Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Coulomb’s Law • Coulomb’s Law: force on charge i due to charge j is qiq j 1 qiq j ˆ 1 (r − rj ) = Fij = r 3 i 2 ij 4πε o rij 4πε o r − r i j rij = ri − rj rij = ri − rj rˆij = ri − rj ri − rj • 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 ri direction shown • Inverse square law of force O Fij qi ri-rj rj qj Principle of Superposition • Total force on one charge i is Fi = qi ∑ j≠i 1 qj 4πε o rij 2 rˆij • 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 Fi 1 qj ˆ Ei (ri ) = = ∑ r 2 ij qi j≠i 4πε o rij Electric Field • Field lines give local direction of field qj +ve • 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 Flux of a Vector 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) da2 • For current density j flux through surface S is ∫ j.dS Cs-1 closed surface S dS da1 Ψ dS` dS = da1 x da2 |dS| = |da1| |da2|sin(π/2) Flux of Electric Field • Electric field is vector field (c.f. fluid velocity x density) • Element of flux of electric field over closed surface E.dS da1 = r dθ θˆ n da 2 = r sinθ dϕ ϕˆ da2 θ φ da1 dS = da1 x da 2 = r 2 sinθ dθ dϕ nˆ nˆ = θˆ x ϕˆ q rˆ 2 . r sinθ dθ dϕ nˆ rˆ.nˆ = 1 E.dS = 2 4πε o r = q 4πε o q ∫ E.dS = ε S o sinθ dθ dϕ = q 4πε o dΩ Gauss’ Law Integral Form Integral form of Gauss’ Law • Factors of r2 (area element) and 1/r2 (inverse square law) cancel in element of flux E.dS q1 + q2 E.dS = dΩ • E.dS depends only on solid angle dΩ 4πε o ∫ E.dS = n da2 da1 θ q1 φ q2 S ∑q i i εo Point charges: qi enclosed by S ∫ ρ (r )dv V E .d S = ∫ S εo ∫ ρ (r )dv = total charge within v V Charge distribution ρ(r) enclosed by S Differential form of Gauss’ Law • Integral form ∫ E.dS = ∫ ρ (r )dr V S εo • Divergence theorem applied to field V, volume v bounded by surface S ∫ V.n dS = ∫ V.dS = ∫ ∇.V dv S S V V.n dS “.V dv • Divergence theorem applied to electric field E ∫ E.dS = ∫ ∇.E dv V S 1 ∫ ∇.E dv = ε ∫ V o V ρ (r )dv ∇.E(r ) = ρ (r ) εo Differential form of Gauss’ Law (Poisson’s Equation) Apply Gauss’ Law to 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 dA E By symmetry, E is perpendicular 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.dS = S Q encl εo ⇒ E.2dA = σ .dA σ ⇒E = εo 2ε o E Apply Gauss’ Law to charged plate σ’ = Q/2A surface charge density Cm-2 (c.f. Q/A for sheet) E 2dA = 2σ’ dA/εo (surface encloses whole plate) E = σ’/εo (outside left surface shown) E = 0 (inside metal plate) why?? E • Outside E = σ’/2εo + σ’/2εo = σ’/εo = σ/2εo • Inside fields from opposite faces cancel + + + + + + + + dA + + + + + + + + + + + + + + + + + + + + + + + + • • • • • Work of moving charge 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 dW = −q W = −q q1 1 dr 4πε o r 2 q1 4πε o = −q ∫ r1 1 dr r2 q1 ⎛ 1 1 ⎞ ⎜ − ⎟ 4πε o ⎝ r1 r2 ⎠ B = −q∫ E.dl A • r2 B E r2 θ q A r r1 dℓ ∫ E.dl = 0 any closed path q1 W is independent of the path (electrostatic E field is conservative) Potential energy function • Path independence of W leads to potential and potential energy functions • Introduce electrostatic potential φ (r ) = q1 1 4πε o r • Work done on going from A to B = electrostatic potential energy difference WBA = PE(B) - PE( A ) = q(φ (B) - φ ( A )) B • Zero of potential energy is arbitrary – choose φ(r→∞) as zero of energy = −q∫ E.dl A Electric field from electrostatic potential q1 • 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 r E= 4πε o r 3 q1 1 φ (rB ) = 