GAUSS S LAW FOR SPHERICAL SYMMETRY

Transcription

1 MISN012 GAUSS S LAW FOR SPHRICAL SYMMTRY by Pete Signell GAUSS S LAW FOR SPHRICAL SYMMTRY G.S. R 1. Intoduction a. The Impotance of Gauss s Law b. Usefulness of the Law Gauss s Law a. Intoduction b. Gaussian Sufaces c. Nomal Component of d. Statement of Gauss s Law e. Choosing the Gaussian Suface Spheically Symmetic Chage A Point Chage: Coulomb s Law A Sphee of Unifom Chage a. Oveview b. Field Inside the Chage Distibution c. Field Outside the Distibution R Acknowledgments Glossay Poject PHYSNT Physics Bldg. Michigan State Univesity ast Lansing, MI 1

2 ID Sheet: MISN012 Title: Gauss s Law fo Spheical Symmety Autho: P. Signell, Dept. of Physics, Mich. State Univ Vesion: /29/2000 valuation: Stage 0 Length: 1 h; 24 pages Input Skills: 1. Vocabulay: cylindical symmety, spheical symmety (MISN0 15). 2. Fo a chage distibution with spheical symmety, indicate the diection of the electic field at any given point in space and daw the suface that includes all othe points with the same magnitude of the electic field (MISN015). Output Skills (Knowledge): K1. Vocabulay: Gauss s law, Gaussian suface, integation suface, suface integal, volume chage density. K2. State the two ules fo constucting a useful Gaussian suface. K. Deive Coulomb s law fom Gauss s law, including all pats of the agument. K4. State the basic physics diffeence between Gauss s law and Coulomb s law. Output Skills (Poblem Solving): S1. Given a spheically symmetic chage distibution, use Gauss s law to detemine the electic field at any specified point. PostOptions: 1. Gauss s Law Applied to Cylindical and Plana Chage Distibutions (MISN01). 2. lectic Fields and Potentials Acoss Chage Layes and In Capacitos (MISN014). THIS IS A DVLOPMNTALSTAG PUBLICATION OF PROJCT PHYSNT The goal of ou poject is to assist a netwok of educatos and scientists in tansfeing physics fom one peson to anothe. We suppot manuscipt pocessing and distibution, along with communication and infomation systems. We also wok with employes to identify basic scientific skills as well as physics topics that ae needed in science and technology. A numbe of ou publications ae aimed at assisting uses in acquiing such skills. Ou publications ae designed: (i) to be updated quickly in esponse to field tests and new scientific developments; (ii) to be used in both classoom and pofessional settings; (iii) to show the peequisite dependencies existing among the vaious chunks of physics knowledge and skill, as a guide both to mental oganization and to use of the mateials; and (iv) to be adapted quickly to specific use needs anging fom singleskill instuction to complete custom textbooks. New authos, eviewes and field testes ae welcome. PROJCT STAFF Andew Schnepp ugene Kales Pete Signell Webmaste Gaphics Poject Diecto ADVISORY COMMITT D. Alan Bomley Yale Univesity. Leonad Jossem The Ohio State Univesity A. A. Stassenbug S. U. N. Y., Stony Book Views expessed in a module ae those of the module autho(s) and ae not necessaily those of othe poject paticipants. c 2001, Pete Signell fo Poject PHYSNT, PhysicsAstonomy Bldg., Mich. State Univ.,. Lansing, MI 48824; (517) Fo ou libeal use policies see: 4

