Exam 1. Exam 1 is on Tuesday, February 14, from 5:00-6:00 PM.

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1 Exam 1 Exam 1 is on Tuesday, Febuay 14, fom 5:00-6:00 PM. Testing Cente povides accommodations fo students with special needs I must set up appointments one week befoe exam Deadline fo submitting accommodation lettes: Monday, Feb 6 by noon student must make sepaate appointment with testing cente if you have othe time conflicts, I need to know now! (they ae the student s esponsibility unless the student paticipates in a majo univesity o intecollegiate event)

2 Couse Leaning Assistance Physics Leaning Cente (Monday+Wednesday 2:00-4:30 and 6:00 to 8:30 in ooms 129 and 130 Physics) LEAD/tutoing sessions ( Student Success Cente, 198 Toomey Hall Monday + Wednesday 2:00 to 4:00pm Tuesday and Thusday 9:00 to 11:00am

3 Today s agenda: Announcements. Gauss Law Examples. You must be able to use Gauss Law to calculate the electic field of a high-symmety chage distibution. Cultual enlightenment time. You must be cultually enlightened by this lectue. Conductos in electostatic equilibium. You must be able to use Gauss law to daw conclusions about the behavio of chaged paticles on, and electic fields in, conductos in electostatic equilibium.

4 Gauss Law Last time we leaned that Φ E = E da = q enclosed ε 0 Gauss Law Always tue, not always easy to apply. and used Gauss Law to calculate the electic field fo spheically-symmetic chage distibutions Today we will calculate electic fields fo chage distibutions that ae non-spheical but exhibit a high degee of symmety, and then conside what Gauss Law has to say about conductos in electostatic equilibium.

5 Example: calculate the electic field outside a long cylinde of finite adius R with a unifom volume chage density ρ spead thoughout the volume of the cylinde. To be woked at the blackboad in lectue. Long cylinde with finite adius means neglect end effects; i.e., teat cylinde as if it wee infinitely long. Moe details of the calculation shown hee: E ρ R = 2 ε 0 2

6 Example: calculate the electic field outside a long cylinde of finite adius R with a unifom volume chage density ρ spead thoughout the volume of the cylinde. Cylinde is looooooong. I m just showing a bit of it hee. I don t even want to think of tying to use de=k dq / 2 fo this.

7 Example: calculate the electic field outside a long cylinde of finite adius R with a unifom volume chage density ρ spead thoughout the volume of the cylinde. ρ >0 R E

8 E da Looking down the axis of the cylinde.

9 da ρ >0 R L E

10 Inside the chaged cylinde, by symmety E must be adial.

11 da E E da = 0 because E da E da = 0 because E da da E

12 E da E da = E da because E da E = E L Also must be constant at any given.

13 E da Φ = E da = E da = E da = E da E L tube tube tube = E ( cicumfeence of Gaussian cylinde )( length of GC) ( )( L) = E 2π

14 E da ρ >0 R L E 2 L ρv E da = E 2π L = ρ q enclosed ε o ( )( ) ( 2 A length ρ πr )( L) enclosed base π = = = ε ε ε o o o

15 E da ρ >0 R L ( 2 )( ) E 2π L = ρ πr L ε 2 2 R R E ρπ ρ = = 2 πε 2 ε o o o

16 E da ρ >0 R L Fo positive ρ: E 2 ρr = 2 ε o, adially out In geneal: E ρ R = 2 ε o 2 Why does this vay as 1/ instead of 1/ 2?

