Electricity and Magnetism Electric Flux Gauss s Law

Transcription

1 Electricity and Magnetism Electric Flux Gauss s Law Lana heridan De Anza College Jan 18, 2018

2 Last time conductors in electric fields electric flux

3 Overview electric flux Gauss s law Gauss s law applied to various cases

4 ple ways of solving ics Gauss s in attaining Law this basic idea ctric field E : of the, we considered The electric the field around a charge is stronger for charges with Then we larger simplified magnitudes. dicular components To make this observation useful, we need to quantify it. save far more work matician and physing the fields de : pherical of Gaussian siders a hypothetical surface is Gaussian surface, s our calculations of distribution. For exse the sphere with a en, as we discuss in t that? ian surface to the E field on a Gaussian s a limited example, Fig A spherical Gaussian surface. If the electric field vectors

5 Flux Flux is a quantity that makes the idea of the electric field through some region precise. AU LAW Flux is a flow rate through an area. tream of veplane of a omponent of the loop : etween v he area vecane of the : h v. (d) The e area of the Air flow v θ v

6 Flux GAU LAW Imagine air blowing directly through a square loop of wire of area A. rstream of vee plane of a Air flow component v e of the loop θ : between v The area veclane of the : th v. (d) The he area The of volume the of air that passes through in 1 s is V = A v (1 s), where v is the speed of the air. (a) (b) v The rate of flow would be dv dt = Av.

7 Flux Now consider a more general situation: the air does not blow directly through the loop, but at some angle θ. Air flow v θ θ A v v (a) (b) If θ = 90, what is the flow rate (flux) through the loop? (c) (d) Before we discuss a flux involved in electrostatics, we need to rewrite Eq in terms of vectors. To do this, we first define an area vector A : as being a vector whose magnitude is equal to an area (here the area of the loop) and whose direction is normal to the plane of the area (Fig. 23-2c). We then rewrite Eq as the scalar (or dot) product of the velocity vector : v of the airstream and the area vector A : of the loop: va cos : v A :, (23-2) where u is the angle between v and. : A :

8 Flux Now consider a more general situation: the air does not blow directly through the loop, but at some angle θ. Air flow v θ θ A v v (a) (b) (c) If θ = 90, what is the flow rate (flux) through the loop? Zero! (d) Before we discuss a flux involved in electrostatics, we need to rewrite Eq in terms of vectors. To do this, we first define an area vector A : as being a vector the whose loop. magnitude is equal to an area (here the area of the loop) and whose direction is normal to the plane of the area (Fig. 23-2c). We then rewrite Eq as the scalar (or dot) product of the velocity vector : v of the airstream and the area vector A : of the loop: In that case there is no flow through the loop. The air goes around The flux depends on the angle that the flow makes to the loop / va cos : v A :, (23-2) area. where u is the angle between v and. : A :

9 Flux 726 Chapter 24 Gauss s Law A The number of field lines that go through the area A is the same as the number that go through area A. w u A w, Normal u E From th coulomb lines pe If the through the norm that the cross th lar to th A,w. by w areas ar

10 Flux The area A = A cos θ. For other values of θ the flux of air that move through is va cos θ. ow v θ θ A v v (a) We can define flux: (b) (c) (d) Before we discuss a flux involved in electrostatics, we need to rewrite Eq. in terms of vectors. To do this, we first define an area vector A : as being a or whose magnitude is equal to an area (here Φ the = va area cos of the θ loop) and whose ction is normal to the plane of the area (Fig. 23-2c). We then rewrite Eq : e scalar (or dot) product of the velocity vector v of the airstream and the area

11 Electric Flux The electric flux, Φ E, through an area A is Φ E = EA cos θ where θ is the angle between the electric field vector at the surface and the normal vector to the surface. This can be written: Φ E = E A The direction of A is to the surface, and the magnitude is the area of the surface.