3 MISN012 1 GAUSS S LAW FOR SPHRICAL SYMMTRY by Pete Signell 1. Intoduction 1a. The Impotance of Gauss s Law. Gauss s law is geneally thought of as being the integal fom of one of the fou geat laws of electicity and magnetism. Gauss s law has exactly the same content as Coulomb s law, but, while Coulomb s law is fomulated in tems of point chages, Gauss s law is fomulated in tems of continuous chage distibutions. Although eithe of these laws can be deived fom the othe, it is easie to deive Coulomb s law fom Gauss s law than the othe way aound, and that makes us think of Gauss s law as the moe basic fom. Futhemoe, it is easy to tansfom Gauss s law fom its usual integal fom to a diffeential fom which is also widely used. In that fom it is known as one of the fou diffeential Maxwell s equations that goven all of electicity and magnetism. 1 1b. Usefulness of the Law. Gauss s law is geneally the option of choice fo finding the electic field at vaious space points when the chages poducing the electic field ae symmetically distibuted. This situation often occus in the design of new electonic components. ven when the chages ae not symmetically distibuted, Gauss s law can still be used to give a ough estimate fo design exploation o fo checking the esult fom a compute pogam. Fo an example of the use of Gauss s law, look at the cosssectional view of a coaxial cable shown in Fig. 1. Hee the chage distibutions have cylindical symmety. Using Gauss s law we can easily detemine the electic field at any point inside o outside this cable, then use that knowledge to detemine the way the cable affects signals that pass down it. By vaying the cable paametes we can quickly optimize the design. 2 Fo a wildly diffeent example, Gauss s law povides a quick and exact poof of an impotant deivation in gavitation, a deivation that eluded 1 See The AmpeeMaxwell quation; The Displacement Cuent (MISN0145) and Maxwell s quations (MISN0146). 2 The coaxial cable is teated in geate detail in Gauss s Law Applied to Cylindical and Plana Chage Distibutions (MISN01). MISN012 2 R 1 R 2 Newton fo nealy 20 yeas. Figue 1. Coss section of a coaxial cable, showing the chages at some instant. 2. Gauss s Law 2a. Intoduction. Ou appoach to Gauss s law will be to: (1) develop some ideas about the Gaussian suface that occus in the law; (2) eview the ules fo finding the component of the electic field that is nomal to the Gaussian suface, since this is what occus in the law; () give a pecise statement of the law; and then (4) give pecise ules fo choosing the Gaussian suface. 2b. Gaussian Sufaces. Befoe we poceed to the statement of Gauss s law, we intoduce some notions about the Gaussian suface occuing in the mathematical statement of the law; futhe study of the law will then convet those notions into pecise statements. The suface that occus in Gauss s law, called a Gaussian suface, is a closed imaginay suface that passes though the space point at which we want to know the electic field and which has a shape detemined by the symmety of the chage distibution. In all of the examples we will be dealing with, at least pat of the Gaussian Suface will be the lectic Field quimagnitude Suface (MS) that goes though the point at which we wish to know the field. 4 The only diffeence is that additional suface aeas must be added to the MS, if necessay, in ode to make it into a closed suface (one that totally encloses some volume). Hee ae typical Gaussian sufaces: (1) a sphee (see Fig. 2); (2) a cylinde with flat See The Gavitational Field Outside a Homogeneous Spheical Mass (MISN0 109). 4 Fo the necessay discussion of quimagnitude Sufaces, see lectic Fields Fom Symmetic Chage Distibutions (MISN015). 5 6

4 MISN012 MISN012 4 G.S. q ` P ` n^ G.S. ` n^ q s n^ ` R G.S. Figue 2. A Spheical Gaussian Suface. incement of G.S. Figue. Illustation of an electic field vecto not nomal to the Gaussian suface. cicula ends that ae nomal to the axis of the cylinde; and () two flat sheets that ae paallel and identical in shape, with the volume between the paallel sheets closed by a suface nomal to the planes of the sheets. 2c. Nomal Component of. Gauss s law uses only the component of the electic field that is nomal (pependicula) to the Gaussian suface. This component is denoted n. Whee the electic field is not nomal to the Gaussian suface we must find its nomal component (see Fig. ): n = ˆn = cos θ, whee ˆn is the outwaddiected unit vecto nomal to the Gaussian suface at the point unde consideation. 2d. Statement of Gauss s Law. Gauss s law states that the integal of the nomal component of the electic field ove any closed suface is popotional to the net chage contained inside that suface. Thus Gauss s law contains the following quantities: 1. S n ds the integal of the nomal component of the electic field, n, ove any closed suface S ; 5 and 5 The cicle on the integal sign eminds you that the suface must be a closed one, meaning that it must completely enclose some egion of space. When you actually use Gauss s law, the suface integals will educe to just a constant times the suface aeas of sphees, cylindes, o ectangles. ` n^ Figue 4. A fanciful Gaussian suface fo illustating Gauss s law. Figue 5. The Gaussian suface (G.S.) fo applying Gauss s law to a spheically symmetic chage distibution that extends to adius R. 2. q S the total (net) electic chage contained within that same suface S. The mathematical statement of Gauss s law is: n ds = 4πk e q S. (1) S This law holds fo any suface whateve, even one as wild as that shown in Fig. 4. 2e. Choosing the Gaussian Suface. Thee ae two ules fo constucting a useful Gaussian suface: 1. the suface must pass though the point at which you wish to know a paticula component of the electic field, and the suface must be nomal to that component at the point in question; 2. the suface must be a completely closed one and, at evey point on the suface, the nomal component of the electic field should eithe have the same value as at the point in question o be zeo. Using the above two ules, the electic field due to a symmetic chage distibution can geneally be detemined in a few lines in a few seconds. 7 8

