Module 05: Gauss s s Law a

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1 Module 05: Gauss s s Law a 1

2 Gauss s Law The fist Maxwell Equation! And a vey useful computational technique to find the electic field E when the souce has enough symmety. 2

3 Gauss s Law The Idea The total t flux of field lines penetating ti any of these closed sufaces is the same and depends only on the amount of chage inside 3

4 Gauss s Law The Equation q Φ = E d A= in E closed suface S ε 0 Electic flux Φ E (the suface integal of E ove closed suface S) is popotional to chage inside the volume enclosed by S 4

5 Now the Details 5

6 Electic Flux Φ E Case I: E is constant vecto field pependicula to plana suface S of aea A Φ E d = E A Φ = +EA E Ou Goal: Always educe poblem to this 6

7 Electic Flux Φ E Case II: E is constant vecto field diected at angle θ to plana suface S of aea A n ˆn Φ = E d E A d A = danˆ Φ E = EAcosθ E 7

8 Concept Question: Flux The electic flux though the plana suface below (positive unit nomal to left) is: ˆn +q -q 1. positive. 2. negative. 3. zeo. 4. I don t know 8

9 Gauss s Law Φ = E d d A = E q in ε closed 0 sufaces Note: Integal must be ove closed suface 9

10 Open and Closed Sufaces A ectangle is an open suface it does NOT contain a volume A sphee is a closed suface it DOES contain a volume 10

11 Aea Element da: Closed Suface Fo closed suface, da is nomal to suface and points outwad (f fom inside id to outside) Φ E > 0 if E points out Φ E <0if E points in 11

12 Electic Flux Φ E Case III: E not constant, suface cuved da E d Φ = E da E Φ E Φ = d Φ E S 12

13 Concept Question: Flux thu Sphee The total flux though the below spheical suface is +q 1. positive (net outwad flux). 2. negative (net inwad flux). 3. zeo. 4. I don t know 13

14 Electic Flux: Sphee Point chage Q at cente of sphee, adius E field at suface: E() = 4 Q πε 0 2 ˆ Electic flux though sphee: Φ = E da E S Q = 2 4πε 0 S Q = ˆ daˆ 2 4πε S da = Q 0 4πε π2 = Q ε 0 d A = daˆ 14

15 Abitay Gaussian Sufaces Φ = E da = E closed suface S ε 0 Q Tue fo all sufaces such as S 1, S 2 o S 3 Why? As A gets bigge E gets smalle 15

16 Choosing Gaussian Suface Q Φ = E d A = E ε 0 closed suface S Tue fo ALL sufaces Useful (to calculate E) fo SOME sufaces Desied E: Pependicula to suface and constant on suface. Flux is EA o -EA. Othe E: Paallel l to suface. Flux is zeo 16

17 Symmety y & Gaussian Sufaces Desied E: pependicula to suface and constant on suface. So Gauss s Law useful to calculate E field fom highly symmetic souces Souce Symmety y Spheical Cylindical Plana Gaussian Suface Concentic Sphee Coaxial Cylinde Gaussian Pillbox 17

18 Applying Gauss ss Law 1. Based on the souce, identify egions in which to calculate E field. 2. Choose Gaussian sufaces S: Symmety y 3. Calculate Φ = E da E S 4. Calculate q in, chage enclosed by suface S 5. Apply Gauss s Law to calculate E: Φ = E d A = E closed sufaces q ε in 0 18

19 Examples: Spheical Symmety Cylindical Symmety y Plana Symmety 19

20 Gauss: Spheical Symmety +QQ unifomly distibuted thoughout non-conducting solid sphee of adius a. Find E eveywhee 20

21 Gauss: Spheical Symmety Symmety is Spheical E = E ˆ Use Gaussian Sphees 21

22 Gauss: Spheical Symmety Region 1: > a Daw Gaussian Sphee in Region 1 ( > a) Note: is abitay but is the adius fo which you will calculate the E field! 22

23 Poblem: Outside Sphee Region 1: > a Use Gauss s Law in Region 1 ( > a) Again: Remembe that is abitay but is the adius fo which you will calculate the E field! 23

24 Gauss: Spheical Symmety Region 2: < a Total chage enclosed: q in = 4 3 π 3 Q = Q a 3 πa3 Gauss s s law: OR q in = ρv Φ E = E( 4π 2 )= q in = 3 3 Q ε 0 a ε 0 E = Q Q E = ˆ 4πε 0 a 3 3 4πε a 0 24

25 Concept Question: Spheical Shell We just saw that in a solid sphee of chage the electic field gows linealy with distance. Inside the chaged spheical shell at ight (<a) what does the electic field do? 1. Constant and Zeo 2. Constant but Non-Zeo 3. Still gows linealy a Q 4. Some othe functional fom (use Gauss Law) 5. Can t detemine with Gauss Law 25

26 Demonstation Field Inside Spheical Shell (Gass Seeds): 26

27 Gauss: Plana Symmety y Infinite slab with unifom chage density σ Find E outside the plane 27

28 Gauss: Plana Symmety Symmety is Plana E = ±E xˆ Use Gaussian Pillbox Note: A is abitay (its size and shape) and should divide out xˆx Gaussian Pillbox 28

29 Gauss: Plana Symmety y Total chage enclosed: q in = σa NOTE: No flux though side of cylinde, only endcaps Φ = E da= E da= EA E S S Endcaps x A = E( 2A)= q in = σ A ε ε E + E A σ σ { xˆ to ight } + E = E = + 2ε0 2 -xˆ to left ε σ 29

30 E fo Plane is Constant???? 1) Dipole: E falls off like 1/ 3 2) Point chage: E falls off like 1/ 2 3) Line of chage: E falls off like 1/ 4) Plane of chage: E constant 30

31 Concept Question: Slab of Chage Conside positive, semi-infinite (in x & y) flat slab z-axis is pep. to the sheet, with cente at z = 0. At the plane s cente (z = 0), E 2d ρ z = 0 z 1. points in the positive z-diection. 2. points in the negative z-diection. 3. points in some othe (x,y) diection. 4. is zeo. 5. I don t know 31

32 Poblem: Chage Slab Infinite slab with unifom chage density ρ Thickness is 2d (fom x=-d to x=d). Find E fo x > 0 (how many egions is that?) xˆ 32

33 Gauss: Cylindical Symmety Infinitely long od with unifom chage density λ Find E outside the od. 33

34 Gauss: Cylindical Symmety Symmety is Cylindical i l E = E ˆ Use Gaussian Cylinde Note: is abitay but is the adius fo which you will calculate the E field! l is abitay and should divide out 34

35 Gauss: Cylindical Symmety Total chage enclosed: q in = λl Φ E = E da= E da= EA ( π ) S S q λ in = E 2 l π l = = ε ε 0 0 λ λ E = E= 2πε0 2πε 0 ˆ 35

36 MIT OpenCouseWae SC Physics II: Electicity and Magnetism Fall 2010 Fo infomation about citing these mateials o ou Tems of Use, visit:

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