12 Questions What are the units of electric flux? (A) N m 2 /C (B) N/C (C) N C 1 m 2 (D) N m 2

13 Questions What are the units of electric flux? (A) N m 2 /C (B) N/C (C) N C 1 m 2 (D) N m 2

14 Questions A surface has the area vector A = (2i + 3j) m 2. What is the flux of a uniform electric field through the area if the field is: E = 4 i N/C? (A) 0 Nm 2 /C (B) 2 Nm 2 /C (C) 4 Nm 2 /C (D) 8 Nm 2 /C

15 Questions A surface has the area vector A = (2i + 3j) m 2. What is the flux of a uniform electric field through the area if the field is: E = 4 i N/C? (A) 0 Nm 2 /C (B) 2 Nm 2 /C (C) 4 Nm 2 /C (D) 8 Nm 2 /C

16 Questions A surface has the area vector A = (2i + 3j) m 2. What is the flux of a uniform electric field through the area if the field is: E = 4 k N/C? (A) 0 Nm 2 /C (B) 2 Nm 2 /C (C) 4 Nm 2 /C (D) 8 Nm 2 /C

17 Questions A surface has the area vector A = (2i + 3j) m 2. What is the flux of a uniform electric field through the area if the field is: E = 4 k N/C? (A) 0 Nm 2 /C (B) 2 Nm 2 /C (C) 4 Nm 2 /C (D) 8 Nm 2 /C

18 The electric field makes an angle Electric Flux u i with the vector A i, defined as being normal to the surface element. What about the electric flux through an arbitrary curved surface? u i and the n the compo the surfac ponent of We assu eral situat nition of f over which into a larg a vector D surface an shown in F The angle θ between the surface normal and the field varies along the surface. E i A i Figure 24.3 A small element of olution: break up surface into small areas A i and add up all the contributions surface area DA i in an electric field. Φ E = E i ( A i ) cos θ i i angle u i wi

19 u i with the vector A i, defined as being normal to the surface To makes this approximation element. exact, take the limit as the areas A i 0. Electric Flux E i u i A i Figure 24.3 A small element of Total flux through surface the area surface: DA i in an electric field. Φ E = E da A the component the surface can t ponent of the el We assumed eral situations, t nition of flux gi over which the f into a large num a vector D A i wh surface and who shown in Figure angle u i with the where we have ( A? B ; AB co gives an approxi The electric flux Φ E through a surface is proportional to the net number of electric field lines passing through that surface.

20 Gaussian urface Gaussian surface An imaginary boundary (closed surface) drawn around some region of space in order to study electric charge and field. The surface can be any shape you like, but must be closed (have an interior and exterior). It is just a tool for calculating charge or field.

21 The word flux comes from the Latin word mean makes sense if we talk about the flow of air volume thro 23-2 can be regarded in a more abstract way.to see this can assign a velocity vector to each point in the airstrea (Fig. 23-2d). Because the composite of all those vectors terpret Eq as giving the flux of the velocity field thr terpretation, flux no longer means the actual flow of so rather it means The the fluxproduct is positive of an area where and the field acro Electric Flux through Gaussian urfaces A Gaussian surface Pierce inward: negative flux θ 1 1 Φ < 0 2 E A Φ = 0 2 E 3 A E θ 3 Φ > 0 kim: zero flux Pierce outward: positive flux the field vector points out of the surface Flux of an Electric Field To define the flux of an electric field, consider Fig (asymmetric) TheGaussian flux is negative surface immersed where in a nonun divide the the surface fieldinto vector small squares pointsof into area A,each to permit us to neglect any curvature and to consider flat. We represent the surface. each such element of area with an a nitude is the area A.Each vector A : is perpendicul and For directed a closed away from surface: the interior of the surface. Because the squares have been taken to be arbitra E : may be taken as constant over any given square. T each square then make some angle u with each ot enlarged view of Φthree E = squares E da on the Gaussian surfac A provisional definition for the flux of the elec surface of Fig is : :. This equation instructs us to visit each square on the Ga : : : :

22 both decrease. (c) The flux increases, and the field x Electric decreases, Flux and the through field increases. Gaussian (e) The urface flux remains Example ld increases. (f) The flux decreases, and the field remains Consider a uniform electric field E = E i in empty space. A cube of edge length l, is placed in the field, oriented as shown. Find the net electric flux through the surface of the cube. y ion in empty d as shown in f the cube. d A 1 d A 3 E t the electric e parallel to e categorize z d A 4 d A 2 x nnumbered Figure 24. (Example 24.1) A closed surface in the shape of a cube in a uniform electric field ori-

23 Electric Flux through Gaussian urface Example Find the net electric flux through the surface of the cube. Φ E = i E ( A i ) cos θ i

24 Electric Flux through Gaussian urface Example Find the net electric flux through the surface of the cube. Φ E = i E ( A i ) cos θ i For sides 3, 4,, and 6, A i E, so Φ E,i = 0. For side 1: For side 2: Φ E,1 = E (l 2 ) cos(180) = El 2 Φ E,2 = E (l 2 ) cos(0) = El 2