4πε o r q1 r ∇φ (rB ) = − 4πε o r 3 Electrostatic potential of point dipole • • +/- charges, equal magnitude, q, separation a axially symmetric potential (z axis) 2 z q+ ⎛a⎞ 2 2 r± = r + ⎜ ⎟ m a r cosθ ⎝2⎠ r+ r a/2 θ p a/2 q- ϕ (r ) = q ⎛ 1 1⎞ ⎜ − ⎟ 4πε o ⎝ r+ r- ⎠ 2 ⎛ ⎞ a a ⎞ ⎛ 2⎜ = r 1+ ⎜ ⎟ m cosθ ⎟ ⎜ ⎝ 2r ⎠ ⎟ r ⎝ ⎠ rx ⎞ 1 1 ⎛⎜ ⎛ a ⎞ a = 1+ ⎜ ⎟ m cosθ ⎟ ⎟ r± r ⎜⎝ ⎝ 2r ⎠ r ⎠ 1 a = ± 2 cosθ r 2r qa cosθ p cosθ ϕ (r ) = = 2 4πε o r 4πε o r 2 2 −1 2 Electrostatic potential of point dipole • • Equipotential lines for an electric dipole || z axis Contours on which electric potential is constant 1 0.5 0 -0.5 -1 -2 -1 ϕ (r ) = p cosθ 4πε o r 2 0 1 2 • Equipotential lines perp. to field lines Electrostatic potential of point dipole • Field lines for a point dipole || z axis • Generated from E = -∇φ r k φ θ θ j i ⎛ ∂ ∂ ∂ ⎞ (i, j, k ) ∇ Cart. = ⎜⎜ , , ⎟⎟ ⎝ ∂x ∂y ∂z ⎠ ⎛∂ 1 ∂ ∂ ⎞ ˆ ˆ ˆ 1 ⎟⎟ (r,θ ,φ ) ∇ Sph.Pol. = ⎜⎜ , , ⎝ ∂r r ∂θ r sinθ ∂φ ⎠ φ ϕ (r ) = p 4πε o cos θ r2 E(r,θ ) = −∇ϕ (r,θ ) = p ⎛ 2cos θ sin θ ⎞ , 3 , 0⎟ ⎜ 3 4πε o ⎝ r r ⎠ ← NB diagram not a point dipole Electrostatic potential of point dipole • More generally, dipole direction given by p vector ϕ (r ) = 1 4πε or 3 p.r p = (p x , p y , p z ) r = (x, y, z) ∂ ⎛ p.r ⎞ ⎛ 1 3x 2 ⎞ 3xz 3xy ⎜ ⎟ p p pz = − − − ⎜ 3 ⎟ ⎜ 3 y 5 ⎟ x 5 5 ∂x ⎝ r ⎠ ⎝ r r r r ⎠ 1 E(r,θ ) = −∇ϕ (x, y, z) = T.p 4πε o (T )ij = 3x i x j - r 2δ ij r5 Dipole field propagation tensor Electrostatic energy of point charges • Work to bring charge q2 to r2 from ∞ when q1 is at r1 W2 = q2 φ2 q1 q1 q2 r12 q2 r12 r13 r1 r2 r1 q1 1 ϕ2 = 4πε o r12 • r2 q1 1 q2 1 ϕ3 = + 4πε o r13 4πε o r23 r3 O • • r23 O 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 1 qj 1 1 • In general W = q = ∑ i ∑j r 2 4πε 4πε o i< j ij o qj ∑q ∑ r i≠ j i j ij Electrostatic energy of charge distribution • For a continuous distribution of charge 1 W= dr ρ (r )φ (r ) ∫ 2 all space φ (r ) = W= 1 4πε o 1 1 2 4πε o ∫ dr' all space ρ (r' ) r − r' ∫ dr ρ (r ) all space ∫ all space dr' ρ (r' ) r − r' Energy in vacuum in terms of E • • Gauss’ law relates ρ to electric field and potential Replace ρ in energy expression using Gauss’ law ρ and E = −∇φ εo ρ ⇒ ∇ 2φ = − ⇒ ρ = −ε o∇ 2φ εo ε ∴ W = 1 ∫ φ ρ dv = − o ∫ φ ∇ 2φ dv 2 ∇.E = v • 2 v Expand integrand using identity: ∇.ψF = ψ∇.F + F.∇ψ 2 ∇ . φ ∇ φ = φ ∇ φ + (∇φ ) Exercise: write ψ = φ and F = ∇φ to show: 2 ⇒ φ∇ 2φ = ∇.φ∇φ − (∇φ ) 2 Energy in vacuum in terms of E εo ⎡ ⎤ W = − ⎢ ∫ ∇.φ∇φ dv − ∫ (∇φ ) dv ⎥ 2 ⎣v v ⎦ ⎤ εo ⎡ 2 = − ⎢ ∫ (φ∇φ ).dS − ∫ (∇φ ) dv ⎥ (Green' s first identity ) 2 ⎣S v ⎦ Surface integral replaces volume integral (Divergenc e theorem) 2 For pair of point charges, contribution of surface term f ~ 1/r “f ~ -1/r2 dA ~ r2 overall ~ -1/r Let r → ∞ and only the volume term is non-zero W= Energy density εo 2 ∫ (∇φ ) dv = 2 εo all space dW ε o 2 = E (r ) dv 2 2 2 E ∫ dv all space Energy of charge distribution in external potential • Energy of localised charge distribution ρ(r) in external potential φ(r) U= ∫ dr ρ (r )φ (r ) all space f(a + h) = f(a) + hf' (a) + ... 1- D Taylor expansion φ (r ) = φ (0) + r.∇φ (0) + ... U= 3 - D Taylor expansion about origin ∫ dr ρ (r )(φ (0) + r.∇φ (0) + ...) all space = φ (0 ) ∫ dr ρ (r ) + ∇φ (0). all space ∫ dr ρ (r )r + ... all space U = Qφ (0) + p.∇φ (0) + ... = Qφ (0) − p.E(0) + ... Q= ∫ dr ρ (r ) all space p= ∫ dr ρ (r )r all space Energy of electric dipole in electric field Up = -p.E