Lecture 9 Electric Flux and Its Density Gauss Law in Integral Form

Transcription

1 Lecture 9 Electric Flux and Its Density Gauss Law in Integral Form ections: 3.1, 3.2, 3.3 Homework: ee homework file

2 Faraday s Experiment (1837), Electric Flux ΨΨ charge transfer from inner to outer sphere electric induction: charge deposition without contact observations on charge on outer electrode it is of the same magnitude but opposite sign ( Q) as that on inner electrode it is the same regardless of the insulating material used (ε r ) it is the same regardless of electrode s shape b a E +Q insulator ε r Q A displacement flux electric charge flux Ψ=Q, C flux depends on enclosed charge but does not depend on medium LECTURE 9 slide 2

3 Flux Density D the E field vector depends on the medium s permittivity 1 Q point charge: E( r) = a 4 2 r πε r What vector field would be independent of the permittivity? 1 Q Ψ point charge at origin: D( r) = εe( r) = a 4 2 r = a π r 4πr2 D is the density of Ψ (units C/m 2 ) 2 4π r D() r Q total flux Ψ through fictitious surface r a r s sph area of fictitious spherical surface surrounding point charge at origin LECTURE 9 slide 3 r

4 Flux Density D 2 in vacuum D= ε 0 E the principle of superposition applies to D as well volume charge distribution: Dr ( ) 1 ρv( r ) a 4π R2 = v R dv LECTURE 9 slide 4

5 TRUE OR FALE? 1. The flux density of a uniform infinitely long line charge is 1 ρl D( ρ) = aρ 2π ρ 2. The flux density of an infinitely sheet of uniform charge is D= a n ρ s LECTURE 9 slide 5

6 Total Flux ΨΨ and Flux Density D total flux is associated always with a closed surface point charge: Ψ= Q= 4 r Dr π 2 area of closed sphere What is the general relation between Ψ and D such that for a given D the result for Ψ does not depend on the choice of closed surface? FLUX INTEGRAL Ψ= d s D [campfirephilosopher.ca] LECTURE 9 slide 6

7 Total Flux ΨΨ and Flux Density D example Verify that the flux integral Ψ= d s D results in Ψ = Q in the case of a point charge where Q D() r = a 4 2 r π r LECTURE 9 slide 7

8 Partial Flux ΨΨ: Example 1 partial flux is the flux through an open surface A 60-μC point charge is located at the origin. Find the partial flux through the surface defined by: (a) r 0 = 26 cm, 0 θ 90, 0 ϕ 90. (b) z 0 = 26 cm. LECTURE 9 slide 8

9 Partial Flux ΨΨ: Example 2 A linear charge ρ l = 2 μc/m lies in the y-z plane as shown in figure. Find the partial flux passing through the planar surface extending from 0 to 1 m along x and from to + along y. [Bakr, He, Matlab Experiments Manual for EE2FH3] LECTURE 9 slide 9

10 Partial Flux ΨΨ: Example 2, cont. LECTURE 9 slide 10

11 Gauss Law the electric flux over a closed surface is equal to the total charge enclosed by the surface Ψ= D d s = Qenclosed = ρvdv V flux equals enclosed charge no enclosed charge means no flux! LECTURE 9 slide 11

12 Is the Total Flux Zero or Not? #1 #3 Ψ=0? #2 #4 LECTURE 6 slide 12

13 Gauss Law: Applications Gauss law makes solutions to problems with planar, cylindrical or spherical symmetry easy procedure: choose integration surface so that D is everywhere either normal or tangential to surface if normal: D ds= Dds if tangential: D ds= 0 when normal to surface, D is also constant on surface D ds= D ds = D LECTURE 9 slide 13

14 D Gauss Law Applications: Field of Infinite Line Charge 2π d s = Q= ρ l due to symmetry D= D ρ a ρ l φ= 0z= 0 l D ρdφdz = D 2πρl = ρ l D= a ρ ρ D E= = a ε ρl, C/m 2πρ ρ 2 ρ ρl, V/m 2περ This result was already obtained in Lecture 8 by the superposition principle. l l LECTURE 9 slide 14 x z φ ds ρ y = ρdφdza ρ ρ l

15 Flux of Infinite Line Charge: Example Calculate the flux density D of a line charge of uniform density ρ l = 10 6 C/m along the z-axis at the points P(3,4,10) mm and Q(3,4, 4) mm. What is the total flux Ψ through a cylinder of height h = 10 m? LECTURE 9 slide 15

16 Gauss Law Applications: Field of Coaxial Cable problem has cylindrical symmetry Gaussian surface chosen as cylinder of radius ρ solution analogous to that of line charge for a ρ b ρl 2 D= aρ, C/m 2πρ ρ for ρ > b, D = 0 for 0 ρ a ρv D ρ = ρ 2 (proof is part of your homework) LECTURE 9 slide 16

17 D d s = Q= ρ A top z s D dxdy D dxdy + d = ρ A D s z z s top bottom sides D Gauss Law Applications: Field of heet Charge A D bottom z A z y x ρ s flux through is 0 top bottom s due to symmetry: Dz = Dz = D 2 DA = ρsa, D = ρ 2 ρs D ρs D=± az E= =± az 2 ε 2ε This result was already obtained in Lecture 8 by the superposition principle. l y A= ll D bottom D top x y a n l x a n LECTURE 9 slide 17 δ

18 Gauss Law Applications: Field of pherical Charge A sphere of radius a has uniformly distributed charge of volume density ρ v, C/m 3. Determine the electric flux density inside and outside the sphere. spherical symmetry of the source implies spherical symmetry of field choose integration surface as sphere (1) inside the sphere D = () = ρv = ρ v r ( ) vr ( ) vr ( ) d Q r dv dv Dr()4 r πr = ρv πr ρv Dr () r = r 3 3 vr ( ) LECTURE 9 slide 18

19 Gauss Law Applications: Field of pherical Charge (2) outside the sphere 4 D d s = Q = ρvdv = ρv πa 3 ( r) 4 Dr()4 r πr = Q = ρ πa 3 v 2 3 v 3 ρv a 2 2 Q Dr () r = = 4π r 3 r 3 ρ v a 3 0 D r a 2 1/ R R outside the sphere the field is the same as that of a point charge LECTURE 9 slide 19

20 You have learned about: the flux density vector D and how it relates to the electric flux Ψ, the charge Q and the E vector Gauss law of electrostatics in integral form Ψ= D d s = Q = ρvdv the application of Gauss law to the solution of symmetrical problems infinite planar charge infinite line charge infinite cylinder (inside and outside) coaxial cable uniformly charged sphere (inside and outside) v LECTURE 9 slide 20