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Gauss law 1
Gauss' law relates the electric flux through a closed surface to the enclosed charge. It can be written in both integral and differential forms. The integral form states that the total flux is equal to the enclosed charge divided by the permittivity of free space. The differential form is Poisson's equation, which relates the divergence of the electric field to the charge density. Gauss' law can be applied to problems involving point charges, charge sheets, and continuous charge distributions. The electrostatic potential and electric field can be derived from each other using calculus operations. The potential energy of a system of charges can be expressed in terms of either the potentials or the electric fields.
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Gauss law 1
- 1. 2). Gauss’ Lawand Applications • Coulomb’s Law: force on charge i due to charge j is • Fij is force on i due to presence of j and acts along line of centres rij. If qi qj are same sign then repulsive force is in direction shown • Inverse square law of force ( ) ˆ ˆ ji ji ijjiijjiij ij2 ij ji o ji3 ji ji o ij r r qq 4 1qq 4 1 rr rr rrrrrr rrr rr F − − =−=−= =− − = πεπε O ri rj ri-rj qi qj Fij
- 2. Principle of Superposition •Total force on one charge i is • i.e. linear superposition of forces due to all other charges • Test charge: one which does not influence other ‘real charges’ – samples the electric field, potential • Electric field experienced by a test charge qi ar ri is ∑≠ = ij ij2 ij j o ii r q 4 1 q rF ˆ πε ( ) ∑≠ == ij ij2 ij j oi i ii r q 4 1 q r F rE ˆ πε
- 3. Electric Field • Fieldlines give local direction of field • Field around positive charge directed away from charge • Field around negative charge directed towards charge • Principle of superposition used for field due to a dipole (+ve –ve charge combination). Which is which? qj +ve qj -ve
- 4. Flux of aVector Field • Normal component of vector field transports fluid across element of surface area • Define surface area element as dS = da1 x da2 • Magnitude of normal component of vector field V is V.dS = |V||dS| cos(Ψ) • For current density j flux through surface S is Cm2 s-1 da1 da2 dS dS = da1 x da2 |dS| = |da1| |da2|sin(π/2) Ψ dS` ∫ Ssurfaceclosed .dSj
- 5. • Electric fieldis vector field (c.f. fluid velocity x density) • Element of flux of electric field over closed surface E.dS da1 da2 n θ φ Flux of Electric Field ϕ ϕ ϕϕ ˆˆˆ ˆ ˆ ˆ θn naaS a θa x ddθsinθrdxdd dsinθrd dθrd 2 21 2 1 = == = = o oo 2 2 o q .d d 4 q ddθsinθ 4 q 1ddθsinθr. r4 q .d ε πε ϕ πε ϕ πε ∫ = Ω== == S SE n.rn r SE ˆˆˆ ˆ Gauss’ Law Integral Form
- 6. • Factors ofr2 (area element) and 1/r2 (inverse square law) cancel in element of flux E.dS • E.dS depends only on solid angle dΩ da1 da2 n θ φ Integral form of Gauss’ Law o i i o 21 q .d d 4 qq .d ε πε ∑ ∫ = Ω + = S SE SE Point charges: qi enclosed by S q1 q2 vwithinchargetotal)d( )dv( .d V o V = = ∫ ∫ ∫ vr r SE ρ ε ρ S Charge distribution ρ(r) enclosed by S
- 7. Differential form ofGauss’ Law • Integral form • Divergence theorem applied to field V, volume v bounded by surface S • Divergence theorem applied to electric field E ∫∫∫ ∇== V SS dv.ddS. VSV.nV V.n dS .V dv o V )d( .d ε ρ∫ ∫ = rr SE S ∫∫ ∫∫ =∇ ∇= VV V )dv( 1 dv. dv.d rE ESE. ρ εo S oε ρ )( )(. r rE =∇ Differential form of Gauss’ Law (Poisson’s Equation)
