(34) Applications of Gauss's Law

# (34) Applications of Gauss's Law - Lecture 34 Applications...

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Unformatted text preview: Lecture 34 Applications of Gauss ’s Law ACT: Crossed planes Which diagram corresponds to the E-field lines for these two uniformly charged infinite sheets that intersect each other as shown? + σ-σ + σ-σ + σ-σ Each sheet produces a uniform electric field.-σ + σ r t ot al E + r E- r E The total E field is uniform in each quadrant. + σ-σ EXAMPLE: E for a charged sphere EXAMPLE: E for a charged sphere Find the magnitude of the electric field produced by a uniformly charged sphere with charge density ρ and radius R. First thing: Assess the symmetry of the problem. In this case, we clearly have a spherical symmetry. The magnitude of the electric field should only depend on the distance to the center of the sphere. Same E Direction of E : Must be radial (in for -, out for +) If we want to find the electric field at a point located at distance r (>R) from the center of the sphere, the Gaussian surface to use is a sphere of radius r. r On the surface, the electric field and the differential area vector are parallel. r E (r) da Φ = = r r E da Eda Also, the electric field has the same magnitude all over the Gaussian surface: π = = 2 4 Eda E da E r 2 4 E r π Φ = Eqn. 1 The second way is using Gauss’s law. The charge enclosed by this surface is all the total charge in the sphere: 3 enclosed t ot al 4 3 q Q R ρ π = = So the flux through the Gaussian surface is also: 3 4 3 R ρ π ε Φ = Eqn. 2 Now we put everything together: 3 2 4 3 4 R E r ρ π π ε = It looks like the point-charge electric field...
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## This note was uploaded on 03/27/2008 for the course PHYS 221 taught by Professor Herrera-siklody during the Fall '08 term at Iowa State.

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(34) Applications of Gauss's Law - Lecture 34 Applications...

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