Physics 212 Lecture 3

Transcription

1 Physics 212 Lecture 3 Today's Concepts: lectric Flux and Field Lines Gauss s Law Physics 212 Lecture 3, Slide 1

2 Music Who is the Artist? A) Professor Longhair B) Henry Butler C) David gan D) Dr. John ) Allen Toussaint WHY? Mardi Gras is 4 weeks from today Physics 212 Lecture 13

3 Our Comments SmartPhysics scores will not be visible in the main gradebook for a few weeks You will need to understand integrals in this course!!! Forces and Fields are Vectors Always Draw a Picture First What do the Forces/Fields Look Like? Prelecture Lecture Worked xample Infinite line of charge dq 1 dq = k rˆ = rˆ r 4πε r WORKS FOR ALL!! Finite line of charge (constant λ) Homework Arc of charge (constant λ) Finite line of charge (non-constant λ) The 1/4 shell problem is not special!! It s the same as all the others! Physics 212 Lecture 3, Slide 3

4 lectric Field Lines Direction & Density of Lines represent Direction & Magnitude of Point Charge: Direction is radial Density α 1/R 2 07 Physics 212 Lecture 3, Slide 4

5 lectric Field Lines Dipole Charge Distribution: Direction & Density much more interesting 08 simulation Physics 212 Lecture 3, Slide 5

6 Checkpoint 3.1 Preflight 3 The following three questions pertain to the electric field lines due to the two charges shown. Compare the magnitude of the two charges A. Q 1 > Q 2 B. Q 1 = Q 2 C. Q 1 < Q 2 D. There isn t enough information to determine the relative magnitude of the charges more field lines emanating from the charge 1, so must be greater in magnitude. 09 simulation Physics 212 Lecture 3, Slide 6

7 Checkpoint 3.3 What do we know about the signs of the charges from looking at the picture? A. Q 1 and Q 2 have the same sign B. Q 1 and Q 2 have the opposite sign C. There isn t enough information to determine the relative signs of the charges The field lines leave one and go to the other.. 10 simulation Physics 212 Lecture 3, Slide 7

8 Preflight 3 Checkpoint 3.5 Compare the magnitude of the electric field at point A and B. A. A > B B. A = B C. A < B D. There isn t enough information to determine the relative magnitude of the electric field at points A and B Since the electric field lines are denser near point B the magnitude of the electric field is greater at point B than at point A. 12 Physics 212 Lecture 3, Slide 8

9 Point Charges Telling the difference between positive and negative charges while looking at field lines. Does field line density from a certain charge give information about the sign of the charge? -q +2q What charges are inside the red circle? -Q -Q -2Q -Q +Q +Q +2Q +Q A B C D 13 Physics 212 Lecture 3, Slide 9

10 Which of the following field line pictures best represents the electric field from two charges that have the same sign but different magnitudes? A B C D 15 simulation Physics 212 Lecture 3, Slide 10

11 lectric Flux Counts Field Lines I still feel a little unclear about flux integrals. Can you go over some more example problems? Flux through surface S Φ = S S Integral of on surface S 18 Physics 212 Lecture 3, Slide 11

12 Checkpoint 1 The flux is directly proportional to the charge enclosed by the Gaussian surface, therefore Case 1 will have 2 times the flux of of Case 2 because it has two charges enclosed, versus 1 in Case 2. The flux is the measure of field lines passing through the surface. Once calculated, it can be shown that the surface area of both cylinders are the same. An infinitely long charged rod has uniform charge density λ and passes through a cylinder (gray). The cylinder in Case 2 has twice the radius and half the length compared TAK s TO B RADIUS! with the cylinder in Case 1. L/2 Φ 1 =2Φ 2 Φ 1 =Φ 2 (A) (B) Φ 1 =1/2Φ 2 (C) none (D) 23 Physics 212 Lecture 3, Slide 12

