Download Electromagnetism Lecture 11: Electric Potential and Potential Energy and more Study notes Physics in PDF only on Docsity! Lecture 11 Carl Bromberg - Prof. of Physics PHY481: Electromagnetism Potential V(x) Potential energy U Potential energy density u(x) Lecture 11 Carl Bromberg - Prof. of Physics 1 Electric potential differences Stokes’s theorem: E ⋅ d C∫ = ∇ × E( )S∫ ⋅ n̂ dA ∇ × E = 0 E ⋅ d C∫ = 0Integral over closed path: E ⋅ d I a b ∫ + E ⋅ d II b a ∫ = 0 V (xb) −V (xa ) = − E ⋅ d a b ∫ Integral is independent of path E ⋅ d I a b ∫ = − E ⋅ d II b a ∫ = E ⋅ d III a b ∫ Integral around loop Reverse direction of II Integral depends only on end points Let V(x) be “potential” at x but Integral is work done (by an external agent) in moving a unit charge from a to b. Lecture 11 Carl Bromberg - Prof. of Physics 4 Potential due to spherical shell of charge z R x = zk̂ ′x Rsinθ Rcosθ θ x − ′x y ρ( ′x )d3 ′x = σR2 sinθ dθ dφ ′x = Rcosθ k̂ + Rsinθ ĵ x − ′x = z − Rcosθ( )k̂ − Rsinθ ĵ x − ′x = z2 + R2 − 2zRcosθ⎡⎣ ⎤ ⎦ 1 2 V (x) = 1 4πε0 ρ( ′x )d3 ′x x − ′x∫ V (z) = σR2 4πε0 dφ 0 2π ∫ sinθdθ 0 π ∫ z2 + R2 − 2zRcosθ⎡⎣ ⎤⎦ −1 2 Spherical shell, charge Q, radius R, at distance z (or r) integrals.wolfram.com = 2πσR 4πε0z z + R( ) z − R( )[ ] V (z)= Q 4πε0z , z > R V (z) = Q 4πε0R , z < RPotential Inside Potential Outside Point charge Constant (−) (+) z > R z < R 4πR 2σ = Q Lecture 11 Carl Bromberg - Prof. of Physics 5 integrals.wolfram.com Lecture 11 Carl Bromberg - Prof. of Physics 6 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 density ρ, 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 11 Carl Bromberg - Prof. of Physics 9 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 11 Carl Bromberg - Prof. of Physics 10 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 11 Carl Bromberg - Prof. of Physics 11 Energy of an electric field Express U as a function of E only ! 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