5 MISN012 5 Advice: People leaning Gauss s law fo the fist time often use a pope imaginay Gaussian Suface on the left side of q. (1) but then incoectly compute the chage within a eal suface, diffeent fom the Gaussian Suface, fo the ight side of that equation. You must compute the chage within the imaginay Gaussian Suface fo the ight side of the equation, and it must be the same Gaussian Suface you used to evaluate the left side of the equation.. Spheically Symmetic Chage Fo chage distibutions having spheical symmety, we choose ou Gaussian suface to be a sphee of adius centeed on the cente of the chage distibution (see Fig. 5). Then is always nomal to the suface: n = ±, the sign depending on whethe the electic field points outwad o inwad. Also, because the chage distibution looks the same fom any point on the spheical Gaussian suface, the magnitude of is constant ove that suface. That is, n can be a function of adius but it will not be a function of position fo fixed adius: n = (). This constancy ove a spheical suface S enables us to easily pefom the integation: n ds = () ds = () 4π 2. S S Then by Gauss s law, q. (1), the electic field is: 4π 2 () = 4πk e q S (), (spheical symmety), (2) whee q S () is the net amount of chage inside the spheical suface of adius. 4. A Point Chage: Coulomb s Law The application of Gauss s law to the case of a single point chage at the oigin of some coodinate system poduces the electic field appopiate to Coulomb s law. The single point chage is an especially simple case because any spheical suface centeed on the chage will enclose all of the chage. Witing the value of the point chage as, q. (2) immediately becomes: () = k e 2. MISN012 6 Inseting the adial diection of the field gives us: () = k e ˆ. Help: [S1]. () 2 If we now place a chage q at the spacepoint, it expeiences a foce F = q () due to the chage at the oigin. Using q. (), we obtain an expession fo the foce between the two chages: which is Coulomb s law. F = k e q 2 ˆ, 5. A Sphee of Unifom Chage 5a. Oveview. Given a sphee with electic chage distibuted unifomly thoughout its volume, we can use Gauss s law to easily find the electic field at any point outside o even inside the sphee. All we need do is evaluate the net amount of chage inside the Gaussian suface at the adius of the point, then inset that chage into q. (2). In the next two sections we will detemine the electic field inside and outside a sphee of chage, but fist we will give a quick qualitative unthough. Imagine stating with a vey small Gaussian suface, one with adius R whee R is the adius of the sphee of chage. If we then incease, the amount of chage enclosed by the Gaussian suface at will incease like until R is eached. As inceases beyond R, the amount of chage enclosed by the suface will stay constant since we ae now outside the sphee. Howeve, the Gaussian suface s aea [in the left side of q. (2)] will continually incease like 2. Combining these adial dependencies, we find that the electic field should incease linealy with adius inside R and decease as the invese squae of the adius outside R (see Fig. 6). 5b. Field Inside the Chage Distibution. We will hee use Gauss s law to compute the electic field at a adius that is less than the adius R of the suface of a sphee of unifomly distibuted chage totaling (see Fig. 5). Since the chage distibution is spheically symmetic we can immediately pass fom q. (1) to q. (2). Fo ou case, the chage enclosed by the imaginay integation suface of adius is: q S = (/R) ( Help: [S6]). Then q. (2) becomes: 4π 2 () = 4πk e q S = 4πk e /R (fo R). 9 10

