Electromagnetism Electric Potential V Electric Field Energy Multipoles | PHY 481, Study notes of Physics

Download Electromagnetism Electric Potential V Electric Field Energy Multipoles | PHY 481 and more Study notes Physics in PDF only on Docsity! Lecture 14 Carl Bromberg - Prof. of Physics PHY481: Electromagnetism Electric potential V Electric field energy Multipoles Lecture 14 Carl Bromberg - Prof. of Physics 1 Potential for sphere with uniform charge density ′x = ′r cosθ k̂ + ′r sinθ ĵ x − ′x = z − ′r cosθ( )k̂ − ′r sinθ ĵ V (x) = 1 4πε0 ρ( ′x )d3 ′x x − ′x∫ V (z) = 2πρ 4πε0 ′r 2d ′r 0 R ∫ sinθdθ 0 π ∫ z2 + ′r 2 − 2z ′r cosθ⎡⎣ ⎤⎦ −1 2  Sphere, uniform charge Q, radius R, at distance z (or r) ρ( ′x )d 3 ′x = ρ ′r 2d ′r sinθ dθ dφ = 2πρ 4πε0z ′r d ′r 0 R ∫ z + ′r( )2⎡⎣ ⎤⎦ 1 2 − z − ′r( )2⎡⎣ ⎤ ⎦ 1 2{ } V (z) = 2πρ 4πε0z ′r d ′r 0 R ∫ z + ′r( ) − z − ′r( ){ } = 4πρ 4πε0z ′r 3 3 ⎡ ⎣⎢ ⎤ ⎦⎥0 R z > R 4πR 3ρ 3 = Q = Q 4πε0z – Outside the sphere Lecture 14 Carl Bromberg - Prof. of Physics 4 Potential energy for point charges V x( ) = − E ⋅ d ∞ x ∫ =W Work done (by an external agent) moving a unit charge from to x  Electric potential (should be a – sign in front of integral) U (x) = qV (x) Potential energy U of charge q . V1(x2 ) = q1 4πε0 x1 − x2  Potential energy Potential energy U12 of the charge q2 due the potential V1 U12 x2( ) = q2V1 x2( ) Put charge q2 at point x2. Potential at x2 due to charge q1 at point x1 ∞ Lecture 14 Carl Bromberg - Prof. of Physics 5 Potential energy for charge distributions  Potential energy for multiple point charges U = 1 2 qiVj i≠ j ∑ (x i ,x j ) = 1 2 qiq j 4πε0 x i − x ji≠ j ∑ i ≠ j, do not count charge with itself. 1/2 corrects for qi qj and qj qi  Generalize for charge distributions U = 1 2 ρ(x)d3x∫ ρ( ′x ) 4πε0 x − ′x ∫ d3 ′x U = 1 2 ρ(x)∫ V (x)d3x V (x) = ρ( ′x ) 4πε0 x − ′x ∫ d3 ′x  Where is this energy? Charge produces potential and the same charge gains potential energy? The charge distribution produces an electric field. We interpret the energy as being stored by (in) the field ! Lecture 14 Carl Bromberg - Prof. of Physics 6 Energy of an electric field  Express U as a function of E U = 1 2 ρ∫ Vd3x E = −∇V U = ε0 2 ∇ ⋅E( )∫ Vd3x ∇ ⋅ VE( ) =V∇ ⋅E + E ⋅ ∇V( ) ∇ ⋅E = ρ / ε0 = ε0 2 ∇ ⋅ VE( )∫ d3x + ε0 2 E 2∫ d3x ∇ ⋅ V E( )d 3 x Vol∫ = V E ⋅ dAS∫ Identity Gauss’s theorem = ε0 2 V E ⋅ dA S∫ + ε0 2 E 2 d 3 x Vol∫ → S ,Vol→∞ U = ε0 2 E 2 d 3 x all ∫ uE (x) = ε0E 2 (x) 2 Field energy Energy density = ε0 2 ∇ ⋅ VE( )∫ d3x − ε0 2 E ⋅ ∇V( )∫ d3x Lecture 14 Carl Bromberg - Prof. of Physics 9 Point dipole V (x) = p ⋅ r̂ 4πε0r 2  Potential of a point dipole p (units are C•m) Exact  Electric field of point dipole E(x) = −1 4πε0 ∇ p ⋅ r̂ r 2 ⎛ ⎝⎜ ⎞ ⎠⎟ = −1 4πε0 ∇ p ⋅r r 3 ⎛ ⎝⎜ ⎞ ⎠⎟ = −1 4πε0 p ⋅r∇ 1 r 3 ⎛ ⎝⎜ ⎞ ⎠⎟ + 1 r 3 ⎛ ⎝⎜ ⎞ ⎠⎟ ∇ p ⋅r( )⎡ ⎣⎢ ⎤ ⎦⎥ E(x) = −∇V (x) ∇ p ⋅r( ) = êi ∂ ∂xi p jx j( ) = êi pi = p = −1 4πε0 p ⋅r −3 r 4 r̂ ⎛ ⎝⎜ ⎞ ⎠⎟ + p r 3 ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣⎢ ⎤ ⎦⎥ = 1 4πε0r 3 3 p ⋅ r̂( )r̂ − p[ ] with p = p k̂ V (r,θ) = pcosθ 4πε0r 2 p = lim d→0 qd( ) qd constant d -> 0, with qd constant Limit of finite dipole as Lecture 14 Carl Bromberg - Prof. of Physics 10 Torque and energy of dipole in electric field + E p θ qE –qE x − y Ν = p × E = pE sinθ (−k̂) U = −p ⋅E = − pE cosθ F(x) = −∇U (x) = ∇ p ⋅E(x)( ) = p ⋅∇( )E(x)  Torque on electric dipole  Force on electric dipole  Energy of electric dipole If E(x) is constant, F(x) = 0 Fj (x) = pi ∂E j (x) ∂xi