A r. (Can you see that this just gives the formula we had above?)


 Sabrina O’Brien’
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1 241 (SJP, Phys 1120) lectic flux, and Gauss' law Finding the lectic field due to a bunch of chages is KY! Once you know, you know the foce on any chage you put down  you can pedict (o contol) motion of electic chages! We'e talking manipulation of anything fom DNA to electons in cicuits... But as you've seen, it's a pain to stat fom Coulomb's law and add all those dan vectos. Fotunately, thee is a emakable law, called Gauss' law, which is a univesal law of natue that descibes electicity. It is moe geneal than Coulomb's law, but includes Coulomb's law as a special case. It is always tue... and sometimes VRY useful to figue out fields! But to make sense of it, we eally need a new concept, lectic Flux (Called Φ). So fist a "flux intelude": Imagine an field whose field lines "cut though" o "piece" a loop. Define θ as the angle between and the "nomal" o "pependicula" diection to the loop. We will now define a new quantity, the electic θ flux though the loop, as Flux, o Φ = A = A cosθ is the component of pependicula to the loop: = cosθ. A, the "nomal" to the loop Fo convenience, people will often chaacteize the aea of a small patch (like the loop above) as a vecto instead of just a numbe. The magnitude of the aea vecto is just... the aea! (What else?) But the diection of the aea vecto is the nomal to the loop. That way, we can wite Φ = A. (Can you see that this just gives the fomula we had above?) Flux is a useful concept, used fo othe quantities besides, too..g. if you have sola panels, you want the flux of sunlight though the panel to be lage. House #2 has pooly designed panels. Although the ARA of the (Lots of flux) Sola panel 1 2 House (Less flux) Sola panel 2, same aea, difffeent tilt. panels is the exact same, and the sunshine bightness is the exact same, panel 2 is less useful: fewe light ays "piece" the panel, thee is less FLUX though that panel.
2 242 (SJP, Phys 1120) xamples of calculating flux: A Aea, A Hee (pictue to the left) Φ = A, because is paallel to the aea vecto. (θ=0) (If pointed the opposite diection, you'd have a "negative" flux!) Note: I've dawn an aea "A" vecto fo you: emembe, that's defined as pependicula to the suface in question. Hee (pictue to the ight), Φ =0, because is pependicula to the aea. (θ=90. ) No flux: the field lines don't "piece" this loop, they "skim" past it... θ Hee, (pictue to the left), Φ = A cosθ. The flux is educed a bit because it's not pefectly pependicula. Again, the A vecto is (by definition) pependicula to ou A suface. (So θ is also the angle between and A, and Φ = A = A cosθ) What if is not constant? O, if the aea cuves? You can still find the flux, you just beak you suface up into little "patches", and add up (integate!) all the small fluxes though each little patch. Fo Gauss' law (that's what we'e afte, emembe!) we'e going to have to evaluate the electic flux though a CLOSD SURFAC: like the suface of a balloon, o a soda can, o a cube... o just some imaginay suface in space. (Closed means thee's an inside and an outside. So the patches shown above ae not closed, but a cube is, o the funny ball shown below...) Of couse, any suface can be thought of as the sum of a bunch of little patches! Just imagine taking any suface and dawing a little "gid" on it, completely coveing it. ach tiny patch will have tiny aea (da). So, the flux though a closed suface is the SUM (eally, INTGRAL) of the flux though all the patches coveing the suface. (The convention is always that da points eveywhee out of a closed suface) We wite this flux fomally as Φ = da (The weid cicle aound the integal is to emind you that we'e talking about an integal, o sum, all the way aound some closed suface!) Aea, A In a cude sense, flux counts how many "field lines" piece out a suface! da da A da
