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

Transcription

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

2 Introduce a new constant: 0 q k r r 2 ˆ k k = 9 x 10 9 N m 2 / C 2 0 = 8.85 x C 2 / N m q r 2 rˆ 04 Physics 212 Lecture 3, Slide 2

3 Plan for Today A little more about electric field lines lectric field lines and flux An analogy Introduction to Gauss s Law Gauss Law will make it easy to calculate electric fields for some geometries. 06 Physics 212 Lecture 3, Slide 3

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

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

6 Checkpoint Preflight 3 Field lines are are denser near Q 1 so Q 1 > Q 2 09 Physics 212 Lecture 3, Slide 6

7 Checkpoint The electric field lines connect the charges. A test charge will move towards one charge and away from the other. So charges 1 and 2 have opposite signs. 10 Physics 212 Lecture 3, Slide 7

8 Checkpoint Preflight 3 Density of lines is greater at B than at A. Therefore, magnitude of field at B is greater than at A. 12 Physics 212 Lecture 3, Slide 8

9 Point Charges -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 Physics 212 Lecture 3, Slide 10

11 lectric Flux Counts Field Lines S S Flux through surface S Integral of on surface S 18 Physics 212 Lecture 3, Slide 11

12 lectric Field/Flux Analogy: Velocity Field/Flux S Flux through surface S S v Integral of v on surface S flowrate 20 Physics 212 Lecture 3, Slide 12

13 Checkpoint 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 TAK s TO B RADIUS! with the cylinder in Case 1. L/2 1 =2 2 1 = 2 (A) (B) 1 =1/2 2 (C) none (D) 23 Physics 212 Lecture 3, Slide 13

14 Checkpoint Definition of Flux: surface 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. L/2 constant on barrel of cylinder perpendicular to barrel surface ( parallel to ) Case 1 l 2 s A (2 s) L 26 barrel ll A barrel A 2 1 =2 2 1 RSULT: = 2 GAUSS 1 =1/2 2 LAW none proportional to charge enclosed (D)! (A) Case 2 l 20(2s) (2 (2s)) L / 2 2sL 2 (B) (C) l( L / 2) 2 0 Physics 212 Lecture 3, Slide 14

15 Direction Matters: For a closed surface, A points outward 29 S S 0 Physics 212 Lecture 3, Slide 15

16 Direction Matters: For a closed surface, A points outward 30 S S 0 Physics 212 Lecture 3, Slide 16

17 Trapezoid in Constant Field Label faces: 1: x = 0 2: z = +a 3: x = +a 4: slanted Define n = Flux through Face n y 1 2 x 3 0 ˆ x 0 z A 1 < 0 A 2 < 0 A 3 < 0 A 4 < 0 B 1 = 0 B 2 = 0 B 3 = 0 B 4 = 0 C 1 > 0 C 2 > 0 C 3 > 0 C 4 > 0 31 Physics 212 Lecture 3, Slide 17

18 Trapezoid in Constant Field + Q y x 0 ˆ Label faces: 1: x = 0 2: z = +a 3: x = +a Define n = Flux through Face n = Flux through Trapezoid +Q x Add a charge +Q at (-a,a/2,a/2) z How does Flux change? A 1 increases A 3 increases A increases B 1 decreases B 3 decreases B decreases C 1 remains same C 3 remains same C remains same 36 Physics 212 Lecture 3, Slide 18

19 Gauss Law Q S closed surface Q enclosed o 41 Physics 212 Lecture 3, Slide 19

20 Checkpoint What happens to total flux through the sphere as we move Q? (A) increases (B) decreases (C) stays same The same amount of charge is still enclosed by the sphere, so flux will not change. 43 Physics 212 Lecture 3, Slide 20

21 Checkpoint (A) d A increases d B decreases (B) d A decreases d B increases (C) d A stays same d B stays same 44 Physics 212 Lecture 3, Slide 21

22 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 22

23 Things to notice about Gauss s Law 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. o 47 Physics 212 Lecture 3, Slide 23

24 Things to notice about Gauss s Law 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 closed surface A Q enclosed o So - if we can figure out Q enclosed and the area of the surface A, then we know! Q This is the topic of the next lecture o Q enclosed A 0 48 Physics 212 Lecture 3, Slide 24

25 Prelecture 4 and Checkpoint 4 due Thursday Homework 2 due next Monday 50 Physics 212 Lecture 3, Slide 25