6 MISN012 7 MISN012 8 Reseach, though Gant #SD to Michigan State Univesity. R R G.S. Glossay Gauss s law: a statement that the integal of the nomal component of the electic field ove any closed suface is a known constant times the net chage inside that suface. The law is the integal fom of one of Maxwell s equations. Figue 6. Radial dependence of the electic field due to a homogeneously chaged sphee of adius R. Since s diection is adial, () = k e ˆ (fo R). R Figue 7. Gaussian Suface (G.S.) fo applying Gauss s law outside a spheical chage distibution. This says that the field is zeo at the cente and inceases linealy as we go out towad the edge (see Fig. 6). 5c. Field Outside the Distibution. In the egion beyond a spheically symmetic distibution of chage, any integation suface (see Fig. 7) encloses the entie chage and thus q. (2) educes to Coulomb s law. If the total chage is, then the chage enclosed by the suface in q. (2) is q S = (fo R). Then putting in the diection gives us: () = k e ˆ (fo R). 2 This field is identical to the one that would be poduced by a point chage located at the cente of ou spheical chage distibution (see Fig. 6). Note also that the expessions fo the field inside and outside the sphee give the same answe at the suface of the chage sphee ( = R). They d bette! Gaussian suface: a suface enclosing a chage distibution, chosen so that: (1) the suface passes though the point at which you wish to know a paticula component of the electic field, and the suface is nomal to that component at the point in question; and (2) the suface is a completely closed one and, at evey point on the suface, the nomal component of the electic field eithe has the same value as at the point in question o is zeo. a suface ove which some function is inte integation suface: gated. suface integal: the integal of some function ove a suface S, witten: I = S F ( ) ds, o I = F ( ) ds, depending on the phenomenon being descibed. If the suface is bounded by a line L, then S the integal can be witten I = F ( ) ds whee should be ead SL SL as the integal ove the suface S that is bounded by the line L. The function F ( ) vaies accoding to whee one is on the suface S. The infinitesimal element of aea ds has a diection defined as the local nomal to the suface: ds = ˆndS. closedsuface integal: the integal of some function ove a closed suface S, called the integation suface; witten as I = F ( ) ds, S whee the cicle on the integal sign denotes a closed suface, F ( ) is some function of coodinate space, and ds is an infinitesimal element of aea on the suface S. volume chage density: the amount of chage pe unit volume, usually expessed in coulombs pe cubic mete. Acknowledgments Pepaation of this module was suppoted in pat by the National Science Foundation, Division of Science ducation Development and 11 12

7 MISN012 PS1 MISN012 PS2 PROBLM SUPPLMNT Note: Poblems, 4, and 5 also occu in this module s Model xam. k e = N m 2 C 2 1. Detemine the electic field fo a constant chage density thoughout a spheical volume of adius R, as shown. valuate at: a. = 2.00 cm b. = 4.00 cm R. Given a spheical shell that has a constant chage density within the shell (between the two sufaces of the shell). Assume thee ae no chages inside the inne suface of the shell o outside its oute suface. Use Gauss s law to detemine the electic field inside, outside, and within the spheical shell. Give you answes in tems of the shell s total chage, inne adius a, oute adius b, and the adius at which the electic field is to be evaluated. 4. At a distance of 2.00 m fom the suface of a sphee of chage, what foce would each of these paticles feel if it was at est with espect to the sphee? a. an electon b. a neuton (zeo chage) The sphee s chage density is 16 C/m and the adius of the sphee is 5.00 cm. 5. What is the atio of the magnitude of the gavitational foce to the magnitude of the electostatic foce on an electon due to a solid sphee with a volume chage density of C/m and a adius of 2.0 cm? The distance between the electon and the cente of the sphee is 2.0 m. (Take the gavitational foce as F = N.) whee R =.00 cm and the volume chage density is C/m. Help: [S7] 2. Given a hollow spheical shell of chage: Region III: 0 < < R 1 : no chage; Region II: R 1 < < R 2 : unifom chage distibution, total chage ; Region I: R 2 < < : no chage. a. Daw a coss sectional view, labeling adii and egions. b. Detemine in Region I Help: [S], Region II Help: [S5] and Region III Help: [S2] in tems of the given quantities. c. Check that I, II, and III agee at thei common boundaies. Help: [S4] d. Sketch () fo this case and fo R

8 MISN012 PS MISN012 PS4 Bief Answes: 5. F G /F = a. = N/C Help: [S8] (The answe is NOT ) b. = N/C Help: [S9] 2. a. I II III R 1 R 2 G.S. b. I = k e ˆ/ 2 ; II = k e ˆ ( R 1 ) [ ( )] / 2 R2 R1 ; III = 0. c. III (R 1 ) = II (R 1 ) = 0; II (R 2 ) = I (R 2 ) = k e ˆ/R2 2. d. 0 R 1 R 2 0 R 2. Outside: = ke 2 ˆ Within: = ke ( a ) 2 (b a ) ˆ Inside: = 0 4. a. F = N, away fom the sphee. (The answe is NOT N.) b. F =