3 243 (SJP, Phys 1120) asy example #1: is unifom (constant) fom left to ight, and ou suface is a cube with face aea A. What's the total flux though the (closed) cube shown? We need to add up the flux though all 6 faces. Now, the top and bottom faces, (3 and 4), have an aea vecto which is pependicula to. So the flux though them is zeo. ( "skims" them, doesn't "poke though" them). Convince youself! The exact same agument is tue fo the font and back faces (not numbeed in the pictue). Again, each has zeo flux. What about face 2? Hee, flux is d A (though wall 2) = A = A What about face 1? This time, the flux is A. The minus sign comes because aea vectos always point OUT. That means A(face 1) points left, but points ight, and the dot poduct is thus A cos(180) = A. The total flux though entie suface is thus Flux(1)Flux(2)... Flux(6) = A A = 0. Thee is NO net flux though this suface! That's coect: the numbe of lines "poking in" on the left is equal to the numbe "poking out" on the ight, the total is thus zeo. Notice: If the field and/o the suface ae simple enough, you don't eally have to DO any integal  just think of the SUM of fluxes though all pats of the suface, and add 'em up. A 1 1 A A 2
4 244 (SJP, Phys 1120) asy example #2: You have a point chage q at the cente of a pefect sphee of adius R. What is the flux though the sphee? q da Hee's the key thing you have to ealize. Since is eveywhee "adially outwad", and the little aea vectos da at any patch on a sphee ae also "adially outwad", then on ANY little patch, and da ae paallel! That means.da = da (the cosine in the dot poduct ALWAYS gives one, at evey point eveywhee aound the sphee!) Fom Coulomb's law, we know that at the suface of this sphee, = kq/r^2 (It's the same in magnitude eveywhee aound the sphee!) So we want to compute Φ = d kq A, which is thus just R da. 2 But now notice that those ae all constants, and constant come out of integals. So Φ = kq da. R 2 Now think what that funny integal means. It means "sum up da fo all patches eveywhee aound the sphee". But that's just the total aea of a sphee, 4 π R^2! So the R^2 cancels, and Flux = 4 π k q. It's a constant (independent of adius. ) It's positive (if q is ): thee is a net flux pointing OUT of the sphee! Remembe, flux is basically "counting flux lines that poke out". If we moved q a little to the side, the flux would be bigge on that side, but smalle on the othe. But still the total numbe of flux lines poking out would be the same! The total flux would be the same no matte whee q was, as long as it is somewhee inside the sphee. If you added a second chage, q2, it would add moe flux = 4 π k q2. So the total flux would just be 4 π k (q q2). (No matte whee pecisely those chages wee inside the sphee.) This leads us to the big idea of this chapte. If you have a suface with chages inside it, the total flux though that suface will just be 4 π k * (q total, inside) 1 Since we defined k = back in Chapte 22, we can wite this as 4π Φ = da This equation is called "Gauss' law". We just showed it fo spheical sufaces.
5 245 (SJP, Phys 1120) Gauss' law his actually comes fom expeiment, and is tue no matte WHAT the shape of the closed suface is! Ut involves measuable quantities, and as fa as we know, thee ae NO XCPTIONS to it. Again: if you have a bunch of chages enclosed by any suface (a closed suface, which has an inside an an outside) then: da Gauss' law, simple but cyptic! Let's eview what it means: The left side is the electic flux though ou suface. q(enclosed) = the total amount of chage inside. = C^2/(N m^2) is a constant of natue (defined in Ch. 22). You can pick a suface, any suface you want. It can be a eal physical suface (like a balloon) o an IMAGINARY suface in space. Doesn't matte. Inside that suface will be some chages. They can be distibuted abitaily. (Could be a mix of and ) Togethe, they add up to a total chage. Those chages will poduce an electic field (all thoughout the univese...!) and those field lines will "piece" you suface... We've just seen that you can always (in pinciple) compute the FLUX of field though any suface...: that's the LFT SID of Gauss' law: total electic flux though ou suface. It tells you "how many flux lines poke OUT the suface" (eally, # out  # in) The ight side is simple as can be: popotional to total chage. Gauss' law is amazing! It's going to take a while to sink in what it eally says. We'll stat by looking at ou simple examples (whee we find both left and ight sides). We'll see how Gauss' law gives us Coulomb's law. We'll see some useful consequences when we conside metals in the eal wold. We'll look at some situations whee Gauss' law can be used to figue out fields fom useful (but athe complicated) chage distibutions whee "integating" Coulomb's law to find, like in the last chapte, would be tuly nightmaish. Gauss' law is one of fou "maste equations" we'll lean this semeste. These equations ae togethe called "Maxwell's quations" (Maxwell tied them togethe). With them, all classical electomagnetic phenomena can be descibed, pedicted, and explained!