25 Electric Flux through Gaussian urface Example Find the net electric flux through the surface of the cube. Φ E = i E ( A i ) cos θ i For sides 3, 4,, and 6, A i E, so Φ E,i = 0. For side 1: For side 2: Φ E,1 = E (l 2 ) cos(180) = El 2 Φ E,2 = E (l 2 ) cos(0) = El 2 In total: Φ E = i Φ E,i = 0

26 ics in attaining this ctric Gauss s field E : of Law the, we considered the Then we simplified dicular Gauss s components law relates the electric field across a closed surface (the flux) to the amount of net charge enclosed by the surface. save far more work matician and physing the fields de : of siders a hypothetical is Gaussian surface, s our calculations of distribution. For exse the sphere with a en, as we discuss in t that? pherical Gaussian surface ian surface to the E field on a Gaussian s a limited example, lly outward from the ediately tells us that Fig A spherical Gaussian surface. If the electric field vectors are of uniform magnitude and point radially outward at all surface points,

27 Gauss s Law The net flux through a surface is directly proportional to the net charge enclosed by the surface. ɛ 0 Φ E = q enc This can also be written: E da = q enc ɛ 0 This is called the integral form of Gauss s Law.

28 Gauss s Law E da = q enc ɛ 0 General definition of divergence of u at point p: 1 u = lim u da V {p} V Differential form of Gauss s Law: E = ρ ɛ 0 where ρ is the charge density.

29 Divergence Differential form of Gauss s Law: E = ρ ɛ 0 where ρ is the charge density. Divergence of a vector field at a point v = [v x, v y, v z ]: v = x v x + y v y + z v z Intuitively, the divergence is a measure of the outgoingness of a vector field at each point.

30 Electrical Permittivity Gauss s Law 1 indicates what the permittivity, ɛ, is: Φ E = q enc ɛ It relates the amount of charge required to generate one unit of electric flux in the vacuum or in a particular medium. 1 assuming the material is homogeneous, isotropic, and linear, and the field is static

31 Electrical Permittivity Gauss s Law 1 indicates what the permittivity, ɛ, is: Φ E = q enc ɛ It relates the amount of charge required to generate one unit of electric flux in the vacuum or in a particular medium. Different materials have different values of ɛ, depending on how they become polarized in response to an electric field. Confusingly, a larger permittivity indicates a larger resistance to an electric field. ɛ = κɛ 0 where κ is the dielectric constant or relative permittivity of the material (see Ch. 26) and ɛ 0 is the vacuum permittivity. 1 assuming the material is homogeneous, isotropic, and linear, and the field is static

32 Gauss s Law Question CHECKPOINT 2 Three Gaussian cubes sit in electric fields. The arrows and the values indicate the directions of the field lines and the magnitudes (in Nm 2 /C) of the flux through the six sides of each cube. (The lighter arrows are for the hidden faces.) The figure shows three situations in which a Gaussian cube sits in an electric field. The arrows and the values indicate the directions of the field lines and the magnitudes (in N m 2 /C) of the flux through the six sides of each cube. (The lighter arrows are for the hidden faces.) In which situation does the cube enclose (a) a positive net charge, (b) a negative net charge, and (c) zero net charge? (1) (2) (3) The cube (1) encloses: (A) positive charge (B) negative charge (C) zero charge

33 Gauss s Law Question CHECKPOINT 2 Three Gaussian cubes sit in electric fields. The arrows and the values indicate the directions of the field lines and the magnitudes (in Nm 2 /C) of the flux through the six sides of each cube. (The lighter arrows are for the hidden faces.) The figure shows three situations in which a Gaussian cube sits in an electric field. The arrows and the values indicate the directions of the field lines and the magnitudes (in N m 2 /C) of the flux through the six sides of each cube. (The lighter arrows are for the hidden faces.) In which situation does the cube enclose (a) a positive net charge, (b) a negative net charge, and (c) zero net charge? (1) (2) (3) The cube (1) encloses: (A) positive charge (B) negative charge (C) zero charge

34 Gauss s Law Question CHECKPOINT 2 Three Gaussian cubes sit in electric fields. The arrows and the values indicate the directions of the field lines and the magnitudes (in Nm 2 /C) of the flux through the six sides of each cube. (The lighter arrows are for the hidden faces.) The figure shows three situations in which a Gaussian cube sits in an electric field. The arrows and the values indicate the directions of the field lines and the magnitudes (in N m 2 /C) of the flux through the six sides of each cube. (The lighter arrows are for the hidden faces.) In which situation does the cube enclose (a) a positive net charge, (b) a negative net charge, and (c) zero net charge? (1) (2) (3) The cube (2) encloses: (A) positive charge (B) negative charge (C) zero charge

