Gauss Law. Physics 231 Lecture 21


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1 Gauss Law Physics 31 Lectue 1
2 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 though a given suface that is elated to the electic field Physics 31 Lectue 
3 Flux How much of something passes though some suface Numbe of paticles passing though a given suface Two ways to define Numbe pe unit aea (e.g., 10 paticles/cm ) Numbe passing though an aea of inteest Physics 31 Lectue 3
4 lectic Flux The electic flux is defined to be Φ = A Whee is the electic field and A is the aea Physics 31 Lectue 4
5 lectic Flux If suface aea is not pependicula to the electic field we have to slightly change ou definition of the flux Φ = Acosφ Whee φ is the angle between the field and the unit vecto that is pependicula to the suface Physics 31 Lectue 5
6 lectic Flux We can see that the elationship between the flux and the electic field and the aea vecto is just the dot poduct of two vectos Φ Φ = A = Anˆ nˆ is a unit vecto pependicula to the suface Physics 31 Lectue 6
7 A Convention The diection of a unit vecto fo an open suface is ambiguous Fo a closed suface, the unit vecto is taken as being pointed outwad Physics 31 Lectue 7
8 lectic Flux Whee flux lines ente the suface, the suface nomal and the electic field lines ae antipaallel Whee the flux lines exit the suface they ae paallel Physics 31 Lectue 8
9 lectic Flux Is thee a diffeence in the net flux though the cube between the two situations? No! It is impotant to emembe to popely take into account the vaious dot poducts Physics 31 Lectue 9
10 lectic Flux The equation we have fo flux is fine fo simple situations the electic field is unifom and the suface aea is plane What happens when eithe one o the othe o both is not tue Physics 31 Lectue 10
11 lectic Flux We poceed as we did in the tansition fom discete chages to a continuous distibution of chages We beak the suface aea into small pieces and then calculate the flux though each piece and then sum them In the limit of infinitesimal aeas this just becomes an integal Φ = da Physics 31 Lectue 11
12 lectic Flux of a Point Chage What is electic flux that comes fom a point chage? We stat fom Φ = da The electic field is given by 1 4πε The poblem has spheical symmety, we theefoe use a sphee as the Gaussian suface Since is adial, its dot poduct with the diffeential aea vecto, which is also adial, is always one Also is the same at evey point on the suface of the sphee = 0 q Physics 31 Lectue 1
13 lectic Flux of a Point Chage Fo these easons, can be pulled out fom the integal and what emains is Φ = da The integal ove the suface aea of the sphee yields A = 4π Pulling all this togethe then yields Φ = A; Φ = q ε Φ 0 = 1 4πε 0 q 4π Notice that this is independent of the adius of the sphee Physics 31 Lectue 13
14 ds xample 1 ds 1 A positive chage is contained inside a spheical shell. How does the diffeential electic flux, dф, though the suface element ds change when the chage is moved fom position 1 to position? dф a) inceases b) deceases c) doesn t change Physics 31 Lectue 14
15 xample 1  continued ds ds 1 dф a) inceases b) deceases c) doesn t change The total flux of a chage is constant, with the density of flux lines being highe the close you ae to the chage Theefoe as you move the chage close to the suface element, the density of flux lines inceases Multiplying this highe density by the same value fo the aea of ds gives us that the incemental flux also inceases Physics 31 Lectue 15
16 ds xample ds 1 A positive chage is contained inside a spheical shell. How does the total flux, Ф, though the entie suface change when the chage is moved fom position 1 to position? a) Ф inceases b) Ф deceases c) Ф doesn t change As we peviously calculated, the total flux fom a point chage depends only upon the chage Physics 31 Lectue 16
17 Gauss Law The esult fo a single chage can be extended to systems consisting of moe than one chage One epeats the calculation fo each of the chages enclosed by the suface and then sum the individual fluxes Φ = 1 q i ε0 i Gauss Law elates the flux though a closed suface to chage within that suface Physics 31 Lectue 17
18 Gauss Law Gauss Law states that The net flux though any closed suface equals the net (total) chage inside that suface divided by ε 0 Qnet Φ = da = ε 0 Note that the integal is ove a closed suface Physics 31 Lectue 18
19 A B xample 3 A blue sphee A is contained within a ed spheical shell B. Thee is a chage Q A on the blue sphee and chage Q B on the ed spheical shell. The electic field in the egion between the sphees is completely independent of Q B the chage on the ed spheical shell. Tue False Physics 31 Lectue 19
20 Sufaces Choose suface appopiate to poblem It does not have to be a sphee xploit symmeties, if any Physics 31 Lectue 0
21 xample 4 Thin Infinite Sheet of Chage A given sheet has a chage density given by σ C/m By symmety, is pependicula to the sheet Use a suface that exploits this fact A cylinde A Gaussian pillbox Physics 31 Lectue 1
22 Thin Infinite Sheet of Chage A A da = σa ε + 0 A left + cuved ight = σa ε But and A cuved ae pependicula to each othe so thei dot poduct is zeo and the middle tem on the left disappeas A σa = ε = 0 σ ε 0 0 Physics 31 Lectue 
