Hopefully Helpful Hints for Gauss s Law

Size: px

Start display at page:

Download "Hopefully Helpful Hints for Gauss s Law"

Brook Clarke
5 years ago
Views:

1 Hopefully Helpful Hints fo Gauss s Law As befoe, thee ae things you need to know about Gauss s Law. In no paticula ode, they ae: a.) In the context of Gauss s Law, at a diffeential level, the electic flux is elated to the amount of electic field E that passes though a diffeential suface da. It is defined as: dφ e = E d A, whee da is a contived vecto whose diection is pependicula to and away fom the suface (that is, it s at ight angles to the face of the suface) and whose magnitude it equal to the diffeential suface aea of the suface. b.) What Gauss obseved was that if thee is a closed suface, the amount of electic flux though the suface will be popotional the amount of chage inside the suface. 1.)

2 In othe wods, if you sum up all the diffeential fluxes though all of the diffeential sufaces ove the entie closed stuctue, you will find that that numbe you end up with will be popotional to the amount of chage inside the stuctue. Mathematically, this is stated as: Φ e α q enclosed In expanded fom, this comes out as: To make this into an equality, you have to multiply by a popotionality constant. In this case, it is the pemittivity of fee space, o. With that, Gauss s Law is witten as: E da α q enclosed S E da = q enclosed S 2.)

3 b.) Gauss s Law is ALWAYS TRUE, whethe the integal is easy to do o not (a chage sitting next to the edge of a sphee will satisfy the law, but with the electic field due to the chage vaying fom point to point on the suface, and with the angle between the electic field vecto and the aea vecto vaying, God help the poo soul who ties to evaluate the integal). Q The tick to using Gauss s Law is to find a geomety that is symmetic with the chage. That is, you want a geomety that eithe has the electic field vectos paallel to the aea vectos, o one in which the dot poduct between the two paametes is zeo (that is, the two ae at ight angles with one anothe this will be impotant when we get to cylindical symmety). d A E E Q d A d A E 3.)

4 c.) As an inteesting side note: Although the integal on the left side of Gauss s Law looks like the nasty pat of a poblem (note the nastiness), E da S = q enclosed fo spheical symmety that integal will ALWAYS be of the same fom (and the same is tue of cylindical symmety). In othe wods, when you stat doing moe complicated poblems, the had pat is detemine how much chage is inside the imaginay Gaussian suface. d.) Just to be complete, the left side will always look like: E( 4π 2 ) fo spheical symmety, whee "" is the Gaussian suface adius E( 2πL) fo cylindical symmety, whee "" is the Gaussian suface adius NOTE: Fo both the AP test and the Gauss s Law test, you should be able to deive both of these elationships fom scatch! 4.)

5 2.) As was said above, the difficult pat of most Gauss s Law poblems is detemining the chage enclosed pat of the elationship. You have to include ALL the chage involved, and in some cases you have to do some petty fancy dancing to execute that feat. The following ae examples of poblems with spheical symmety and cylindical symmety. 3.) Poblem with Spheical Symmety: Assume you have a thick skinned, insulating (non-conducting), hollow sphee of inside adius a and outside adius b (a coss-section is shown on the next page). Assume also that the insulato has a volume-chage-density in it that is non-linea and is defined as: ρ = k 1, whee "k 1 " is a constant (not 1 4π ). 5.)

6 Detemine: a.) E b.) E c.) E ( ) fo < a; ( ) fo a < < b; ( ) fo > b; a b (Note that this is, quite liteally, how most Gauss s Law poblems ae stated on AP tests.) ρ = k ai a.) E( ) fo < a; If we place an imaginay, spheical, Gaussian suface of adius less than a centeed at the configuation s cente, the chage enclosed within the Gaussian suface will be zeo. That mean the electic field in that egion will be zeo inside that egion. Gaussian sphee with no chage enclosed 6.)

7 a b ai ρ = k 7.)

8 b.) E( ) fo a < < b; Define a Gaussian suface between a and b. Call it s adius c and its diffeential thickness dc. a b Being a sphee, the shell s suface aea will be 4πc 2. c dc Again, being a sphee, the shell s diffeential volume will equal the poduct of the shell s suface aea and thickness (dc). That is: diffeentially thin spheical shell Gaussian suface ( )( diffeential thickness) ( ) dv = suface aea ( ) dc = 4πc 2 8.)

9 The diffeential chage dq inside will be: ( )( diffeential volume) = ( ρ) 4πc 2 dc = ( kc) 4πc 2 dc dq = volume chage density evaluated at "c" ( ) = ( 4πk) c 3 dc ( ) ( ) a b c dc Gaussian suface diffeential chage dq unifomly distibuted thoughout shell 9.)

