Lect_34 - ACT: Crossed planes Which diagram corresponds to...

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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 + + Each sheet produces a uniform electric field. + G total E + G G The total E field is uniform in each quadrant. + 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 .
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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 . On the surface, the electric field and the differential area vector are parallel. ( ) da Φ= = ∫∫ G G Eda Also, the electric field has the same magnitude all over the Gaussian surface: π == 2 4 E da 2 4 Er 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 total 4 3 q QR ρ So the flux through the Gaussian surface is also: 3 0 4 3 ρπ ε Eqn. 2
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Now we put everything together: 3 2 0 4 3 4 R Er ρπ π ε = It looks like the point-charge electric field , but it’s a “coincidence” ( only for spherical symmetries and r > )!
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Lect_34 - ACT: Crossed planes Which diagram corresponds to...

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