6 246 (SJP, Phys 1120) Remembe the cube example? The total flux was zeo. (Numbe of lines poking out = numbe of lines poking in). Is that consistent with Gauss' law? Yes it is  the cube was empty! field lines came in and went back out. (If thee wee any chages in thee, field lines would stat o end on the chages: if they pass on though, thee's no chage in thee!) Suppose we only knew Gauss's law. I claim we can DDUC Coulomb's law fom it! Hee's how: Place a chage q at the oigin. I'd like to figue out the field at some abitay point, a distance R away. (Coulomb tells me the answe, but let's wok it out fom Gauss' law). I do need to make an additional agument, based on symmety. A point chage is totally symmetic: no DIRCTION in space can be any diffeent than any othe. Noth, South, ast, West... the field must be the same in diffeent diections. So by SYMMTRY, I can ague that whateve is, it must i) point adially away fom the chage, and ii) must depend ONLY on the distance away fom the chage. We call this a spheically symmetic field. I could say = () ˆ (That means the the field might depend on adius, but at a q given adius, it has a definite value and points outwads) So let's daw an IMAGINARY sphee, centeed on the chage, with adius R. Let's see what Gauss' law says. da da. The ight side is obviously just q/. What about the left? If we don't KNOW yet, how can we do the integal? Well, the point is, I don't know what is, but I DID ague that = () ˆ, i.e. it's the SAM at all points on the sphee, and always points out (i.e. paallel to da). That means eveywhee on ou suface,.da = (R) da. Since (R) is just a constant, pull it out of the integal: (R) da= q /. Once again, that funny integal is just the aea of the sphee, 4 π R^2, which means (R) 4π R 2 = q/, o, q solving, =. That's Coulomb's law! 2 4π R Let me summaize: I could "do" the integal in Gauss' fomula even without knowing what exactly (R) was, by clevely CHOOSING a nice Gaussian suface and invoking symmety. And then I came back with Coulomb's law! So Coulomb's law can be thought of as a consequence of Gauss' law, which is the deepe and moe fundamental of the two...
7 247 (SJP, Phys 1120) Let's ty anothe example. Suppose you have a spheical ball of chage (NOT a point). What's the field outside this ball? Coulomb's law doesn't tell us the answe (it's only valid fo point chages). But Gauss' law does: follow XACTLY the same steps as the pevious example, and conclude that the field is the same as if all the chage was located at a point at its cente! What if we had a spheical shell of chage Q (not a sphee: a thin shell, like the skin of a balloon). What's outside that? The exact same agument says it's the same as if all the chage was located at a point at the cente of the shell. Let's do anothe one: suppose I have this spheical shell of chage Q, spead evenly. What's inside the shell, a distance R fom the oigin?? R Since the chage is spheically symmetic, my old symmety agument still holds: can only depend on R (not on angle), and it must point adially. (Might be adially out, o in, that I can't tell, but it can't point at some funny othe angle: that would be nonsymmetic.) It's the same game as befoe! Daw an imaginay spheical suface with adius R. Note: that's a new suface, it's NOT the same as the shell that has the chage on it. It's inside that one! Now apply Gauss' law: da. R We'e going to be integating ove ou new imaginay sphee! We've seen what to do with the left side:.da = (R) da, the (R) comes out of the integal, and we'e left with (R) 4π R 2 / But what is q(enclosed)? Think about it : the chage Q of the shell is outside my imaginay suface. q(enclosed) is zeo, thee's no chage INSID the dashed line. That means (R) = 0. Thee is NO field ANYWHR inside that shell. Not even ight up close to all the chages just inside the shell! This has useful physical consequences: we'll come back to it late. (It's why you'e safe in you ca if lightning hits it and dumps chage all aound you. If you'e INSID, =0!) But we need to come back and see how this is still tue even with a "nonspheical" ca, and "nonunifom" chage distibution!