10 Knowing the diffeential chage dq = ( k 1 c) ( 4πc 2 dc) = 4πk 1 c 3 dc, we can sum up all the diffeential chage amounts inside all the diffeential volumes between a and (that is, eveywhee thee s chage inside the Gaussian suface) and wite Gauss s Law fo E between a and as: a b dc c Gaussian suface diffeentially thin spheical shell c=a E da = q encl This is evaluated fully on the next page. 10.)

11 E da = q encl S S E da cos0 o = E da = S c=a E 4π 2 ( ) = 4πk 1 E = 1 4π 2 dq 4πk 1 c 3 dc c=a 4πk 1 c 4 ε o 4 E = k a4 4 E = k a 4 2 c 3 dc c = c = a a b dc c Gaussian suface diffeentially thin spheical shell 11.)

12 c.) E( ) fo > b; Now the Gaussian adius is geate than b. The ONLY a b diffeence between this and the pevious section is that the chage integal will go c fom c=a to c=b instead of dq fom c=a to c= (this is because is now outside the sphee, so the chage enclosed is ALL of the chage inside the sphee). By simply putting a b wheeve we see an in the chage pat of the equation, we get: E = k 1 4 b 4 a 4 2 dc Gaussian suface Note: The in the denominato is fom the flux pat (it s the Gaussian adius), not the chage enclosed pat, so it doesn t change! 12.)

13 ADDED TWIST: Let s assume we still wanted the field outside of b, but in addition to the volume chage density in the insulato section thee is a negative chage -Q at the cente? How would things have changed? dq a -Q c b dc The only diffeence would have been on the ight side of Gauss s Law in the chaged enclosed tem. That would have been: Q + dq = Gaussian suface The math fo this situation is shown on the next page. What impotant thing to obseve hee is that ALL the chage inside the Gaussian suface that has be accounted fo!!! 13.)

14 It s now time to explain why I always assume the electic field to be in the diection of da. Look at the solution below. In the numeato, thee ae two tems, one positive, one negative. When you stat the poblem, YOU DON T KNOW which one will be lage. And as a consequence, you won t know whethe the electic field though the Gaussian suface will be inwad o outwad. In cases like this, you have to chose one o the othe. Fo simplicity, I ALWAYS choose outwad along the line of da. IF I M WRONG, all that happens is that when I put numbes in, I get a negative sign in font of the electic field magnitude. Because magnitudes shouldn t be negative, that sign lets me know I ve assumed the wong diection fo E. E da = q encl S Q + = b dq = Q + 4πk 1 c3 dc c=a = Q + 4πk 1 b c=a = Q + 4πk 1 c 4 c = b 4 = Q + 4πk 1 b 4 ε o c 3 dc 4 a4 4 = Q + πk 1 ε b 4 a 4 o E( 4π 2 ) = Q + πk 1 ε b 4 a 4 o E = c = a Q 4π + k b 4 a )

15 4.) Cylindical symmety: Conside now a thick-skinned insulating cylinde with inside adius a and outside adius b. Assume thee is λ's woth of chage on a vey thin wie down its axis, and in the insulating mateial a volume chage density of ρ = k 1, whee "k 1 " is a positive constant. thee diffeent views of ou system ae shown below. side view 3-d view λ end view λ a a wie λ b wie ρ = k wie b ρ = k a b ρ = k 15.)

16 side view 3-d view λ end view λ a a wie λ b wie ρ = k wie b ρ = k a b ρ = k Ou poblem is to deive an expession fo: a.) E b.) E c.) E ( ) fo < a; ( ) fo a < < b; ( ) fo > b; 16.)

17 ( ) fo < a : a.) E As always, we begin by geneating an imaginay Gaussian suface that has one of two popeties: eithe the magnitude of the electic field is the same at evey point on the suface with a diectional elationship with da that doesn t vay, o the dot poduct of E and da evaluated on the suface is zeo (that is, E uns along the face, not pependicula to the face). Fo a cylindical symmety, that suface will be... wait fo it... a cylinde! The Gaussian suface that is appopiate fo cylindical symmety is shown on the next thee pages fom all thee views. 17.)

18 CROSS-SECTION Fo the 3-d view: ed section is the pat of the chaged wie that is INSIDE Gaussian suface λ wie 3-d view a b h Gaussian cylinde of abitay length h ρ = k 18.)

19 CROSS-SECTIONS side view a b λ end view h ρ = k a λ ρ = k Gaussian cylinde of abitay length h and adius Gaussian cylinde of abitay length h 19.)

