Gauss s Law Flux Flux in Physics is used to two distinct ways. The fist meaning is the ate of flow, such as the amount of wate flowing in a ive, i.e. volume pe unit aea pe unit time. O, fo light, it is the amount of enegy pe unit aea pe unit time. Let s look at the case fo light: Aea Vecto Repesent an aea as a vecto A, of length equal to the aea, and diection of the outwad nomal to the suface. The flux of light though a hole of aea A is popotional to the aea, and the cosine of the angle between the light diection and this aea vecto. If we use a vecto L to epesent the light enegy pe unit time, then the light out of the hole is L A cos L A. In this case it is negative ( 9) which means the light flux is into the hole. Flux of Electic Field Like the flow of wate, o light enegy, we can think of the electic field as flowing though a suface (although in this case nothing is actually moving). We epesent the flux of electic field as (geek lette phi), so the flux of the electic field though an element of aea A is E A E Acos When, the flux is positive (out of the suface), and when, the flux is negative. When we have a complicated suface, we can divide it up into tiny elemental aeas: d E da E dacos
Gauss Law Flux of Electic Field We ae going to be most inteested in closed sufaces, in which case the outwad diection becomes self-evident. We can ask, what is the electic flux out of such a closed suface? Just integate ove the closed suface: d E da Flux positive => out Flux negative => in The symbol has a little cicle to indicate that the integal is ove a closed suface. The closed suface is called a gaussian suface, because such sufaces ae used by Gauss Law, which states that: Gauss Law The flux of electic field though a closed suface is popotional to the chage enclosed. 1. Which of the following figues coectly shows a positive electic flux out of a suface element? A. I. B. II. C. III. D. IV. E. I and III. I. A II. E III. E IV. A A A E E Mathematical Statement of Gauss Law The constant of popotionality in Gauss Law is ou old fiend. qenc E da q We can see it now by integating the electic flux of a point chage ove a spheical Gaussian suface. E da E da E4 qenc enc Solving fo E gives Coulomb s Law. 1 qenc E 4 E da E da q enc Example of Gauss Law Conside a dipole with equal positive and negative chages. Imagine fou sufaces S 1, S, S, S 4, as shown. S 1 encloses the positive chage. Note that the field is eveywhee outwad, so the flux is positive. S encloses the negative chage. Note that the field is eveywhee inwad, so the flux though the suface is negative. S encloses no chage. The flux though the suface is negative at the uppe pat, and positive at the lowe pat, but these cancel, and thee is no net flux though the suface. S 4 encloses both chages. Again thee is no net chage enclosed, so thee is equal flux going out and coming in no net flux though the suface.
Field At the Suface of a Conducto Field Inside a Conducto E E E Imagine an electic field at some abitay angle at the suface of a conducto. Thee is a component pependicula to the suface, so chages will move in this diection until they each the suface, and then, since they cannot leave the suface, they stop. Thee is also a component paallel to the suface, so thee will be foces on chages in this diection. Since they ae fee to move, they will move to nullify any paallel component of E. In a vey shot time, only the pependicula component is left. We can use Gauss Law to show that the inside of a conducto must have no net chage. Take an abitaily shaped conducto, and daw a gaussian suface just inside. Physically, we expect that thee is no electic field inside, since othewise the chages would move to nullify it. Since E = eveywhee inside, E must be zeo also on the gaussian suface, hence thee can be no net chage inside. Hence, all of the chage must be on the suface (as discussed in the pevious slide). If we make a hole in the conducto, and suound the hole with a gaussian suface, by the same agument thee is no E field though this new suface, hence thee is no net chage in the hole. Field Inside a Conducto We have the emakable fact that if you ty to deposit chage on the inside of the conducto... The chages all move to the outside and distibute themselves so that the electic field is eveywhee nomal to the suface. This is NOT obvious, but Gauss Law allows us to show this! Chage Distibution on Conductos conducto chages collect nea tip balanced foces Fo a conducting sphee, the chages spead themselves evenly aound the suface. Fo othe shapes, howeve, the chages tend to collect nea shap cuvatue. To see why, conside a line of chage. Thee ae two ideas hee Electic field is zeo inside conductos Because that is tue, fom Gauss Law, cavities in conductos have E = unbalanced foces (pushed on by one chage fom left, but by 5 chages fom ight) edistibuted chages (pushed on by one neaby chage fom left, but by 5 moe distant chages fom ight)
Spheical cavity Conducting sphee A Chage Inside a Conducto Positive point chage. What will happen when we add a chage inside a conducto? A. E field is still zeo in the cavity. B. E field is not zeo in the cavity, but it is zeo in the conducto. C. E field is zeo outside the conducting sphee. D. E field is the same as if the conducto wee not thee (i.e. adial outwad eveywhee). E. E field is zeo in the conducto, and negative (adially inwad) outside the conducting sphee. E Field of Chage In Conducto What happens when we move the inne chage off-cente? It induces an off-cente chage distibution on the Inne wall. Note that the field lines distoted, so they emain pependicula to the inne wall. What happens to the oute positive chage distibution? Daw a gaussian suface inside the conducto to find out. The net chage enclosed is zeo, so E =, which we aleady knew because it is inside the conducto. The inne chage is shielded by the induced chage distibution, so the oute chages will be evenly distibuted. Field Lines and Conductos 4. Why cannot the dawing of the conducto and its field lines be coect? A. The field lines ae in both diections, coming to and leaving fom the conducto. B. One of the field lines loops aound, with both end points on the conducto. C. The field lines ae not all pependicula to the suface of the conducto. D. All of the above. E. B and C only. Othe Geometies Always use the symmety of the poblem to detemine what shape to make you gaussian suface. Hee is a plate (plane) geomety, whee the chages ae evenly distibuted on a flat suface. If the total chage on the plate is Q, and the plate has a total aea A tot, then the suface chage density is Q / A tot C/m The E field is eveywhee pependicula to the plate (again, if not, the chages will move until the pat paallel to the suface is nullified). What is? Use a gaussian suface that is paallel to E E on the sides (so no flux though side sufaces), and closes inside the conducto (no flux though that end). On the emaining side, the aea vecto A is paallel to the E field, so E da EA A o E Conducting Suface