35 Gauss s Law Question CHECKPOINT 2 Three Gaussian cubes sit in electric fields. The arrows and the values indicate the directions of the field lines and the magnitudes (in Nm 2 /C) of the flux through the six sides of each cube. (The lighter arrows are for the hidden faces.) The figure shows three situations in which a Gaussian cube sits in an electric field. The arrows and the values indicate the directions of the field lines and the magnitudes (in N m 2 /C) of the flux through the six sides of each cube. (The lighter arrows are for the hidden faces.) In which situation does the cube enclose (a) a positive net charge, (b) a negative net charge, and (c) zero net charge? (1) (2) (3) The cube (2) encloses: (A) positive charge (B) negative charge (C) zero charge

36 Gauss s Law Question CHECKPOINT 2 Three Gaussian cubes sit in electric fields. The arrows and the values indicate the directions of the field lines and the magnitudes (in Nm 2 /C) of the flux through the six sides of each cube. (The lighter arrows are for the hidden faces.) The figure shows three situations in which a Gaussian cube sits in an electric field. The arrows and the values indicate the directions of the field lines and the magnitudes (in N m 2 /C) of the flux through the six sides of each cube. (The lighter arrows are for the hidden faces.) In which situation does the cube enclose (a) a positive net charge, (b) a negative net charge, and (c) zero net charge? (1) (2) (3) The cube (3) encloses: (A) positive charge (B) negative charge (C) zero charge

37 Gauss s Law Question CHECKPOINT 2 Three Gaussian cubes sit in electric fields. The arrows and the values indicate the directions of the field lines and the magnitudes (in Nm 2 /C) of the flux through the six sides of each cube. (The lighter arrows are for the hidden faces.) The figure shows three situations in which a Gaussian cube sits in an electric field. The arrows and the values indicate the directions of the field lines and the magnitudes (in N m 2 /C) of the flux through the six sides of each cube. (The lighter arrows are for the hidden faces.) In which situation does the cube enclose (a) a positive net charge, (b) a negative net charge, and (c) zero net charge? (1) (2) (3) The cube (3) encloses: (A) positive charge (B) negative charge (C) zero charge

38 the flux through the sphere and the magnitude of the elecrface Example: of the sphere? uniform (a) The field flux and field both increase. field both decrease. (c) The flux increases, and the field e flux Return decreases, to this and example: the field increases. (e) The flux remains e field Consider increases. a (f) uniform The flux electric decreases, field E and = Ethe i infield empty remains space. A cube of edge length l, is placed in the field, oriented as shown. Find the net electric flux through the surface of the cube. y irection in empty ented as shown in ace of the cube. da 1 da 3 E that the electric d are parallel to so we categorize z da 4 da 2 x he unnumbered Figure 24. (Example 24.1) A closed surface in the shape of a cube in a uniform electric field ori-

39 t happens to the flux through the sphere and the magnitude of the elecield at the surface of the sphere? (a) The flux and field both increase. he flux and field both decrease. (c) The flux increases, and the field eases. (d) The flux decreases, and the field increases. (e) The flux remains ame, and the field increases. (f) The flux decreases, and the field remains ame. Electric Flux through Gaussian urface Example We found the net electric flux through the surface of the cube: Φ E = i Φ E,i = 0 Cube (,, and the unnumbered e four faces and therefore pery ted in the x direction in empty the field, oriented as shown in ough the surface of the cube. da 1 da 3 E efully. Notice that the electric ndicularly and are parallel to ts definition, so we categorize z da 4 da 2 x gh faces F E 3 E? d A 1 3 E? d A 1 2 Figure 24. (Example 24.1) A closed surface in the shape of a cube in a uniform electric field oriented parallel to the x axis. ide is the bottom of the cube, and side is opposite side. From Gauss s Law ɛ 0 Φ E = q enc, we know: q enc = 0 This is always true for any Gaussian surface in a uniform electric field. ward but d A 1 3 E? d A 3 E 1cos da 2E 3 da 2EA 2E, 2

40 ummary electric flux Gauss s law Quiz tomorrow. Homework Collected homework 1, posted online, due on Monday, Jan 22. erway & Jewett: Ch 24, Obj Qs: 3; Conc. Qs: 1, ; Probs: 3, 7, 17, 21