20 The amount of chage found inside the Gaussian suface is equal to the chage-pe-unit-length λ times the length h of the Gaussian suface (that is, q enclosed = λh ). Noting that the cylindical pat of the cylinde has flux though it but the endcaps don t (the angle between E and da at the end-caps is ninety degees), the sum of all the diffeential da s ove the cylindical pat of the Gaussian suface will be equal to the cicumfeence of that Gaussian cylinde ( 2π) times its length h, o ( 2π)h. Assuming (as always) that da and E ae in the same diection (even though we know they aen t in this case as the field-poducing chage is negative emembe, we ae going fo consistency hee with ou diection assumptions), we can wite Gauss s Law as shown to the ight: ( ) E da = q enclosed E da cosθ = λh E d A E 2π h = ( λh) ( ) ( ) = ( λh) E = λ 2π Remembe, a negative electic field magnitude means only that the assumed diection fo E was wong no big deal! In fact, we expected that fo this elatively simple configuation.) 20.)

21 b.) detemine E( ) fo a < < b : The imaginay, cylindical, Gaussian suface now must extend into the insulating mateial. That means we ae going to have two chage souces within the Gaussian suface. Getting the insulato s chage (inside the Gaussian suface) is a little ticky. Follow along: We begin by defining a diffeentially thin cylindical shell of adius c located inside both the insulato and inside the Gaussian adius. If we can detemine how much diffeential chage dq is in that diffeentially thin shell, we can integate ove all the shells fom c=a (whee the insulating mateial stats) to c= (whee the Gaussian suface is) to get the insulato chage inside the Gaussian suface. The sketch on the next page highlights all of this. 21.)

22 The diffeential volume dv of a cylindical shell of length h, adius c and thickness dc is equal to the cicumfeence times the thickness times the length, o: ( )( length) ( thickness) ( ) ( h) ( dc) ( )( c dc) dv = cic. = 2πc = 2πh a c end view dc The diffeential chage dq in that diffeentially thin cylindical shell (again, thickness dc) is equal to: dq = ( chage pe unit volume) ( volume of shell) = ρ dv = k 1 c ( ) c 2 dc = 2πk 1 h ( ) ( 2πh) ( c dc) ( ) Gaussian suface 22.)

23 The total chage inside the Gaussian suface will be the chage on the wie (i.e., λh ) added to the net chage inside the insulating, diffeential, cylindical shells between a and. In othe wods, it will be the evaluation of the integal ρ dv. With all this, we can wite Gauss s Law as: E d A = q enclosed E E d A cosθ = λh d A = λh ( ) + ρ dv ( ) + ( k1c )( 2πh c dc) c=a E ( 2π h) ( = λh) + 2πhk 1 c 2 dc c=a E ( 2π h) = c ( λh) 3 + 2πhk 1 3 c=a E ( 2π h) = ( λh) πhk 1 3 a 3 E = ( λ) πk 1 3 a 3 2π 23.)

24 ( ) fo > b : c.) E Just as was the case with the spheical symmety poblem, the only diffeence between the magnitude of E in the outside egion and in the insulating egion is the limits used when doing the integal that detemines the chage enclose. As befoe, those limits should be c=a to c=b instead of c=a to c=. Making that alteation, you can make the appopiate adjustments to the solution fom the pevious section and end up with the solution given below. E = ( λ) πk 1 b 3 a 3 2π Note that the only time you eally can t make simple adjustments like this is when you ve done some canceling that is no longe applicable. Fo that you have to be caeful. 24.)

25 5.) Thee is one moe thing we need to say something about when it comes to both spheical and cylindical symmety. That has to do with conductos. If you will emembe, conductos in electostatic situations cannot have electic fields within them. Othewise, fee chage will espond to those electic fields and move aound until the motion-poducing fields ae neutalized. So conside the following poblem: You have a positive point chage Q suspended fom an ignoable sting and located at the cente of a hollow conducting sphee of inside adius a and outside adius b. Use Gauss s Law on the egion inside the conducto. conducto ai a Q b 25.)

26 Dutifully, we begin the Gauss s Law pocess by geneating an imaginay Gaussian suface o adius between a and b (see sketch). With the sketch, we execute Gauss s Law: E d A = q enclosed E E d A cosθ = Q d A E 4π 2 = Q ( ) = Q E = Q 4π 2 a Q b Oopsie! This suggests thee s a nonzeo electic field inside the conducto, but we know that can t happen. So what gives? ai Gaussian suface 26.)

