Gauss s Law. The first Maxwell Equation A very useful computational technique This is important!

Transcription

1 Gauss s Law The first Maxwell quation A very useful computational technique This is important! P05-7

2 Gauss s Law The Idea The total flux of field lines penetrating any of these surfaces is the same and depends only on the amount of charge inside P05-8

3 Gauss s Law The quation Φ = d A = closed surfaces q ε in 0 lectric flux Φ (the surface integral of over closed surface S) is proportional to charge inside the volume enclosed by S P05-9

4 Now the Details P05-10

5 lectric Flux Φ Case I: is constant vector field perpendicular to planar surface S of area A Φ d = A Φ =+ A Our Goal: Always reduce problem to this P05-11

6 lectric Flux Φ Case II: is constant vector field directed at angle θ to planar surface S of area A d Φ = A Φ = Acosθ P05-12

7 PRS Question: Flux Thru Sheet P05-13

8 Gauss s Law Φ = d A = closed surfaces q ε in 0 Note: Integral must be over closed surface P05-14

9 Open and Closed Surfaces A rectangle is an open surface it does NOT contain a volume A sphere is a closed surface it DOS contain a volume P05-15

10 Area lement da: Closed Surface For closed surface, da is normal to surface and points outward ( from inside to outside) Φ > 0 if points out Φ < 0 if points in P05-16

11 lectric Flux Φ Case III: not constant, surface curved da dφ = da Φ = d Φ S P05-17

12 xample: Point Charge Open Surface P05-18

13 xample: Point Charge Closed Surface P05-19

14 PRS Question: Flux Thru Sphere P05-20

15 lectric Flux: Sphere Point charge Q at center of sphere, radius r field at surface: = Q 4πε r 0 2 rˆ lectric flux through sphere: Φ = d A S Q 4πε r = 2 0 S da Q = ˆ r dar 2 4πε S 0r = Q 4πε r 0 2 4π r 2 = ˆ Q ε 0 d A = darˆ P05-21

16 Arbitrary Gaussian Surfaces Φ = da = closed surface S Q ε 0 For all surfaces such as S 1, S 2 or S 3 P05-22

17 Applying Gauss s Law 1. Identify regions in which to calculate field. 2. Choose Gaussian surfaces S: Symmetry 3. Calculate Φ = da 4. Calculate q in, charge enclosed by surface S 5. Apply Gauss s Law to calculate : Φ = d A = S closed surfaces q ε in 0 P05-23

18 Choosing Gaussian Surface Choose surfaces where is perpendicular & constant. Then flux is A or -A. Choose surfaces where is parallel. Then flux is zero A A OR xample: Uniform Field Flux is A on top Flux is A on bottom Flux is zero on sides P05-24

19 Symmetry & Gaussian Surfaces Use Gauss s Law to calculate field from highly symmetric sources Symmetry Spherical Cylindrical Planar Gaussian Surface Concentric Sphere Coaxial Cylinder Gaussian Pillbox P05-25

20 PRS Question: Should we use Gauss Law? P05-26

21 Gauss: Spherical Symmetry +Q uniformly distributed throughout non-conducting solid sphere of radius a. Find everywhere P05-27

22 Gauss: Spherical Symmetry Symmetry is Spherical = rˆ Use Gaussian Spheres P05-28

23 Gauss: Spherical Symmetry Region 1: r > a Draw Gaussian Sphere in Region 1 (r > a) Note: r is arbitrary but is the radius for which you will calculate the field! P05-29

24 Gauss: Spherical Symmetry Region 1: r > a Total charge enclosed q in = +Q Φ = d A = S ( 2 4π r ) = da = A q 2 in Φ = 4π r= = ε0 ε0 S Q Q Q = = 2 4πε 0r 4πε 2 0r rˆ P05-30

25 Gauss: Spherical Symmetry Region 2: r < a Total charge enclosed: q in 4 πr 3 4 πa 3 3 Q = Gauss s law: r a = 3 3 ( 2 4π ) Φ = r Q r 3 Q OR qin = = Q r = = 3 4πε0 a 4πε 3 0 a = ρv 3 qin r Q 3 ε0 a ε0 rˆ P05-31

26 PRS Question: Field Inside Spherical Shell P05-32

27 Gauss: Cylindrical Symmetry Infinitely long rod with uniform charge density λ Find outside the rod. P05-33

28 Gauss: Cylindrical Symmetry Symmetry is Cylindrical = rˆ Use Gaussian Cylinder Note: r is arbitrary but is the radius for which you will calculate the field! is arbitrary and should divide out P05-34

29 Gauss: Cylindrical Symmetry Total charge enclosed: q in = λ Φ = da = da= A S S = qin λ 2π r = = ( ) ε ε 0 0 = λ 2πε r 0 = λ 2πε r 0 rˆ P05-35

30 Gauss: Planar Symmetry Infinite slab with uniform charge density σ Find outside the plane P05-36

31 Gauss: Planar Symmetry Symmetry is Planar = ± xˆ Use Gaussian Pillbox Note: A is arbitrary (its size and shape) and should divide out xˆ Gaussian Pillbox P05-37

32 Gauss: Planar Symmetry Total charge enclosed: q in = σa NOT: No flux through side of cylinder, only endcaps Φ = da = da= A = S in = 2A = = σ 2ε 0 ( ) = q ε S σ A ε 0 0 σ 2ε 0 ndcaps { xˆ to right} ˆ -x to left x σ A P05-38

33 PRS Question: Slab of Charge P05-39

34 Group Problem: Charge Slab Infinite slab with uniform charge density ρ Thickness is 2d (from x=-d to x=d). Find everywhere. xˆ P05-40

35 PRS Question: Slab of Charge P05-41

36 Potential from P05-42

37 Potential for Uniformly Charged Non-Conducting Solid Sphere From Gauss s Law V Use B Q rˆ, r > R 2 4πε0r = Qr rˆ, r < R 3 4πε0R Region 1: r > a B VB VA = d V ( ) = 0 A 0 s r Q = 2 dr 4πε r Point Charge! = 1 4πε 0 Q r P05-43

38 Potential for Uniformly Charged Non-Conducting Solid Sphere Region 2: r < a VD V ( ) = 0 R R = > < dr r R dr r R Q r Qr = dr dr 4πε r 4πε R r ( ) ( ) 2 R = 1 Q 1 Q 1 πε πε R 4 0 R 2 R ( 2 2 r R ) 1 Q = 3 8πε0 R r R 2 2 P05-44

39 Potential for Uniformly Charged Non-Conducting Solid Sphere P05-45

40 Group Problem: Charge Slab Infinite slab with uniform charge density ρ Thickness is 2d (from x=-d to x=d). If V=0 at x=0 (definition) then what is V(x) for x>0? xˆ P05-46

41 Group Problem: Spherical Shells These two spherical shells have equal but opposite charge. Find everywhere Find V everywhere (assume V( ) = 0) P05-47