Lecture 5. Electric Flux and Flux Density, Gauss Law in Integral Form

Transcription

1 Lecture 5 Electric Flux and Flux ensity, Gauss Law in Integral Form ections: 3.1, 3., 3.3 Homework: ee homework file LECTURE 5 slide 1

2 Faraday s Experiment (1837), Flux 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 flux Ψ=Q, C flux depends on the charge and does not depend on the medium LECTURE 5 slide

3 Flux ensity and Flux 1 charge densities on inner and outer spheres Q Q in ou s = 4 s = Q > 0 πa 4πb flux has direction and density corresponding to the enclosed charge density (flux density points from +Q to Q) Q= ds Ψ= ds s on the surface of the inner sphere Ψ Q Ψ= in r 4 πa r( a) = = = 4 4 s πa πa n on the surface of the outer sphere Ψ Q Ψ= r 4 πb r( b) = = = 4 4 s πb πb in spherical coordinates (charged sphere +Q at origin) n ou LECTURE 5 slide 3 r b n = r a+ Q Q

4 Flux ensity and Flux flux density at any distance r (point charge at origin) Ψ= 1 4 () Q Q r π r r r = () r 4 4 r πr = π a r compare with E of a point charge at origin 1 Q E() r = 4πε a r r E depends on the permittivity, does not: describes the sources regardless of the medium in vacuum = ε 0 E r b n = + Q r Q the principle of superposition applies to as well 1 v( r ) ar r ( ) = dv 4π R v LECTURE 5 slide 4

5 Flux ensity of Fundamental Charge Configurations multiple point charges N N 1 Qn( rn ) 1 Qn( rn ) Er ( ) = a ( ) n r = a n 4πε r r 4π r r n= 1 n n= 1 line charge l 1 E = cosθ1 cosθ 4πε E z ( ) l 1 = 4πε l = cosθ1 cosθ 4π ( sinθ sinθ1) ( ) dz L z 0 x P θ dq = ldz θ α z R y P l 1 z = ( sinθ sinθ1) 4π s infinite sheet of uniform charge: E= an =? ε θ 1 R R 1 n a z a P( φ,,0) 5

6 Total Flux through Closed urface total flux through closed surface depends neither on the surface shape nor on the mutual positioning of the charge and its enclosing surface it depends solely on the enclosed charge analogies: water flow, electrical current, etc. find the mathematical expression for flux so that the above is true Ψ (, ) =? procedure: consider a point charge at the origin and its electric flux through: (a) sphere, (b) any surface if the expression is true for a point charge it will be true for any collection of charges as per the superposition principle LECTURE 5 slide 6

7 Background: olid Angle and 1 teradian 1 radian of a angle (φ = 1 rad) cuts out an arc from a circle of length equal to the radius of the circle larc = r ϕ 1 steradian of a solid angle (Ω = 1 sr) cuts out an area from a sphere equal to (radius) ssph = r Ω if Ω= 1, ssph = r if Ω= 4 π, s = 4πr sph if ϕ = 1, larc = r if ϕ = π, l = πr arc full circle full sphere LECTURE 5 slide 7

8 ifferential olid Angle and ifferential urface Area differential solid angle and differential surface area on a sphere ds = r dω in C: ds = r sinθdθdφ dω example: the area and the solid angle of a full sphere (radius is r 0 ) ππ π sph = = 0 sinθ θ φ = 0 π ( cos θ) 0 = 4 π 0, m A ds r d d r r sph phere 0 0 ππ Ω = dω= sinθdθdφ = 4 π, sr phere 0 0 solid angle of full 3 space LECTURE 5 slide 8

9 True or False The solid angle defined by 0 θ π / and 0 φ π is equal to π. The solid angle defined by a rectangular trihedral corner is equal to π/. LECTURE 5 slide 9

