Module 05: Gauss s s Law a 1
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
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
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
Now the Details 5
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
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
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
Gauss s Law Φ = E d d A = E q in ε closed 0 sufaces Note: Integal must be ove closed suface 9
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
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
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
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
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 0 2 2 4π2 = Q ε 0 d A = daˆ 14
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
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
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
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
Examples: Spheical Symmety Cylindical Symmety y Plana Symmety 19
Gauss: Spheical Symmety +QQ unifomly distibuted thoughout non-conducting solid sphee of adius a. Find E eveywhee 20
Gauss: Spheical Symmety Symmety is Spheical E = E ˆ Use Gaussian Sphees 21
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
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
Gauss: Spheical Symmety Region 2: < a Total chage enclosed: q in = 4 3 π 3 Q = 3 4 3 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
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
Demonstation Field Inside Spheical Shell (Gass Seeds): 26
Gauss: Plana Symmety y Infinite slab with unifom chage density σ Find E outside the plane 27
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
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 0 0 + E + E A σ σ { xˆ to ight } + E = E = + 2ε0 2 -xˆ to left ε 0 + + σ 29
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
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
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
Gauss: Cylindical Symmety Infinitely long od with unifom chage density λ Find E outside the od. 33
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
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
MIT OpenCouseWae http://ocw.mit.edu 8.02SC Physics II: Electicity and Magnetism Fall 2010 Fo infomation about citing these mateials o ou Tems of Use, visit: http://ocw.mit.edu/tems.