8 248 (SJP, Phys 1120) What if we lose symmety?.g, suppose you have a dipole nea the oigin. What's a distance R away fom the oigin?? It is still tue that da =0, but now we can't R make the symmety agument, and I no longe can figue out in any easy way! q q I cetainly can NOT conclude that =0, just because the integal is zeo. Thee's no eason to assume is the same eveyone on the suface I dew, so if is not a constant, you can't pull it out of the integal. We'e stuck  I can't simplify the integal. I know that the total flux though that sphee is zeo  the #of lines poking out = # poking in. But that does NOT tell us that those #'s ae each sepaately zeo! We'd have to go back to moe diect (and painful) ways: summing up d fom each chage. Bottom line: Gauss' law is geat, especially when thee's symmety. But, it doesn't ALWAYS tell you what is at any given point in space. (So it's always tue, but not always helpful fo poblem solving! )
9 249 (SJP, Phys 1120) Anothe example: Suppose you have a lage (effectively infinite) flat sheet of unifomly smeaed out chage, with unifom chage density σ = chage/aea. What is the field a distance L away fom the sheet? We have symmety again, but it's not spheical. z^ ^y What I'll ague is that if the sheet is infinite, then cannot ^ depend on z o y. (It might depend on x, we'll have to see..) x Futhemoe, can't point "up" o "down" because of symmety. In fact, the only way it could point is away fom the sheet (in the x diection). L? Symmety aguments ae subtle, and maybe unfamilia. But they'e a eally impotant idea in lots of applications. Think about the agument I just made. Ask youself why can't point any way except "diectly away"! So we've agued that = (x) ˆ i. It's time to apply Gauss' law. But emembe, you fist have to pick a suface ove which you will integate. What suface should we pick? We want on that takes advantage of the symmety. Picking a big sphee is a bad idea, because this poblem doesn't have spheical symmety. (The field would not point paallel to da on a sphee, and wouldn't necessaily be the same eveywhee. ) Hee we need a little ceativity! I'm going to popose the infamous "bee can" suface, also known as a "Gaussian pillbox". It's an imaginay cylinde, endcap aea A, A centeed aound the chaged sheet: Now conside Gauss' law applied to that pillbox. da That integal is a suface integal: we need to figue out.da all aound the pillbox. I popose thinking of the pillbox as having 3 distinct pats: the left endcap, the ight endcap, and the est of the cylinde. To find the total flux, we'll add the flux though all thee of those pats of the suface. L L
10 2410 (SJP, Phys 1120) Conside fist the "cylinde pat". It has little da vectos spiking out, all of which ae pependicula to the x diection! (Do you see that?) So if points puely in the x diection, it's "skimming" the can, not poking it. The flux though that entie cylindical suface is zeo! da What about the ight end cap? Thee, points ight, and so does da. is totally unifom ove that whole end, so integating.da just gives you A, whee A is the aea of the endcap, and =(L) is the value of the field out thee a distance L away fom the sheet. da What about the left end cap? Thee points left, and so does da! You need to think about both of those. points "away" fom the sheet: on the left side, away = left. da We also ageed that da always points OUT of a closed suface, (which means left on the left endcap) That means.da = A again! The sign is impotant: if and A ae paallel, the dot poduct is positive, even if both point left. Thee IS a net flux out of the can, fields poke "out" of BOTH endcaps! Note also that I'm using symmety to claim that (L) = (L) I'm calling them both just "". Bottom line: The left side of Gauss' law is the flux: da = 0 A A = 2A. How about the ight side, q(enclosed)? Ou pillbox has suounded a cicula patch of the sheet, with aea A, so the chage enclosed must just be σ A. Let's put it all togethe: 2 (L) A = σ A/. The aea cancels. (That's good, because A was an atifically intoduced aea of ou imaginay suface!) And (L) = σ /2. Hey, check it out: thee's no dependence on "L"! is the same at ANY distance fom the sheet! = σ /2 i if x>0 = σ /2 i if x<0
11 2411 (SJP, Phys 1120) One moe example: Suppose you have a vey long (infinite) LIN of chage, with linea chage density λ (Coulombs/mete). What is the electic field at some distance away fom this long line? By now pehaps you see the method. Fist use symmety to convince youself that can only depend on "", the distance away fom the line. Also, must point staight away fom the line at all points. Then choose a sensible imaginay Gaussian suface: anothe "pillbox" (bee can!) shape seems pomising Then, wite down Gauss' law fo the pillbox suface: da The left side equies integating ove the two endcaps plus the "can" itself. This time, it is the endcaps that both give zeo contibution (because doesn't "piece" them, it's paallel to them L Fo the est: at ANY little patch of aea,. da is just da (because points staight out fom the line at the middle, and so does da.) Convince youself! Once again, is the same value, () eveywhee on the suface (by symmety) so we can pull it out of the integal, and we have () da. But that funny integal just means "add up the little aea squaes on the cylindical suface", and we know that: the aea of a cylinde is 2 π L. The othe side of Gauss' law equies q(enclosed), which is just λ L (do you see that?) Putting it togethe, ()2 π L = λ L/. The "L"'s cancel (Good thing! Because L was the length of the imaginay suface, not anything physically eal!) and we've got = 2 π L = λ /(2 π ) This field DOS depend on, the distance fom the line. But it doesn't dop off like ^2 (like Coulomb's law), instead it dops off less apidly.
