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Lecture05_Kim

# Lecture05_Kim - 1 Lecture 5-1 Applications of Gauss’s Law...

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Unformatted text preview: 1/25/2012 1 Lecture 5-1 Applications of Gauss’s Law The net electric flux through any closed surface equals the net charge enclosed by that surface divided by ε . Applications to systems with symmetries: 1. Shperical Symmetry: Thin spherical shells, solid spheres 2. Cylindrial Symmetry: Infinite thin wire 3. Planar Symmetry: Infinite non-conducting sheet, Infinite conducting sheet enclosed E Q E ndA ε = Φ = ∫ a ɵ i I Lecture 5-2 Spherial Symmetry Thin , uniformly charged cell (Review) E = 0 inside Discontinuity in E 2 Q E k r = outside r E σ ε ∆ = Lecture 5-3 3 Spherical Symmetry: Uniform Distribution (1) • Let’s start with a Gaussian surface with r 1 < R • From spherical symmetry we know that the electric field will be radial and perpendicular to the Gaussian surface. • Gauss’s Law gives us • Solving for E we find ( 29 3 1 2 1 4 3 4 r q E dA E r ρ π π ε ε ⋅ = = = ∫∫ a a d volume volume volume volume area area area area E inside = ρ r 1 3 ε R Lecture 5-4 4 Spherical Symmetry: Uniform Distribution (2) 1 3 4 3 3 inside r Q E R π ε = 1 3 inside r E ρ ε = In terms of the total charge Q … 1 1 3 3 4 inside Qr kQr E R R πε = = R 1/25/2012 2 Lecture 5-5 5 Spherical Symmetry: Uniform Distribution (3) • Now consider a Gaussian surface with radius r 2 > R • Again by spherical symmetry we know that the electric field will be radial and perpendicular to the Gaussian surface. • Gauss’ Law gives us • Solving for E we find E outside = kQ r 2 2 same as a point charge! same as a point charge! same as a point charge! same as a point charge! total charge total charge total charge total charge area area area area R ( 29 3 2 2 4 3 4 R Q E dA E r ρ π π ε ε ⋅ = = = ∫∫ a a d Lecture 5-6 6 Cylindrical Symmetry (Review) • The electric flux through the ends of the cylinder is zero because the electric field is always parallel to the ends • The electric field is always perpendicular to the wall of the cylinder so • … and now solve for the electric field E = λ 2 πε r = 2 k λ r ( 29 2 / / (Gauss) E dA EA E rL q L π ε λ ε Φ = ⋅ =...
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Lecture05_Kim - 1 Lecture 5-1 Applications of Gauss’s Law...

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