9 MISN012 AS1 MISN012 AS2 SPCIAL ASSISTANC SUPPLMNT S (fom PSpoblem 2b) G.S. S1 (fom TX4, TX5c) The definition of ˆ is: ˆ = /. Then = ˆ and both and have the dimensions of length. S2 (fom PSpoblem 2b) R 1 R 2 I II III R 1 R 2 G.S. xamine the sketch and notice that thee is no chage in Region III ( R 1 ). Gauss s law efes only to the chage contained inside the integation suface. If we pick a spheical Gaussian suface in Region III, so its adius is R 1, it will contain no chage and Gauss s law leads to this fom fo q. (2): Outside the sphee, in Region I ( R 2 ), ou Gaussian suface encloses the entie sphee (see sketch) and so Gauss s law leads to this fom fo q. (2): 4π 2 I () = 4πk e q S = 4πk e, whee is the total chage of the hollow sphee. Then: I () = k e 2. Putting in the adial diection of the field, I () = k e 2 ˆ. 4π 2 III () = 0. We can immediately conclude that III () =

10 MISN012 AS MISN012 AS4 S4 (fom PSpoblem 2c) Simply evaluate I (R 2 ) and II (R 2 ) and show that they ae the same. Then evaluate II (R 1 ) and III (R 1 ) and show that they ae the same (see the Bief Answes). S5 (fom PSpoblem 2b) We fist define the elationship between volume chage density ρ and total chage fo the hollow sphee, then assume ρ is constant: = ρ dv = R 2 R 1 ρ4π 2 d = ρ(4/)π(r2 R1) so ρ = (4/)π(R 2 R 1 ). Now a spheical Gaussian suface of adius inside egion II, whee R 1 R 2, will enclose a net amount of chage: q S = R 1 ρ4π 2 d = ρ(4/)π( R1), = (4/)π(R2 R 1 ) (4/)π( R1) = R1 R2. R 1 So q. (2) gives us: 4π 2 II () = 4πk e q S = 4πk e R1 R2, R 1 o ( R II () = k 1) e 2 (R2 R 1 ). Putting in the adial diection of the field, II () = k e ( R 1) 2 (R 2 R 1 ) ˆ. S7 (fom PSpoblem 1) The volume chage density fo a unifomly distibuted chage is ρ = /V, whee is the total chage and V is the volume ove which it is distibuted. If the distibution is spheical with adius R, then V = 4πR /. S8 (fom PSpoblem 1a) = ( N m 2 C 2 )( C m )(4π)(0.02 m) = N/ C S9 (fom PSpoblem 1b) = ( N m 2 C 2 )( C m )(4π)(0.0 m) (0.04 m) 2 = N/ C S6 (fom TX5b) q S = V = (4/)π V R (4/)πR o q S = 0 ρ4π 2 d = ρ(4/)π = (4/)πR (4/)π 19 20

11 MISN012 M1 k e = N m 2 C 2 electon chage = C MODL XAM 1. See xam Skills K1K4 in this module s ID Sheet. 2. Use Gauss s law to detemine the electic field inside, outside, and within a spheical shell of constant chage density. Give you answes in tems of the shell s total chage, inne adius a, oute adius b, and the adius at which the electic field is to be evaluated.. At a distance of 2.00 m fom the suface of a sphee of chage, what foce would each of these paticles feel if it was at est with espect to the sphee? a. an electon b. a neuton (zeo chage) The sphee s chage density is 16 C/m and the adius of the sphee is 5.00 cm. 4. What is the atio of the magnitude of the gavitational foce to the magnitude of the electostatic foce on an electon due to a solid sphee with a volume chage density of C/m and a adius of 2.0 cm? The distance between the electon and the cente of the sphee is 2.0 m. (Take the gavitational foce as F = N.) Bief Answes: 1. See this module s text. 2. See Poblem in this module s Poblem Supplement.. See Poblem 4 in this module s Poblem Supplement. 4. See Poblem 5 in this module s Poblem Supplement

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