the shape of the suface! D. Acosta Page 4 9//5
5 PHY61 Eniched Physics Lectue Notes Law Law genealizes the pevious example, and states that: q Φ= E d A = enc o the amount of flux passing though any closed suface depends only on the net amount of electic chage enclosed. Why is this geneal case tue? Well, and abitay suface can always be inscibed by a sphee within it. And the flux though that sphee, calculated in the pevious example, must be equal to the flux intecepted by the oute suface. q Law elates the electic flux though a closed suface to the net chage enclosed. It was deived as a consequence of Coulomb s Law, but in fact is completely equivalent to Coulomb s Law. It can fom an altenative stating point fo calculating electic fields. D. Acosta Page 5 9//5
6 PHY61 Eniched Physics Lectue Notes Applications of Law Law is a poweful technique to calculate the electic field fo situations exhibiting a high degee of symmety. Infinite heet of Chage Let s calculate the electic field fom an infinite sheet of chage, with a chage density of σ (measued in C/m ). E E A left A ight By symmety, we expect E to point pependicula to the suface fo a vey lage sheet (and fa fom the edges). We would not expect any othe angle, because why would any paticula diection be pefeed ove any othe fo such a symmetic situation? Now conside a closed suface (which we will call a ian suface) that extends though the sheet of chage. The sides, hee assumed cylindical, ae chosen to be pependicula to the sheet. o E A =. The caps ae paallel to the sheet, so E A = E A left ight though the suface is thus: Φ= E Aleft E A ight = EA By Law, σ A Φ= EA = side since both vectos always point in the same diection. The total flux q enc Φ=, whee qenc = σ A. o: σ E = This is exactly the same solution fo the magnitude of the electic field fom an infinite sheet that we obtained by painstaking integation of ings of chage! And as noted befoe, the electic field is a constant and does not depend on the distance fom the sheet. D. Acosta Page 6 9//5
7 PHY61 Eniched Physics Lectue Notes Infinite Line of Chage Let s calculate the electic field a distance fom a line of electic chage infinite in extent with chage density λ (measued in C/m). y E da x z By symmety, we expect that E points in the adial diection (no pefeed diection): E= E ( )ˆ Choose a ian suface with cylindical geomety, with top and bottom caps aligned such that thei aea vectos point in the same diection as the line of chage (and pependicula to the electic field): E Atop = E A bottom =. o the total flux leaves though the sides: ( ) ( ) Φ= E da= E da = E π h since E is constant at a fixed adius. Thus, qenc λh Φ= E( ) π h= = λ Kλ E( ) = = π ame as deived by integation in the last lectue. D. Acosta Page 7 9//5
8 PHY61 Eniched Physics Lectue Notes phees of Chage Let s calculate the electic field outside a spheically chaged shell. The adius of the sphee is R. A total chage Q is spead unifomly on the suface, so the Q chage density pe unit aea is σ = 4π By symmety we expect that the electic field points adially: E= E ( ) ˆ Fo ou ian suface choose anothe sphee with adius chaged sphee. > R, centeed on the ( ) Φ= E da= E da since E is constant at a fixed adius. The suface integal is just the integal of the suface aea of a sphee, so: Q Φ= = E ( ) 4π Q Q E( ) = = K 4π Exactly the same field stength as if all the chage on the sphee wee placed at the cente of the sphee as a point chage! This esult applies to a solid sphee of total chage Q as well. The situation changes fo the electic field inside a spheical shell. Fo < R, then thee is no net enclosed chage, and E( ) =. No foce acts on any point chage placed anywhee inside a chaged shell (foces fom all infinitesimal chages in the shell balances). Fo the electic field inside a solid sphee of total chage Q, then thee is a net enclosed chage. Fist assume that the total chage of a solid sphee is spead unifomly thoughout its volume. The volume chage density is theefoe: Q ρ = (measued in C/m ). 4 π R The chage enclosed fo < R is: D. Acosta Page 8 9//5
9 PHY61 Eniched Physics Lectue Notes 4 Q 4 q Q π R R enc = ρ π = π = 4 o Law tells us: E A ( ) Φ= d = E 4π = Q E( ) = 4π R Q R That is, the electic field ises linealy fom at the cente of the sphee to the magnitude obtained at the suface of the sphee (which is the same as that fom a point chage at a distance R). D. Acosta Page 9 9//5
10 PHY61 Eniched Physics Lectue Notes Electic Fields inside Conductos The pevious examples of continuous chage distibutions have implicitly assumed that the infinitesimal chages wee not fee to move locked in place within the medium. A conducto, on the othe hand, has some electical chages fee to move (the valence electons in the metal). o a conducto can cay an electic cuent (units: amps, C/s). In electostatic situations, which means that the chages have eached static equilibium inside the conducto, we must have: E =. If the field wee not zeo, that would imply a foce acting on the chage caies ( F= qe= ma), which would cause them to acceleate. Thus the situation would still be dynamically in motion and not have eached equilibium. But once equilibium has been achieved (vey quickly unless an electomotive foce maintains an electic potential diffeence acoss the conducto see next chapte), the field must be zeo. Now let s exploe the consequences of this obsevation. If E = inside the conducto, then the flux of electic field passing though any closed suface inside the conducto is also zeo: Φ= E d A = Chage Q on outside uface inside conducto whee E= and Φ=. o by Law, thee can be no net enclosed chage inside the suface. Theefoe, fo conductos, all deposited chage must be on the outside of the conducto. The same applies even if cavities exist inside the conducting volume. Now electic chage could exist inside a non-conducting cavity within the conducto, but in that case sceening chages (of opposite sign to the oiginal chage) must move to the inne suface of the conducto in ode to sceen the electic field aising fom this inne chage so that E= in the conducto. (Thus fo a ian suface containing this inne suface and the chage, the net chage would be zeo and by Law, the flux and field would be zeo as well). Now because the conducto was pesumably balanced in chage, the distibution of chage on the inne suface will esult in an excess of chage on the oute suface of the conducto equal in magnitude and sign to the enclosed chage. D. Acosta Page 1 9//5