17 Fo a solid cylinde Chage pe volume is Q ρ= πrl 2 Chage pe length is λ= Q L Q 1 Q λ ρ= = = πrl πr L πr So And E λ 2 2 R 2 ρ R πr λ = = = 2ε 2ε 2πε o o o

18 Example: use Gauss Law to calculate the electic field due to a long line of chage, with linea chage density λ. This is easy using Gauss Law (emembe what a pain it was in the pevious chapte). Study the examples in this lectue and othes in you text! To be woked at the blackboad in lectue. E = λ 2πε 0

19 Example: use Gauss Law to calculate the electic field due to a long line of chage, with linea chage density λ. Line is looooooong. E λ >0

20 E da Looking down the line.

21 E da λ >0 L

22 E da λ >0 L ( 2π )( L) enclosed E da = E = = E = λ 2πε 0 q ε 0 λl ε 0

23 E λ >0 Fo positive λ: E = λ 2πε 0, adially out In geneal: E λ Same as outside a = solid cylinde! 2πε0

24 Example: use Gauss Law to calculate the electic field due to an infinite sheet of chage, with suface chage density σ. This is easy using Gauss Law (emembe what a pain it was in the pevious chapte). Study the examples in this lectue and othes in you text! To be woked at the blackboad in lectue. σ E sheet =. 2 ε 0

25 Example: use Gauss Law to calculate the electic field due to an infinite sheet of chage, with suface chage density σ. σ >0 Two views of sheet of chage; side view looking edge on, and top view looking down. Sheet extends infinitely fa in two dimensions.

26 Example: use Gauss Law to calculate the electic field due to an infinite sheet of chage, with suface chage density σ. E σ >0 E Fo this electic field symmety, we usually use a pillbox (cylinde shape) fo ou Gaussian suface. In the views above, it will look like a ectangle and a cicle. You could also use a ectangula box.

27 Example: use Gauss Law to calculate the electic field due to an infinite sheet of chage, with suface chage density σ. E σ >0 H E E da = 2 E = [ ( ) ] 2 π q enclosed ε 0 q = σ enclosed ( 2 π )

28 Example: use Gauss Law to calculate the electic field due to an infinite sheet of chage, with suface chage density σ. E σ >0 H E [ ( ) ] ( 2 ) 2 σ π E π 2 = σ E = 2ε 0 ε 0

29 Example: use Gauss Law to calculate the electic field due to an infinite sheet of chage, with suface chage density σ. E σ >0 H E Fo positive σ: E In geneal: = σ 2ε 0 E =, away fom the sheet σ 2ε 0 That sue was easie than the deivation back in Lectue 2!

30 Gauss Law woks well fo thee kinds of symmety: Chage Symmety spheical cylindical plana Gaussian Suface concentic sphee coaxial cylinde pillbox

31 Today s agenda: Announcements. Gauss Law Examples. You must be able to use Gauss Law to calculate the electic field of a high-symmety chage distibution. Cultual enlightenment time. You must be cultually enlightened by this lectue. Conductos in electostatic equilibium. You must be able to use Gauss law to daw conclusions about the behavio of chaged paticles on, and electic fields in, conductos in electostatic equilibium.

32 Cultual Enlightenment* Time The top 5 easons why we make you lean Gauss Law: 5. You can solve difficult (high-symmety) poblems with it. 4. It s good fo you. It s fun! What moe can you ask! 3. It s easy. Smat physicists go fo the easy solutions. 2. If I had to lean it, you do too. And the numbe one eason will take a couple of slides to pesent *This will not be tested on the exam.

33 You have leaned the integal fom of Gauss Law: E da = q enclosed ε o This equation can also be witten in diffeential fom: E ρ = ε a 3-dimensional deivative opeato 0 Now you can see we ae on the tail of something Really Big This will not be tested on the exam.

34 The Missoui S&T Society of Physics Students T-Shit! This will not be tested on the exam.

35 Today s agenda: Announcements. Gauss Law Examples. You must be able to use Gauss Law to calculate the electic field of a high-symmety chage distibution. Cultual enlightenment time. You must be cultually enlightened by this lectue. Conductos in electostatic equilibium. You must be able to use Gauss law to daw conclusions about the behavio of chaged paticles on, and electic fields in, conductos in electostatic equilibium.

36 Conductos in Electostatic Equilibium Homewok hints buied in the next 3 slides! Electostatic equilibium means thee is no net motion of tne chages inside the conducto. The electic field inside the conducto must be zeo. If this wee not the case, chages would acceleate. Any excess chage must eside on the outside suface of the conducto. Apply Gauss law to a Gaussian suface just inside the conducto suface. The electic field is zeo, so the net chage inside the Gaussian suface is zeo. Any excess chage must go outside the Gaussian suface, and on the conducto suface.

