Topic 7. Electric flux Gauss s Law Divergence of E Application of Gauss Law Curl of E

Transcription

1 Topic 7 Electric flux Gauss s Law Divergence of E Application of Gauss Law Curl of E

2 urface enclosing an electric dipole. urface enclosing charges 2q and q.

3 Electric flux Flux density : The number of field lines per unit area, perpendicular to the field direction (a vector quantity). This quantity is proportional to electric field strength E. Flux (symbol ϕ) : is the total number of field lines passing through a particular area (a scalar quantity). φ E = EA = E A = EAcos θ ( )% N m 2 / C & ' (

4 Area elements We assumed a uniform area and a flat surface area in the previous slide. If we take an infinitesimally small area element, da, the field will be uniform and the area element flat. E da The flux through the area element is dφ E = E da Total flux through the entire surface is φ E = E da

5 A 2 E Gauss s Law r A 1 da q E da = 1 4π % q r ˆr ( ' * r 2 sinθdθdφ ˆr & 2 ) ( ) = 1 q The same number of field lines pass through surface A 2 as through the spherical surface A 1. ince the flux through a surface is proportional to the number of field lines through it, the flux through both surfaces is the same: E da = E da A 2 = q A 1

6 q 1 q 2 Multiple charges q 3 uppose now that we have a bunch of charges scattered around. The total field is the vector sum of all the individual fields: E = n i=1 E i (Principle of superposition) The flux through a surface that encloses them all: E da = For any closed surface, then n ( E i da) = i=1 n i=1 % 1 ( ' q i * & ) E da = 1 Q enc

7 Differential form Apply divergence theorem E da = V ( E) dτ Rewrite Q enc in terms of charge density ρ Q enc = ρdτ V ( E) dτ = V V ' ) ( ρ *, dτ ince this holds for any volume the integrands must be equal E = ρ

8 Example calculation uppose an electric field (in spherical coordinates) in some region is found to be E = kr 3ˆr in spherical coordinates (k is some constant). (a) Find the charged density ρ. ρ = 1 r 2 ρ = E E = 1 ( r 2 r r2 E r ) r r2 kr 3 ( ) = 1 r 2 k 5r 4 ( ) = 5 kr 2 1 rsinθ θ sinθε θ ( ) (b) Find the total charge contained in a sphere of radius R, centered at the origin. 1 Ε φ rsinθ φ Gauss Law ( )( ) = 4π kr 5 Q enc = E da = kr 3 4πR 2 Integration: Q enc = R ρ dτ = 5 kr 2 0 4πr 2 R ( dr) = 20π k r 4 dr = 4π kr 5 0

9 Divergence of E E( r) = 1 4π E = 1 4π where # 1 % $ r - r' 2 allspace ρ ( r' ) r - r' 2 # 1 % $ % r - r' 2 ( r - r' ) r - r' dτ' ( r - r' )& ( r - r' '( ρ ( r' )dτ' ( r - r' )& r - r' ( = 4πδ 3 r - r' ' ( ) E da = $ 1 R ˆr ' & ) % 2 ( E = 1 4πδ 3 ( r - r' ) ρ ( r' )dτ' = 1 ρ r 4π ( Unit vector = r - r' ) r - r' eparation vector = ( r - r' ) Magnitude = r - r' V ( E) dτ ( R 2 sinθdθdφ ˆr ) = 1 r 2 $ 1 ' & r % r2 ) dτ r 2 ( δ( r)dτ = δ( x)δ( y)δ( z) dxdydz = 1 all space Integral form - integrate over volume and apply divergence theorem: 1 Edτ = E da = V V ( ) ρ dτ = 1 Q enc

10 Application of Gauss s Law uppose we have a uniformly charged sphere of radius R possessing a total charge Q. Find the electric field outside the shell. r R A Choose our Gaussian sphere (r > R) so it is concentric with the shell A. E da = 1 Q enc E da = E da E da = E da = E 4πr 2 E 4πr 2 = 1 Q E = 1 Q 4πr 2 r ˆr 2 The field outside is the same as that for a point charge of the same magnitude located at the center!

11 Example. uppose we have a thin spherical shell of radius R possessing a total charge σ that is uniformly distributed over the surface. Find the electric field outside the shell. r R A Choose our Gaussian sphere (r > R) so it is concentric with the shell A. E da = E ( 4πr 2 ) = 1 σ ( 4π R 2 ) dq = σ da = σ R 2 sinθ dθ dφ E = 1 σ ( 4π R 2 ) 4πr 2 ε ˆr = σ R2 0 r 2 ˆr The field outside is the same as if all charge was concentrated at the middle!

12 What is the field inside the shell (r < R)? Q enc = 0 R r E da = E ( 4πr 2 ) = 1 Q enc = 0 What happens to a uniformly charged solid spherical conductor?

13 What is the field inside the sphere? r R A Choose our Gaussian surface of radius r (r < R) so it is concentric with the shell A. Q enc is only a portion of the total charge Q Define charge density as charge per unit volume ρ E = dq dv Charge enclosed in our Gaussian surface is! 4 πr 3 ρ Qenc = # 3 E # 4 " π R 3 ρ 3 E Therefore from Gauss s law: $ & & Q = r3 % R Q 3 E da = E( 4πr 2 ) = Q E ( 4πr 2 ) = Q encl = r3 R Q E = Q 3 4π R r 3 Lets plot this!

14 E Q E = 4π R 2 r=r r Magnitude of the electric field as a function of the distance r from the center of a uniformly charged solid sphere.

15 Example: Cylindrical symmetry A long cylinder carries a charge density that is proportional to the distance from the axis: ρ = ks, for some constant k. Find the electric field inside the cylinder. E Gaussian surface a l E Gauss s Law states : E da = 1 Q enc

16 The enclosed charge is Q enc = ρ dτ Q enc = ( ks) ( sdsdφdz) z s P ẑ ŝ ˆφ Q enc = 2πkl Volume element for cylindrical coordinates 0 a ( s 2 ds) = 2 3 πkla3 x ϕ y ymmetry dictates that E must point radially outward, so for the curved part of the Gaussian surface we have: E da = Eda = E da = E 2πal ince E is perpendicular to da, the two ends contribute nothing. E 2πal = πkla3 E = 1 3 ka 2 ŝ

17 Example: Plane symmetry uppose we want to find the electric field of an infinite plane carrying a uniform surface charge σ. E Draw Gaussian pillbox, extending equal distances above and below the plane. Apply Gauss s Law E da = 1 Q enc E The enclosed charge is Q enc = σ A By symmetry E points away from the plane, with the sides contributing nothing, so that the top and bottom surfaces yield E da = 2A E 2A E = 1 σ A E = σ 2 ˆn

18 ummary Electric field of a sphere falls off as 1/r 2 Electric field of an infinite line as 1/r An infinite plane does not fall off at all

19 Curl of E What is the curl of the field in this figure with a point charge at the center? E = 1 4π q r 2 ˆr uppose we have to calculate the line integral of the field from point a to point b. a b E dl z b In spherical coordinates r b dl = drˆrrdθˆθ rsin dφˆφ q E dl = 1 4π q r 2 dr x a r a y

20 b E dl = 1 a 4π b q r dr = 1 r q b = 1 a 2 4π r ra 4π & q ( - q ' r a r b ) * The integral around a closed path is zero (as then r a = r b ) Applying tokes theorem ( v ) E dl = 0 da = v di P E = 0 From Principle of uperposition E = E 1 E 2 E 3. E = ( E 1 E 2 E 3... ) = 0