27 The key is in the fact that electons can move aound feely inside the conducto. In this case, -Q s woth of chage will migate to the inside suface of the hollow sphee leaving +Q s woth of chage on the outside suface. Then, when the chage enclosed is detemine fo the situation between a and b, the net chage enclosed becomes zeo and we have no electic field in the egion. That is: E d A = q enclosed E ( ) da cosθ = Q + Q = 0 ε o E = 0 +Q's chage on outside suface +Q's chage on outside suface Q +Q's chage on outside suface When you deal with the egion outside b, the total chage enclose becomes +Q at the cente, -Q on the inside of the hollow, +Q on the outside of the sphee fo a net chage of +Q. In othe wods, fom a distance the configuation will look like a point chage. Q's chage on inside suface Gaussian suface +Q's chage on outside suface +Q's chage on outside suface 27.)

28 As an inteesting side-note, the electic field lines fo this poblem would look like: Q 28.)

29 Relationships you should emembe! Suface aea fo a sphee: S = 4π 2 Suface aea fo bael of a cylinde: (the cicumfeence times the length) S = ( 2π)h Cic = ( 2π) h 29.)

30 Diffeential volume dv fo a spheical shell: (this is the sphee s suface aea S times its thickness d) S = ( 4π 2 ) d dv = ( 4π 2 )d coss-section of spheical shell Volume dv of a diffeentially thin cylindical shell: (this is the sphee s suface aea S times its thickness d) d S = ( 2π)h dv = S d = ( 2π)h d h 30.)

31 Diffeential chage dq inside diffeentially thin spheical shell of volume dv: (this is the diffeential volume of the shell times its chage density) dq = ρdv ( ) d = ρ 4π 2 Diffeential chage dq inside diffeentially thin cylindical shell of volume dv: (this is the diffeential volume of the shell times its chage density) dq = ρdv = ρ ( 2πh)d 31.)

32 Lastly, what about a thin sheet of a conducto mateial with a chagepe-unit-aea σ on it? To begin with, when talking about a conducto, denotes the amount of chage pe unit aea on ONE SURFACE of the stuctue. That means thee is σ 's woth of chage-pe-unit-aea on one side of the sheet, and σ 's woth of chage-peunit-aea on the othe side of the sheet. Gaussian sufaces ae chosen so as to have one of two geneal chaacteistics. Eithe: 1.) The magnitude of E is the same eveywhee on the suface, and E and the diffeential aea vecto da have the same angle between them eveywhee on the suface (pefeably zeo degees), OR 2.) The dot poduct of E and da be zeo (that is, the two ae pependicula to one anothe). Fo ou situation, a plain cylindical plug will do the job. 32.)

33 What is shown is a side-view and a 3-d view of the conducto. The Gaussian suface is a plug. Note: a.) If the sheet extends to infinite, o if it is finite but the outside endcap of the Gaussian suface lies vey nea both the cente of the sheet AND nea to the sheet itself, then the electic field nea the cente and in close to the sheet will be diected outwad. This is the same diection as the end-cap s aea vecto A. In shot, all the flux will be though the ight end-cap (the left end-cap is in an aea whee thee is NO electic field) and will equal EA. E A 33.)

34 E infinite, chaged, conducting sheet = σ So with the flux only passing though the ight-hand end-cap, Gauss s Law yields: E ds = q enclosee EA = σa Note that the field is not a function of you distance fom the sheet. This was explained in class. If you don t undestand it, come talk to me. 34.)

35 HAVING DONE THAT, now conside an infinite (o at least vey lage), chaged INSULATING sheet. At least pat of the poblem hee is that the chage-pe-unit-aea function now means something diffeent than it did when used with a conducto. Fo an insulato, it is assumed that chage is shot thoughout the stuctue. That means that the σ function doesn t tell you how much chage is on the suface, it tells you how much chage is below the suface (unde a given aea). What this means is that means we can not have ou Gaussian suface teminate inside the slab! σ Both a side-view and a 3-d view of ou slab and Gaussian plug ae shown on the next page. 35.)

36 The electic field E is still away fom the slab, but because it is not zeo inside the slab we have to extend the Gaussian suface out in both diections. That leaves Gauss s law looking like: E ds = q enclosee 2( EA) = σa E infinite, chaged, insulating sheet = σ 2 A E E A Eithe of these situations (the conducto o insulato) could pop up on an AP test! 36.)

Gauss Law. Physics 231 Lecture 2-1

Gauss Law. Physics 231 Lecture 2-1 Gauss Law Physics 31 Lectue -1 lectic Field Lines The numbe of field lines, also known as lines of foce, ae elated to stength of the electic field Moe appopiately it is the numbe of field lines cossing