Gauss s law - plane symmetry

Transcription

1 Gauss s aw - pane symmetry Submitted by: I.D Find the eectric fied aong the z-axis of an infinite uniformey charged pane at the x y pane (charge density σ) with a hoe at the origin of a radius r. Using the principe of superposition we sha cacuate the eectric fied induced by the missing hoe as being of a disk charged with σ, then combine it with an infinite pane: ( ) z Ez disk = 2πkσ sign(z) () z 2 + r 2 Ez pane = 2πkσ sign(z) (2) z E z = 2πkσ (3) z 2 + r 2

2 Gauss s aw - cyindrica symmetry Gien an infinite cyinder charged with ρ( r) = ρ. find the eectric fied. 2. What shoud be the charge density ρ( r), so that the eectric fied inside the cyinder woud be zero?. For cyindrica symmetry E = E rˆr. The oume eement is d = rdrdθdz. For r < R r E r 2πrh = 4πh ρd = 4πh ρ2πr dr h () V E = 2πkρrˆr (2) For r > R E r 2πrh = 4πh V E = 2πkρR2 ˆr r ρd = 4πh R ρ2πr dr h (3) 2. We soe the probem in two ways - using integra and diferentia ersions of the Gauss s aw: a. integra: E ds = 4πk ρd (5) E 2πrh = 4πk r ρ(r ) 2πr dr h (6) Checking: r b. differentia: ρ(r) = A r ρ(r ) 2πr dr h = 2πArh (8) di E = 4πkρ E = 4πkAˆr = const (9) E = const die = E () r ρ = A r E r = 4πkA (2) r E = 4πkAˆr = const (3) (4) (7) ()

3 Gauss aw - cyindrica symmetry Submitted by: I.D In an infinite cyinder of a radius R charged uniformy with a charge density ρ there is a cyindrica infinite hoe of a radius r which center positioned at b. Find the eectric fied. We use the principe of superposition: the probem is equa to a fu cyinder of a radius R charged with ρ pus a fu cyinder of a radius r charged with ρ.. outside the cyinder E = 2πkρR2 r 2 r + 2πkρr2 r b 2 ( r b) () 2. inside the cyinder, outside the hoe E = 2πkρ r + 2πkρr2 r b 2 ( r b) (2) 3. inside the hoe E = 2πkρ r 2πkρ( r b) = 2πkρ b = const! (3)

4 Gauss aw - spherica symmetry Submitted by: I.D Find the eectric fied of a point charge 2. Find the eectric fied from a spherica she of a radius R charged with a charge Q uniformy. For a point source the charge density is ρ( r) = Qδ( r). Then E d s = 4πk ρd () s V E r 4πr 2 = 4πk Qδ( r)dv = 4πkQ (2) E r = kq r 2 (3) For a spherica she the charge density is ρ( r) = Qδ( r R) where R = Rˆr. V ρdv = Qδ( r R)dV = {, r < R, r > R (4) Then E = {, r < R kq ˆr, r 2 r > R (5)

5 Eectric fied Submitted by: I.D Cacuate directy the eectric fied fux through an infinite pane a distance d from a point charge. Expain why your resut does not depend on d. Let the pane to be at the x y pane, so that any point on it is r = (r cos(ϕ), r sin(ϕ), ) and the point charge is at r = (,, d). By the Cooumb aw d E = kdq( r r ) r r 3 () In our case So that r r = (r cos(ϕ), r sin(ϕ), d) (2) r r = (r 2 + d 2 ) /2 (3) E = The fus is E da = kq (r 2 + d 2 (r cos(ϕ), r sin(ϕ), d) (4) ) 3/2 = 2π E z da dϕ = 2πkqd kqd rdr (6) (r 2 + d 2 ) 3 2 rdr = 2πkq (7) (r 2 + d 2 ) 3 2 The tota fux from a point charge if 4πkq. Since the pain is infinite, haf of the eectric fied ines pass through this pane (not depending on d), that is haf of the fux 2πkq. (5)