Line of Chage In the pevious chapte, we calculated the E field on the axis of a line of chage, but with Gauss Law we can now handle finding E off the line axis. Hee is a line geomety, whee the chages ae evenly distibuted on a long line. If the total chage on the line is Q, and the line has a total length L tot, then the linea chage density is Q / Ltot C/m The E field is eveywhee pependicula to the line (again, if not, the chages will move until the pat paallel to the line is nullified). Use a cylindical gaussian suface that is paallel to E on the top and bottom (so no flux though those sufaces), and is pependicula to elsewhee. The aea vecto da E is paallel to E, and the total aea is h so E da E h h o E Line of Chage Symmety 5. A chaged sphee has an electic field just like that of a point chage of the same total chage, located at its cente. What is the electic field of a long conducting cylinde like? A. Also like that of a point chage at its cente. B. Like a cicula ing of chage at its cente. C. Like a line chage along the cylinde axis. D. Cannot tell fom the infomation given. Unifom Sphee of Chage Hee is a spheical geomety, whee the chages ae evenly distibuted thoughout the volume. If the total chage in the sphee is Q, and the sphee has a adius R, then the volume chage density is Q C/m 4 R By symmety, the E field is eveywhee adial fom the cente of the sphee. Use a spheical gaussian suface, which is pependicula to eveywhee. The aea vecto A E is paallel to E, and the total aea is 4 so when the gaussian suface adius is < R, then 4 E da E4 o E Field Lines and Conductos 6. The dawing shows thee cylindes in coss-section, each with the same total chage. Each has the same size cylindical gaussian suface (again shown in cosssection). Rank the thee accoding to the electic field at the gaussian suface, geatest fist. A. I, II, III B. III, II, I C. All tie. When >R, then the chage enclosed is just Q, so Q o E da E4 Q E 4 Coulomb s Law again I. II. III.
Nonconducting Sheet A nonconducting sheet with a unifom suface chage density has the same geomety as fo the conducting plate, so use the same gaussian suface. The only diffeence is that now one end cannot close in a conducto, so thee is electic flux out both ends. As you may expect, the esulting electic field is half of what we got befoe. E da EA A Two Paallel Conducting Plates When we have the situation shown in the left two panels (a positively chaged plate and anothe negatively chaged plate with the same magnitude of chage), both in isolation, they each have equal amounts of chage (suface chage density ) on both faces. But when we bing them close togethe, the chages on the fa sides move to the nea sides, so on that inne suface the chage density is now. A gaussian suface shows that the net chage is zeo (no flux though sides da pependicula to E, o ends E = ). E = outside, too, due to shielding, in just the same way we saw fo the sphee. E Sheet of Chage Paallel Conducting Plates 7. The sketch shows the case of two paallel conducting plates of equal and opposite chage. If the positively chaged plate has a geate chage than the absolute value of the chage of the negatively chaged plate, what would be the diection of E in the two egions A and B? A. To the ight in egion A, and to the left in egion B B. To the left in egion A, and to the ight in egion B C. To the left in both egions. D. To the ight in both egions. E. The field emains zeo in both egions. A B Two Paallel Nonconducting Sheets The situation is diffeent if you bing two nonconducting sheets of chage close to each othe. In this case, the chages cannot move, so thee is no shielding, but now we can use the pinciple of supeposition. In this case, the electic field on the left due to the positively chaged sheet is canceled by the electic field on the left of the negatively chaged sheet, so the field thee is zeo. Likewise, the electic field on the ight due to the negatively chaged sheet is canceled by the electic field on the ight of the positively chaged sheet. The esult is much the same as befoe, with the electic field in between being twice what it was peviously.
Spheical Symmety Spheical shell 1 q E R 4 E R We ealie said that a shell of unifom chage attacts o epels a chaged paticle that is outside the shell as if the shell s chage wee concentated at the cente of the shell. We can now pove this using Gauss Law. We also said that a shell of unifom chage exets no electostatic foce on a chaged paticle that is located inside the shell. Again, Gauss Law can be used to pove this. Summay Electic flux is the amount of electic field passing though a closed suface. Flux is positive when electic field is outwad, and negative when electic field is inwad though the closed suface. Gauss Law states that the electic flux is popotional to the net chage enclosed by the suface, and the constant of popotionality is. In symbols, it is qenc E da q Thee ae thee geometies we typically deal with: Geomety Chage Density Gaussian suface Electic field Linea = q/l Cylindical, with axis along line of chage Sheet o Plane = q/a enc Cylindical, with axis along E. Spheical = q/v Spheical, with cente on cente of sphee E E q E 4 Line of Chage Conducting E Nonconducting q R E R < R 4 Summay, cont d The electic field is zeo inside a conducto. The electic field is zeo inside a cavity within a conducto, unless thee is a chage inside that is not in contact with the walls. The electic field at the suface of a conducto is always pependicula to that suface. Note, none of this is tue fo insulatos.