13 Checkpoint 1 WHAT IS WRONG WITH THIS RASONING?? Definition of Flux: Φ surface constant on barrel of cylinder perpendicular to barrel surface ( parallel to ) Case 1 λ = 2πε s 1 0 Φ1 = 1 = π 0 A (2 s) L 26 The flux is the measure of field lines passing through the surface. Once calculated, it can be shown that the surface area of both cylinders are the same. TH FILD AT TH BARRL SURFAC IS NOT TH SAM IN TH TWO CASS!! Φ = = A barrel λl ε barrel A 2 An infinitely long charged rod has uniform charge density l and passes through a cylinder (gray). The cylinder in Case 2 has twice the radius and half the length compared with the TAK s TO B RADIUS! cylinder in Case 1. Φ 1 =2Φ 2 Φ 1 RSULT: =Φ 2 ΦGAUSS 1 =1/2Φ 2 LAWnone Φ proportional to charge enclosed (D)! (A) Case 2 λ = 2πε 0 (2 s) = (2π (2s)) L / 2 2πsL 2 = (B) Φ (C) λ( L / 2) = L/2 2 ε0 Physics 212 Lecture 3, Slide 13

14 Direction Matters: For a closed surface, A points outward 29 Φ = > S S 0 Physics 212 Lecture 3, Slide 14

15 Direction Matters: For a closed surface, A points outward 30 Φ = < S S 0 Physics 212 Lecture 3, Slide 15

16 Trapezoid in Constant Field Label faces: 1: x = 0 2: z = +a 3: x = +a 4: slanted Define Φ n = Flux through Face n y = x 3 0 ˆ x < 0 z A Φ 1 < 0 A Φ 2 < 0 A Φ 3 < 0 A Φ 4 < 0 B Φ 1 = 0 B Φ 2 = 0 B Φ 3 = 0 B Φ 4 = 0 C Φ 1 > 0 C Φ 2 > 0 C Φ 3 > 0 C Φ 4 > 0 31 Physics 212 Lecture 3, Slide 16

17 Trapezoid in Constant Field + Q y = x 0 ˆ Label faces: 1: x = 0 2: z = +a 3: x = +a Define Φ n = Flux through Face n Φ = Flux through Trapezoid z +Q x 0 Add a charge +Q at (-a,a/2,a/2) How does Flux change? A Φ 1 increases A Φ 3 increases A Φ increases B Φ 1 decreases B Φ 3 decreases B Φ decreases C Φ 1 remains same C Φ 3 remains same C Φ remains same 36 Physics 212 Lecture 3, Slide 17

18 Gauss Law Q Φ S closed surface = Q enclosed ε o 41 Physics 212 Lecture 3, Slide 18

19 Checkpoint 2.3 (A) Φ increases (B) Φ decreases (C) Φ stays same The flux throughout the entire surface does not change as the charge moves, for it is still in the sphere. However, if it was moved outside of the sphere, the fluxes would be different because one would be zero! 43 Physics 212 Lecture 3, Slide 19

20 Checkpoint 2.1 charge is closer, field is larger, more lines through a so increase in flux. as long as the charge is within the shell, the flux will remain the same (A) dφ A increases dφ B decreases (B) dφ A decreases dφ B increases (C) dφ A stays same dφ B stays same 44 Physics 212 Lecture 3, Slide 20

21 Think of it this way: 1 2 The total flux is the same in both cases (just the total number of lines) The flux through the right (left) hemisphere is smaller (bigger) for case Physics 212 Lecture 3, Slide 21

22 Things to notice about Gauss s Law Φ S = closed surface Q enclosed If Q enclosed is the same, the flux has to be the same, which means that the integral must yield the same result for any surface. ε o 47 Physics 212 Lecture 3, Slide 22

23 Things to notice about Gauss s Law Gausses law and what concepts are most important that we know. = closed surface enclosed In cases of high symmetry it may be possible to bring outside the integral. In these cases we can solve Gauss Law for Q ε o closed surface = A = Q enclosed ε o = Q enclosed Aε 0 So - if we can figure out Q enclosed and the area of the surface A, then we know! This is the topic of the next lecture 48 Physics 212 Lecture 3, Slide 23

24 Keep plugging away Final Thoughts. Prelecture 4 and Checkpoint 4 due Thursday Homework 2 due next Tuesday 7 questions mostly Gauss law try them today! 50 Physics 212 Lecture 3, Slide 24