The electic field just outside a chaged conducto must be pependicula to the conducto s suface. Othewise, the component of the electic field paallel to the suface would cause chages to acceleate.

37 The electic field just outside a chaged conducto must be pependicula to the conducto s suface. Othewise, the component of the electic field paallel to the suface would cause chages to acceleate. The magnitude of the electic field just outside a chaged conducto is equal to σ /ε 0, whee σ is the magnitude of the local suface chage density. A simple application Gauss Law. Diffeent fom infinite sheet of chage because E is zeo inside the conducto.

38 If thee is an empty nonconducting cavity inside a conducto, Gauss Law tells us thee is no net chage on the inteio suface of the conducto. conducto E da = 0 da = 0 qenclosed = 0 Constuct a Gaussian suface that includes the inne suface of the conducto. The electic field at the Gaussian suface is zeo, so no electic flux passes though the Gaussian suface. Gauss Law says the chage inside must be zeo. Any excess chage must lie on the oute suface! The conducto does not have to be symmetic, as shown.

39 If thee is a nonconducting cavity inside a conducto, with a chage inside the cavity, Gauss Law tells us thee is an equal and opposite induced chage on the inteio suface of the conducto. conducto +Q Q I =-Q E da = 0 da = 0 qenclosed = 0 q = 0 =+ Q + Q enclosed Q = Q I I Constuct a Gaussian suface that includes the inne suface of the conducto. The electic field at the Gaussian suface is zeo, so no electic flux passes though the Gaussian suface. Gauss Law says the chage inside must be zeo. Thee must be a Q on the inne suface. If the net chage on the conducto is not Q, any additional chage must lie on the oute suface! The conducto does not have to be symmetic.

40 Example: a conducting spheical shell of inne adius a and oute adius b with a net chage -Q is centeed on point chage +2Q. Use Gauss s law to show that thee is a chage of -2Q on the inne suface of the shell, and a chage of +Q on the oute suface of the shell. -Q E da = q enclosed ε 0 a E=0 inside the conducto! +2Q b Q I Let be infinitesimally geate than a. q ε Q + 2Q ε Q enclosed I 0 = = I = 0 0-2Q

41 Example: a conducting spheical shell of inne adius a and oute adius b with a net chage -Q is centeed on point chage +2Q. Use Gauss s law to show that thee is a chage of -2Q on the inne suface of the shell, and a chage of +Q on the oute suface of the shell. Q I = -2Q -Q +2Q a b Q I Fom Gauss Law we know that excess* chage on a conducto lies on sufaces. Electic chage is conseved: Q = -Q = Q + shell I Q O = -2Q + Q O Q O - Q = -2Q + Q QO O = + Q *excess=beyond that equied fo electical neutality

42 Example: an insulating sphee of adius a has a unifom chage density ρ and a total positive chage Q. Calculate the electic field at a point inside the sphee. Q a E da q enclosed = = ( 2 ) ε E 4π = o 4 ρ π 3 ε ρv ε enclosed This object in this example is not a conducto. See D. Waddill s lectue on Gauss Law fom a few yeas ago. Click on the wod lectue in the pevious sentence to view/download the lectue. Hee is the addess fo you to copy and paste into a web bowse, in case the link in the above paagaph doesn t wok: o 3 o

43 Example: an insulating spheical shell of inne adius a and oute adius b has a unifom chage density ρ. Calculate the electic field at a point inside the sphee. Q E da = enclosed a b q qenclosed =ρ Venclosed =ρ π πa 3 3 ε o Also see exam 1, Fall 2015, poblem 7, fo a elated example. Calculate the electic field at a point outside the sphee. 4 4 qenclosed =ρ Venclosed =ρ πb πa

44 Demo: Pofesso Ties to Avoid Debilitating Electical Shock While Demonstating Van de Gaaff Geneato

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