advertisement

19 Electric Charge 1 static electricity applications: air fresheners, xerography, painting cars, smokestack pollution control, … 2 conductors, insulators, semiconductors, superconductors 3 Coulomb’s law: The force of attraction or repulsion between two point charges q1 and q 2 is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. F =K⋅ K= q1 q 2 r2 (N ⋅ m C ) 9 1 = 8.99 × 10 4πε 0 2 2 permittivity of vacuum ε 0 = 8.85 × 10 −12 (C 2 N ⋅ m2 ) Principle of Superposition v v v v F1,net = F12 + F13 + .... + F1n electric potential V =K⋅ q1 r Coulomb’s law and the Principle of Superposition constitute the physical input for electrostatics. For a conducting sphere of surface charge density ρ R dV = K ⋅ dq r' V = ∫ dV = θ dq = ρ ⋅ r 2 sin θ d θ dφ r K ∫∫ r ' ρr 2 sin θdθdφ r’ φ : 0 → 2π sin θdθ r' 1 r' dr ' = 2πKρr 2 ∫ ⋅ r ' Rr R +r 2πkρr dr ' = R R∫− r = Kρr 2 ⋅ 2π ∫ = r ' 2 = R 2 + r 2 − 2 Rr cos θ 2 r ' dr ' = 2 Rr sin θ d θ r' sin θ d θ = dr ' Rr r ': R − r → R + r 4πKρr 2 K (4πr 2 )⋅ ρ Kq = = R R R Charge is quantized Millikan oil drop experiment When the Electric field is on, and the oil drop is at rest. qE = mg When the Electric field is off, and the oil drop reaches its terminal velocity. mg =bυ qE= bυ bυ q= = ne E the elementary charge e can be obtained by the {n1, n2, … .nm} 20 Electric Field electric field v v E=Fq Electric Field Due to a Point Charge v F= 1 q q0 ∧ r 4πε 0 r 2 v v F 1 q ∧ E= = ⋅ 2r q0 4πε 0 r + The Electric Field Due to an Electric Dipole E = E+ + E− 1 q 1 q = − 2 4πε 0 r+ 4πε 0 r−2 q 1 = 4πε 0 z 2 z 〉〉 d 1 1 − 2 2 d d z 2 1 + 1 − 2z 2z 2d 2d 1 + 2 z − 1 − 2 z 2d 1 q p = = ⋅ 3 2 4πε 0 z z 2πε 0 z for θ ≠0 1 p ⋅ 3 ⋅ f (θ ) E= 2πε 0 r ≅ q 4πε 0 z 2 dipole moment p = qd d 2 d r− = z + 2 r+ = z − f (θ ) = 1 for θ =0 The Electric Field Due to a Line of Charge λ：the charge per unit length ds = Rd φ v λ ds 1 ⋅ 2 dE = 4 πε 0 r v v = d E z + d E || dE z = 1 λ ds cos θ 4πε 0 r 2 1 λ ds sin θ ∫ dE || = 0 4 πε 0 r 2 v v v λ R cos θ E = ∫ d E = ∫ dE z = dφ 4πε 0 r 2 ∫ dE || = = = λ R cos θ ⋅ 2π q cos θ = 2 2 4πε 0 (z + R ) 4πε 0 (z 2 + R 2 ) qz 4πε 0 (z + R 2 for v E = ) 3 2 2 zˆ z 〉〉 R q zˆ 4πε 0 z 2 v z =0 E =0 point charge net force = 0