23 xample 5 Infinite Line having a Chage Density λ y h Apply Gauss Law: On the ends, ds = 0 On the bael, quating these and eaanging yields By Symmety field must be to line of chage and can only depend on distance fom the line Theefoe, choose the Gaussian suface to be a cylinde of adius and length h aligned with the xaxis ds = π h since // is zeo and q = λh λ πε 0 This is the same esult as using the integal fomulation = Physics 31 Lectue 3
24 xample 6 Solid Unifomly Chaged Sphee A chage Q is unifomly distibuted thoughout the volume of an insulating sphee of adius R. What is the electic field fo < R? Calculate aveage chage density Chage Density : ρ = Q 4π R Now select a Gaussian sphee of adius within this lage sphee Chage within this sphee is given by Q Q 4 Q 4π R / 3 R 3 encl = ρvencl = = 3 π /3 Physics 31 Lectue 4
25 xample 6 Solid Unifomly Chaged Sphee lectic Field is eveywhee pependicula to suface, i.e. paallel to suface nomal Gauss Law then gives 4π da = = Q 4πε 0 Q ε Q = ε 0 R encl 3 0 R 3 3 Field inceases linealy within sphee Outside of sphee, electic field is given by that of a point chage of value Q Physics 31 Lectue 5
26 Chages on Conductos Given a solid conducto, on which is placed an excess chage then in the static limit The excess chage will eside on the suface of the conducto and veywhee the electic field due to this excess chage will be pependicula to the suface and The electic field within the conducto will eveywhee be zeo Physics 31 Lectue 6
27 xample 7 A solid conducting sphee is concentic with a thin conducting shell, as shown The inne sphee caies a chage Q 1, and the spheical shell caies a chage Q, such that Q =  3 Q 1 1. How is the chage distibuted on the sphee?. How is the chage distibuted on the spheical shell? 3. What is the electic field at < R 1? Between R 1 and R?At > R? 4. What happens when you connect the two sphees with a wie? (What ae the chages?) Physics 31 Lectue 7
28 1. How is the chage distibuted on the sphee? Remembe that the electic field inside a conducto in a static situation is zeo. By Gauss s Law, thee can be no net chage inside the conducto The chage, Q 1, must eside on the outside suface of the sphee + + Physics 31 Lectue
29 . How is the chage distibuted on the spheical shell? The electic field inside the conducting shell is zeo. Thee can be no net chage inside the conducto Using Gauss Law it can be shown that the inne suface of the shell must cay a net chage of Q 1 The oute suface must cay the chage +Q 1 + Q, so that the net chage on the shell equals Q The chages ae distibuted unifomly ove the inne and oute sufaces of the shell, hence σ inne Q = 4πR 1 and σ oute Q + Q Q = = 4πR 4πR 1 1 Physics 31 Lectue 9
30 3. What is the lectic Field at < R 1? Between R 1 and R? At > R? The electic field inside a conducto is zeo. < R 1 : This is inside the conducting sphee, theefoe = 0 Between R 1 and R : R 1 < < R Chage enclosed within a Gaussian sphee = Q 1 = k Q 1 ˆ > R Chage enclosed within a Gaussian sphee = Q 1 + Q = k Q + Q Q 3Q 1 ˆ = k 1 1 ˆ = Q k 1 ˆ Physics 31 Lectue 30
31 4. What happens when you connect the two sphees with a wie? (What ae the chages?) Afte electostatic equilibium is eached, thee is no chage on the inne sphee, and none on the inne suface of the shell The chage Q 1 + Q esides on the oute suface Also, fo < R = 0 and fo > R = k Q 1 ˆ Physics 31 Lectue 31
32 xample 8 An unchaged spheical conducto has a weidly shaped cavity caved out of it. Inside the cavity is a chage q. q i) How much chage is on the cavity wall? (a) Less than q (b) xactly q (c) Moe than q ii) How is the chage distibuted on the cavity wall? (a) Unifomly (b) Moe chage close to q (c) Less chage close to q iii) How is the chage distibuted on the outside of the sphee? (a) Unifomly (b) Moe chage nea the cavity (c) Less chage nea the cavity Physics 31 Lectue 3
33 xample 8 An unchaged spheical conducto has a weidly shaped cavity caved out of it. Inside the cavity is a chage q. q i) How much chage is on the cavity wall? (a) Less than< q (b) xactly q (c) Moe than q By Gauss Law, since = 0 inside the conducto, the total chage on the inne wall must be q (and theefoe q must be on the outside suface of the conducto, since it has no net chage). Physics 31 Lectue 33
34 xample 8 An unchaged spheical conducto has a weidly shaped cavity caved out of it. Inside the cavity is a chage q. q ii) How is the chage distibuted on the cavity wall? (a) Unifomly (b) Moe chage close to q (c) Less chage close to q The induced chage will distibute itself nonunifomly to exactly cancel eveywhee in the conducto. The suface chage density will be highe nea the q chage. Physics 31 Lectue 34
35 xample 8 An unchaged spheical conducto has a weidly shaped cavity caved out of it. Inside the cavity is a chage q. q iii) How is the chage distibuted on the outside of the sphee? (a) Unifomly (b) Moe chage nea the cavity (c) Less chage nea the cavity The chage will be unifomly distibuted (because the oute suface is symmetic). Outside the conducto, the field always points diectly to the cente of the sphee, egadless of the cavity o chage o its location. Note: this is why you adio, cell phone, etc. won t wok inside a metal building! Physics 31 Lectue 35
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