- 8. Apply Gauss’ Lawto charge sheet • ρ (C m-3 ) is the 3D charge density, many applications make use of the 2D density σ (C m-2 ): • Uniform sheet of charge density σ = Q/A • By symmetry, E is perp. to sheet • Same everywhere, outwards on both sides • Surface: cylinder sides + faces • perp. to sheet, end faces of area dA • Only end faces contribute to integral + + + + + + + + + + + + + + + + + + + + + + + + E EdA ooo ε σ ε σ ε 2 =⇒=⇒=∫ ESE. S .dA E.2dA Q d encl
- 9. • σ’ =Q/2A surface charge density Cm-2 (c.f. Q/A for sheet) • E 2dA = σ’ dA/εo • E = σ’/2εo (outside left surface shown) Apply Gauss’ Law to charged plate ++++++ ++++++ ++++++ ++++++ E dA • E = 0 (inside metal plate) • why?? ++++ ++++ • Outside E = σ’/2εo + σ’/2εo = σ’/εo = σ/2εo • Inside fields from opposite faces cancel
- 10. Work of movingcharge in E field • FCoulomb=qE • Work done on test charge dW • dW = Fapplied.dl = -FCoulomb.dl = -qE.dl = -qEdl cos θ • dl cos θ = dr A B q1 q r r1 r2 E dl θ ∫ ∫ −= −−= −= −= B A 21o 1 r r 2 o 1 2 o 1 .dq r 1 r 1 4 q q dr r 1 4 q qW dr r 1 4 q qdW 2 1 lE πε πε πε 0=∫ pathclosedany lE.d
- 11. Potential energy function •Path independence of W leads to potential and potential energy functions • Introduce electrostatic potential • Work done on going from A to B = electrostatic potential energy difference • Zero of potential energy is arbitrary – choose φ(r→∞) as zero of energy r 1 4 q )( o 1 πε φ =r ( ) ∫−= == B A BA .dq )(-)(q)PE(-)PE(W lE ABAB φφ
- 12. Electrostatic potential • Workdone on test charge moving from A to B when charge q1 is at the origin • Change in potential due to charge q1 a distance of rB from B ( ) Bo 1 r 2 o 1 B r 1 4 q )( dr r 1 4 q .d -)()(-)( B πε φ πε φφφ = −= −= =∞→ ∫ ∫ ∞ ∞ B lE BAB 0 ( ))(-)(q)PE(-)PE(WBA ABAB φφ==
- 13. Electric field fromelectrostatic potential • Electric field created by q1 at r = rB • Electric potential created by q1 at rB • Gradient of electric potential • Electric field is therefore E= – φ 3 o 1 r4 q r E πε = r 1 4 q r o 1 B πε φ =)( 3 o 1 B r4 q r r πε φ −=∇ )(
- 14. Electrostatic energy ofcharges In vacuum • Potential energy of a pair of point charges • Potential energy of a group of point charges • Potential energy of a charge distribution In a dielectric (later) • Potential energy of free charges
- 15. Electrostatic energy ofpoint charges • Work to bring charge q2 to r2 from ∞ when q1 is at r1 W2 = q2 φ2 • NB q2 φ2 =q1 φ1 (Could equally well bring charge q1 from ∞) • Work to bring charge q3 to r3 from ∞ when q1 is at r1 and q2 is at r2 W3 = q3 φ3 • Total potential energy of 3 charges = W2 + W3 • In general O q1 q2 r1 r2 r12 12o 1 2 r 1q πε ϕ 4 = O q1 q2 r1 r2 r12 r3 r13 r23 23o 2 13o 1 3 r 1q r 1q πεπε ϕ 44 += ∑ ∑∑ ∑ ≠< == ji j ij j i ji j ij j i r q q 1 2 1 r q q 1 W oo πεπε 44
- 16. Electrostatic energy ofcharge distribution • For a continuous distribution ∫∫ ∫ ∫ − = − = = spaceallspaceallo spaceallo spaceall )( d)(d 4 1 2 1 W )( d 4 1 )( )()(d 2 1 W r'r r' r'rr r'r r' r'r rrr ρ ρ πε ρ πε φ φρ
- 17. Energy in vacuumin terms of E • Gauss’ law relates ρ to electric field and potential • Replace ρ in energy expression using Gauss’ law • Expand integrand using identity: ∇.ψF = ψ∇.F + F.∇ψ Exercise: write ψ = φ and F = ∇φ to show: ∫∫ ∇−==∴ ∇−=⇒−=∇⇒ −∇==∇ v 2o v 2 o o 2 o dv 2 dv 2 1W and. φφ ε ρφ φερ ε ρ φ φ ε ρ EE ( ) ( )22 22 . . φφφφφ φφφφφ ∇−∇∇=∇⇒ ∇+∇=∇∇
- 18. Energy in vacuumin terms of E For pair of point charges, contribution of surface term 1/r -1/r2 dA r2 overall -1/r Let r → ∞ and only the volume term is non-zero Energy density ( ) ( ) ( ) ( ) theorem)e(DivergencintegralvolumereplacesintegralSurface identityfirstsGreen'dv.d 2 dvdv. 2 W v 2o v 2 v o ∇−∇−= ∇−∇∇−= ∫∫ ∫∫ φφφ ε φφφ ε S S ( ) ∫∫ =∇= spaceall 2o spaceall 2o dvE 2 dv 2 W ε φ ε )(E 2dv dW )( 2o E rr ε ρ ==