6 Eectric fied Submitted by: I.D An infinite ayer of width 2 ( < z < ) is charged non-uniformy, such that the charge density depends on the height of the ayer ony: { ρ(z), < z < ρ( r) = (), otherwise. Expain how to cacuate in genera the eectric fied inside the ayer for any charge density ρ (z) using the Gauss aw. 2. The charge density inside the ayer is: ρ (z) = Q (e z 3 + e ). Cacuate the eectric fied. 3. What is the charge per unit area σ of the box, for the aboe charge density? Write the eectric fied outside the box in terms of σ.. The Gauss aw is E d s = 4πk ρ(z)dv The charge density is ρ = ρ(z) for < z < and ρ = for z >. Using the symmetry considerations we deduce that E = E(z)ẑ. Since we can choose the direction of the z-axis arbitrariy, the boundary conditions are E z ( ) = E z ( ). Choosing the Gaussian she as a cyinder which axis is parae to z and imited by z < z < z 2 we get z2 E d s = (E z (z 2 ) E z (z ))S = 4πkS ρ(z )dz (3) z where S is the area of the cyinder base. Therefore, E z (z 2 ) E z (z ) = 4πk z2 Using the boundary conditions in the infinity we get E z ( ) E z ( ) = 2E z ( ) = 4πk z ρ(z )dz (4) ρ(z )dz = 4πk Substituting into Eq. (4) z 2 = z, z = and E z ( ) from Eq. (5) we obtain E z (z) = 2πk ρ(z )dz + 4πk z (2) ρ(z )dz (5) ρ(z )dz (6) 2. Let us use Eq. (6) for the cacuation with the gien charge density: E z (z) = 2πk Q z 3 (e + e )dz + 4πk z Q z 3 (e + e )dz (7)

7 For < z < E z (z) = 2πk Q ( ) 3 2 e z + e z ) + 4πk Q ( ) z 3 e z + e z ) = 4πk Q 2 + 4πk Q ) (e 3 z e z = 4πk Q ( ) 3 + e z e z For < z < For z > Finay: E z (z) = 2πk Q 3 2 ( e z + e z ) + 4πk Q 3 [ ( e z + e z ) = 4πk Q 3 ( e z e z E z (z) = 2πk Q 3 2 ( e z + e z ) E z (z) = { 4πkQ 3 4πkQ ) ) ( + e z + e z ) ) ) ( e z + z e ) sign(z), z < ) z ] (8) (9) () () (2) = 4πk Q 2 (3) sign(z), z > 2 Another way to find the eectric fied for this particuar charge density is to expoit the fact that this charge density is an een function, i.e. ρ(z) = ρ( z). In this one can construct the Gaussian cyindrica she inside the ayer with its tops at z, z. See the appendix. 3. The surface charge density is σ = ρ(z )dz = 2Q 2 Then the eectric fied outside the ayer is E z (z) = 2πkσ sign(z) (6) as expected. APPENDIX The Gauss aw: E ds = 4πk Cacuating: E ds = 2E (z) s d = s dz (4) (5) ρ (z )d (7) ρ (z ) = Q z 3 (e + e ) (2) ρ (z )d = s = sq 3 z Q z 3 (e [ z (8) (9) z + e )dz (2) (e z + e )dz + z (e z + e )dz ] = sq 3 [ (e z + z e ) z + (e z + z e ) z = 2sQ 3 ( e z + ze ) (24) 2 ] (22) (23)

8 Then 2E (z) s = 8πkQ s 3 ( e z + ze ) (25) Using the fact that E (z) = E ( z) we get: E (z) = sign(z) 4πkQ 3 ( e z + z e )ẑ, < z < (26) Outside of the ayer: z > or z <. We take another simiar cyinder ony this time z > : E ds = 4πk ρ (z )d (27) E ds = 2E (z) s (28) d = s dz ρ (z ) = Q z 3 (e + e ) (3) z [ z ] ρ (z )d = s ρ (z )dz = s ρ (z )dz + ρ (z )dz + ρ (z )dz (3) z z Because ρ = for z >, z < : Then Finay z ρ (z )dz = z ρ (z )d = s (29) ρ (z )dz = (32) = sq 3 [ ρ (z )dz = s (e z + e )dz + Q 3 (e z + e )dz (33) (e z + e )dz ] = sq [ 3 (e z + z e ) z + (e + z e ) 2E (z) s = 8πkQs 2 ] (34) = 2sQ 2 (35) E (z) = sign(z) 4πkQ 2 ẑ, z >, z < (37) (36) 3