IClicker question. We have a negative charge q=-e. How electric field is directed at point X? q=-e (negative charge) X A: B: C: D: E: E=0

Transcription

1 We have a negative charge q=-e. How electric field is directed at point X? IClicker question q=-e (negative charge) X A: B: C: D: E: E=0 1

2 A: q=-e (negative charge) F X E Place positive charge q0 Force E=F/q0 2

3 Recall: The Electric Field A charge creates an electric field around itself and the other charge feels that field. + 3

4 Recall: The Electric Field A charge creates an electric field around itself and the other charge feels that field. + + Test charge q 3

5 Review - Point Charge The electric field created by a point charge, as a function of position r, The force exerted by an electric field on a point charge q located at position r direction: tangent to the field line 42

6 Electric field points away from the positive charge +q E 5

7 Electric field points toward the negative charge E -q 6

8 Electric field points toward the negative charge E -q 6

9 Continuous Distribution of charges Two methods: Method #1: Divide the distribution into infinitesimal elements de and integrate to get the full electric field Method #2: If there is some special symmetry of the distribution, use Gauss Law to derive the field from 46

10 Calculating E-field from Coulomb s law From a point charge From continuos distribution of charges d E = kdq r 2 ˆr Integrate over volume E = k dq r 2 ˆr 8

11 E-field from a uniformly charged rod 9

12 E-field from a uniformly charged rod 10

13 E-field from a uniformly charged rod Simmetry: 11

14 Electric Field from a Ring of Charge Total charge Q is distributed homogeneously over a ring of radius R. Find E-field on the axis at distance z. 79

15 dφ Take section Its charge is dφ dq = dφ 2π Q It produces E-field E dq = k dq R 2 + z 2 = k dφ 2π Q R 2 + z 2 From symmetry: only Ez is important E z,dq = E dq cos θ = k dφ 2π Q R 2 + z 2 z R2 + z 2 13

16 E z,dq = E dq cos θ = k dφ 2π Q R 2 + z 2 z R2 + z 2 Integrate over E dφ E = k Qz (R 2 + z 2 ) 3/ /z z/r 14

17 Field on the axis from a charged disk from previous: E x = k a=r 0 E x = k E x = k Qx (a 2 + x 2 ) 3/2 dqx (a 2 + x 2 ) 3/2 dq =2πσada z 2πσa (a 2 + z 2 da = ) 3/2 1 2πkσ 1 1+(R/x) 2 15

18 Field on the axis from a charged plate E x =2πkσ (R/x) 2 R E x =2πkσ 16

19 Gauss Law Two methods fro calculating E-field from continuous distribution of charges: Method #1: Divide the distribution into infinitesimal elements de and integrate to get the full electric field Method #2: If there is some special symmetry of the distribution, use Gauss Law to derive the field from Gauss Law The flux of electric field through a closed surface is proportional to the charge enclosed by the surface. First we need to define the concept of flux 46

20 Water Flux Let s imagine that we put a ring with area A perpendicular to a stream of water flowing with velocity v The product of area times velocity, Av, gives the volume of water passing through the ring per unit time The units are m 3 /s If we tilt the ring at an angle θ, then the projected area is Acosθ, and the volume of water per unit time flowing through the ring is Av cosθ. 47

21 Electric Flux (1) We call the amount of water flowing through the ring the flux of water We can make an analogy with electric field lines from a constant electric field and flowing water We call the density of electric field lines through an area A the electric flux 48

22 Consider a constant electric field E passing through a given area A The angle θ is the angle between the electric field vector and the area vector Electric Flux (2) The density of electric field vectors passing through a given area A is called the electric flux N.B.: Nothing is actually flowing! 49

23 Surfaces and Normal Vectors 50

24 Electric Flux for Closed Surface Assume an electric field and a closed surface, rather than the open surface associated with our ring analogy In this closed-surface case, the total electric flux through the surface is given by an integral over the closed surface The differential area vectors always point out of the closed surface 51

25 Gauss s Law We can now formulate Gauss s Law as Flux through a closed area equals the enclosed charge q is the total charge inside a closed surface We call this surface a Gaussian surface This surface could be our imaginary box This surface could be any closed surface Usually we chose this closed surface to have symmetries related to the problem we are trying to study 55

26 Gauss s Law (5) Gauss s Law (named for German mathematician and scientist Johann Carl Friedrich Gauss, ) states (q = net charge enclosed by Gaussian surface S). If we add the definition of the electric flux we get another expression for Gauss Law E da = q 0 Gauss s Law: the electric field flux through S is proportional to the net charge enclosed by S 56

27 Gauss s Law and Coulomb s Law Are Equivalent Let s derive Coulomb s Law from Gauss s Law We start with a point charge q We construct a spherical surface with radius r surrounding this charge This is our Gaussian surface 57

28 Gauss Law and Coulomb s Law (2) The electric field from a point charge is radial, and thus is perpendicular to the Gaussian surface everywhere E da = q 0 The electric field has the same magnitude anywhere on the surface, directed radially, along the local normal 58

29 Gauss Law and Coulomb s Law (3) Now we are left with a simple integral over a spherical surface So for Gauss law related to a point charge we get Which gives 59

30 iclicker What is the flux through the spherical surface encompassing two charges q1= 2q, q2=-2q A B C D E 4 q 0 q 0 q 0 2 q 0 0 2q -2q 28

31 29

32 iclicker Total electric flux through a closed surface Φ=0 A: Electric field inside is 0 B: Not necessarily 30

33 iclicker old What is the flux through the spherical surface encompassing two charges q1= 2q, q2=-2q A B C D E 4 q 0 q 0 q 0 2 q 0 0 2q -2q q tot =2q 2q =0 Φ=0 31

34 E A E da > 0 2q -2q E da < 0 E A Note: Zero total flux DOES NOT mean zero E-field! 32

35 What is an electric flux through spheres of radii r1 and r2= 2r1 surrounding a charge of q=-3 e? A :Φ 1 =Φ 2 =0 B :Φ 1 =Φ 2 =0 r2-3e C :Φ 1 =Φ 2 /2 r1 D :Φ 1 = 2Φ 2 E: Φ 1 =Φ 2 /4 33

36 What is a electric flux through spheres of radii r1 and r2= 2r1 surrounding a charge of q=-3 e? A :Φ 1 =Φ 2 =0 B :Φ 1 =Φ 2 =0 r2-3e C :Φ 1 =Φ 2 /2 r1 D :Φ 1 = 2Φ 2 E: Φ 1 =Φ 2 /4 equal: A 34

37 iclicker What is the flux through the spherical surface encompassing two charges q1= 3q, q2=-2q A 5 q 0 q 3q B C D q q 0-2q C 0 35

38 iclicker What is the flux through the spherical surface encompassing two charges q1= 3q, q2=-2q A 5 q 0 q 3q B C D q q 0 q tot =3q 2q = q -2q C 0 36

39 iclicker The total flux of electric field through the closed surface is A: > 0 B: =0 C: <0 +3q +3q +3q +3q +3q -q 37

40 End of lecture 38