11 PHY61 Eniched Physics Lectue Notes Enclosed chage Q Cavity with Q deposited along inne suface A ian suface Conducto with Q on oute suface A Van de Gaff acceleato exploits this phenomenon that the chage inside a conducto always moves to the outside suface, no matte how much chage is aleady thee. By scaping static chage fom a mateial by fiction, we can tansfe this tiny amount of chage via a belt to the inside of a conducting shell (which by the popety above must not have any chage on its suface). A faction of this chage will then be tansfeed to the shell, and then immediately it will move to the outside suface of the shell whee it can accumulate continuously to vey lage values. This phenomenon that E= inside a conducto also has implications fo electomagnetic waves. If a adio tansmitte (such as a cell phone) is placed inside a metal box, the electomagnetic fields it geneates cannot penetate though the metal. It is shielded. In fact, the metal need not be completely solid: a sceen o even chicken wie woks faily well at shielding electic fields. uch a box is known as a Faaday cage. You can make one youself by wapping you cell phone in aluminum foil and then tying to call its phone numbe. The phone will not ing even when the phone is on. You might also notice this effect if you happen to be talking on you cell phone while you ente an elevato that is completely suounded by metal. Once the doos close, call is ove! D. Acosta Page 11 9//5
12 PHY61 Eniched Physics Lectue Notes Electic Fields on the uface of Conductos Thee ae also constaints on the diection of the electic field at the suface of a conducto in electostatic equilibium. Outside the conducto, we must have that the electic field diection is pependicula to the suface of the conducto (paallel to the aea vecto). If it wee not, thee would be a component of the electic field paallel to the suface, and that component would cause the fee chages in the conducto to acceleate (thus not being in equilibium). Afte ashot amount of time, all chages will edistibute themselves and the electic field in this configuation will always be pependicula to the suface. E Conducto suface Chage acceleates Conducting sheet of Chage Let s now calculate the magnitude of the electic field aising fom electic chage applied to the suface of a conducto (we know the diection will be pependicula to the suface). E E A ight We solve this poblem in a manne simila to the static sheet of chage solved ealie, except that now we do not extend the ian suface though the sheet, but stop it inside the conducto (it will always have some thickness). The flux though the sides of the ian suface is zeo because the electic field and aea vectos ae pependicula. The flux though the inne left cap is also zeo because E= inside a conducto. Thus the total flux only aise though the ightmost cap: qenc σ A Φ= E Ai = E A ight = EA = = i o we have fo the magnitude of the electic field: E = σ D. Acosta Page 1 9//5
13 PHY61 Eniched Physics Lectue Notes Note that this is exactly two times lage than the same expession fo a non-conducting sheet of chage with chage density σ. To undestand why this is the case, note that if a chage Q is deposited on the suface of a conducto, it will spead and unifomly cove both sides of the conducting sheet. The chage density on each side will be: ( Q /) Q σ = = A A Now let s examine the contibution to the total electic field fom each side of the sheet sepaately, and then add the two togethe to get the total field. E R E L E R E L E L E R Hee you can conside each suface as a sheet of chage on a non-conducting suface (it is infinitesimally thin). The magnitude of the field fom these sufaces is: σ Q ER = EL = = 4 A If we assume that the sheets ae lage compaed to the thickness of the conducto, the fields fom each suface cancel inside the conducto (oppositely diected fields). Outside the conducto, the fields add in the same diection, so the magnitude is: Q Q σ ETOT = EL ER = = = 4 A A D. Acosta Page 1 9//5
14 PHY61 Eniched Physics Lectue Notes Method of Image Chages We agued that the electic field must always be pependicula to the suface of a conducto in equilibium. That means any chage placed nea the suface of a conducto must have field lines that stop at the suface of the conducto pependicula to the suface. conducto Thee is a poweful tick to solve fo the shape of the field lines fo such electostatic situations known as the method of image chages. The tick is to place an imaginay opposite sign chage of equal magnitude to the fist chage behind the suface of the conducto (obviously whee we cannot actually place a physically chage). Fo the example above, we have actually solved this befoe! It s an electic dipole. Fig. fom HRW 7/e This technique can be applied to othe chage and conducto distibutions. In fact, images of images may even be necessay if a chage is placed between two conductos. D. Acosta Page 14 9//5
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