12 2412 (SJP, Phys 1120) fields and metals: lectostatic equilibium I said is defined eveywhee in space. What is the electic field inside a chunk of metal? I claim, in steady state, "electostatic equilibium", the answe must be zeo! Why? Because metals ae filled with fee electons that can move aound (they conduct)  so if was NOT zeo at some point inside, then the foce on an electon thee would be nonzeo  the electon would stat to move. They would continue to move, building up a "counte" field, a canceling field, until finally eveything settles down, nothing moves, =0 thoughout the metal (so F=0) No moe motion! (Steady state MANS eveything has settled down, no chages ae moving, so the feel no net foce on them any moe) It's a bit of a subtle agument  you have to think about it. But it's a vey impotant idea: WILL ALWAYS B XACTLY 0 eveywhee inside a metal (conducto), in steady state. If you ty to add an extenal field, the chages (conduction electons) inside the chunk of metal will quickly move aound, ext ext until =0 eveywhee inside.  =0 inside   Gauss' law applied to an imaginay suface anywhee INSID the conducto will give da Since =0 eveywhee inside, the left side always vanishes. Thus, q(enclosed) vanishes fo any suface inside the chunk of conducto, which tells you thee ae no net chages anywhee in thee. The sepaated chages must live ight on the suface of the conducto! (xtenal field lines end on one suface, and stat on the othe: stating o stopping on suface chages, but neve going into the metal.) The same agument tells you that, in steady state, if thee's a nonzeo field outside of a chunk of metal, it will always be pependicula ext to the suface ight at the edge. If it wasn't pependicula, thee would be a component of =0 inside paallel to the suface, and electons ight at the suface would then flow (suface cuents), until that piece of the field got cancelled out.
13 2413 (SJP, Phys 1120) Thee ae tons of pactical consequences of the above simple statements. Just one example: if you make a metal box and put it in a egion of lage field, the electons in the metal quickly (essentially instantly) eaange on the suface to make =0 eveywhee INSID the box. If you then hollow out the box, it makes no diffeence, =0 inside, still. If thee's a lightning stom (lage, potentially fatal fields all aound you), a elatively safe place to be is inside a metal ca, because the field INSID the ca (box) is zeo. (It would be bette if the ca was entiely metal) If you'e in a fancy fibeglassbody ca, too bad  fibeglass doesn't conduct, it's not a metal, so the above aguments fail to hold. At least you'll stay dy.. If you look at the suface of a conducto, we've agued above that thee might be a "suface chage density" σ. But inside, =0 and thee won't be any othe net chages in thee. So let's look ight close to the edge of a conducto, any conducto, and apply Gauss' law to a small imaginay cubical suface, half of which is inside and half outside. Since we agued that is pependicula to the suface, thee will only be flux though ON side of the cube, (the ight side in the pictue) and that will simply be A. So Gauss' law says A = q(enc)/ =σ A / Thus = σ/ just outside of a conducto in steady state. That's always tue! The chage density will "adjust itself" to make it tue. It's a vey geneal statement, no appoximations involved.
14 2414 (SJP, Phys 1120) Recall that we found a couple of pages ago that = σ /2 just outside of a sheet of chages. Thee's a facto of two diffeence fom the fomula we just found fo outside a conducto. We need to think about this a bit, it's a puzzle! Because it looks fom the pictue on the ight like what we have is...,well, it looks like a sheet of chage with density σ!? So why do we get a diffeent answe (by 2) than we got befoe? The answe is subtle and woth thinking about: The infinite sheet poblem was a vey special case. We had to assume we had that sheet, and nothing else, no othe chages anywhee in the univese (!) Othewise, the field fom the othe chages would supepose, add in, and we'd get a diffeent answe. So the fomula σ /2 is JUST what you get fom a single sheet and NOTHING LS. But fo the case of the conducto, we'e not making any such assumption. Not only CAN thee be othe chages aound... thee MUST be, in ode to ensue that =0 inside the conducto!! (If the ONLY chages wee the ones shown above, then would NOT be zeo inside, it would be σ /2 to the left) So thee must be othe chages aound, and thee field will add to the σ /2 fom the sheet alone. The magic of Gauss' law is that we know what this sum is going to tun out to be, even without hunting aound and looking fo those othe chages. They have to be thee, they will be thee, and they'll make the total field tun out to be σ / ight outside the conducto..g., pehaps the "lage" pictue looks moe like this: Pehaps we have σ on the ight side, and σ on the left tot side of the metal. The esulting field ove on the ight is eally the supeposition of the fields caused by the two sheets, tot = 12. But both 1 and 2 ae σ/2, aising fom a sheet of chage. The total is σ/, just as we said it has to be: ight outside a conducto. Inside, the fields fom the two sheets will cancel, giving tot=0, again is it has to be. The field outside a conducto aises fom ALL chages eveywhee. The fomula = σ/ outside a conducto just tells me what the total field outside a conducto is, taking into account all chages eveywhee. It's eally petty cool!
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