10 Total and ifferential Flux of Point Charge flux through a sphere (charge at origin) ππ ds Ψ= nds = r r sinθdθdφ = Q sphere 0 0 dψ due to spherical symmetry Q r () r = 4π r dψ= r dω Q dψ= dω, dω= sinθdθdφ 4π r d r Q 4π r 1 dω dψ the differential flux dψ is proportional to (QdΩ) and does not depend on r LECTURE 5 slide 10

11 ifferential Flux of Point Charge flux through arbitrary surface (charge is at origin) differential flux through surface element of an arbitrary surface ds = ds α sphere other cos 0 dω a n a r α we want dψ sphere = dψ other = dψ QdΩ dψ sphere = rdssphere = r cosαdsother = ndsother = ds ds sphere general expression for differential flux d Ψ= ds ds sphere ds other LECTURE 5 slide 11

12 Total Flux of Point Charge Through Arbitrary urface regardless of the chosen surface Ψ= d s = Q as per superposition principle, result holds for any collection of charges Q dω dω P P Q Q 1 Q= Q1+ Q Ψ=Ψ +Ψ 1 P sphere other single charge at origin 1 other any two charges 1

13 Gauss Law the electric flux over a closed surface is equal to the total charge enclosed by the surface Ψ= d s = Q = vdv v flux equals enclosed charge (no enclosed charge means no flux) Q a n P d Ψ< 0 dψ = dψ QdΩ dω ( P ) total flux Ψ= 0 a n P d Ψ > 0 ( P ) LECTURE 5 slide 13

14 Gauss Law: Applications Gauss law makes solutions to problems with planar, cylindrical or spherical symmetry easy procedure: choose integration surface so that is everywhere either normal or tangential to surface if normal: ds= ds if tangential: ds= 0 when normal to surface, is also constant on surface ds= ds = LECTURE 5 slide 14

15 Gauss Law Applications: Field of Infinite Line Charge π d s = Q= l due to symmetry = a l φ= 0z= 0 l dφdz = πl = l = a E= = a ε l, C/m π l, V/m πε This result was already obtained in Lecture 4 by the superposition principle. l l LECTURE 5 slide 15 x z φ ds y = dφdza l

16 Gauss Law Applications: Field of Coaxial Cable problem has cylindrical symmetry Gaussian surface chosen as cylinder of radius solution analogous to that of line charge for a b l = a, C/m π for > b, = 0 Homework: Prove that for a uniformly charged cylinder v of radius a, =, for 0 a. LECTURE 5 slide 16

17 d s = Q= A top z s dxdy dxdy + d = A s z z s top bottom sides Gauss Law Applications: Field of heet Charge A bottom z A z y x s flux through is 0 top bottom s due to symmetry: z = z = A = sa, = s s =± az E= =± az ε ε This result was already obtained in Lecture 4 by the superposition principle. l y A= ll bottom top x y a n l x a n LECTURE 5 slide 17 δ

18 Gauss Law Applications: Field of pherical Charge A sphere of radius a has uniformly distributed charge of volume density v, C/m 3. etermine the electric flux density in the sphere and out of it. spherical symmetry of the source implies spherical symmetry of field choose integration surface as sphere (1) inside the sphere = () = v = v r ( ) vr ( ) vr ( ) d Q r dv dv 4 3 r()4 r πr = v πr v r () r = r 3 3 vr ( ) LECTURE 5 slide 18

19 Gauss Law Applications: Field of pherical Charge () outside the sphere 4 d s = Q = vdv = v πa 3 ( r) 4 r()4 r πr = Q = πa 3 v 3 v 3 v a Q r () r = = 4π r 3 r 3 v a 3 0 r a 1/ R R outside the sphere the field is the same as that of a point charge LECTURE 5 slide 19

20 Gauss Law Applications: Field of pherical Charge E-field magnitude in 3- space LECTURE 5 slide 0

21 You have learned about: the flux density vector and how it relates to the charge Q and the E vector Gauss law of electrostatics in integral form Ψ= d s = Q = vdv 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 5 slide 1