chapter22--PHY131 JAB

# chapter22--PHY131 JAB - 1 Chapter 22 Gauss’s law...

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Unformatted text preview: 1 Chapter 22: Gauss’s law •Vector area •The unit normal •Definition of flux •Gauss’s Law •Applications of Gauss’s Law •Charges on conductors 2 Flux is flow through an area. Define n A ˆ A = A is the area of the surface is the outward directed unit vector normal to the surface n ˆ §22.1-2 Electric flux 3 Examples n ˆ A sheet z y x Here z n ˆ ˆ + = 4 Cylinder 2 ˆ n 3 ˆ n 1 ˆ n z Here ; ˆ ˆ ; ˆ ˆ ; ˆ ˆ 3 2 1 z n r n z n + = + =- = 5 Sphere n ˆ r n ˆ ˆ + = Here 6 Side view of a sheet G is some vector field ⊥ G Flow out from the surface || G Flow along the surface ⊥ G || G θ n ˆ G 7 The flux through an area dA of the surface is ( ) dA G dA G d ⊥ = = Φ θ cos A G d ⋅ = Φ = ∫∫ flux 8 The electric flux through a closed surface is ∫∫ ⋅ = Φ A E d e A closed surface is one that divides space into an inside region and an outside region. You can think of Φ e as proportional to the number of field lines passing through the closed surface. More field lines means the field is stronger. 9 Example: A cube is placed in a uniform electric field that points in the +x direction. What is the net flux through the cube? x E ˆ x E = There are 6 faces, each with a different normal. 10 ∫∫ ∫∫ ∫∫ ∫∫ ⋅ + ⋅ + ⋅ = ⋅ = Φ 6 2 1 e A E A E A E A E d d d d K The electric field is perpendicular to the front face, back face, top face, and bottom face so there is no flux through those surfaces. What about the remaining two faces? Example continued 11 ( ) ( ) ( ) ( ) ˆ ˆ ˆ ˆ right left e = = + ⋅ +- ⋅ = ⋅ + ⋅ = ⋅ = Φ ∫∫ ∫∫ ∫∫ ∫∫ ∫∫ L x x x x A E A E A E dA E dA E d d d x x There is no net flux through the cube. Since E is uniform, the field lines that enter the cube leave the cube....
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## This note was uploaded on 04/29/2008 for the course PHY 131 taught by Professor Fuchs during the Fall '07 term at ASU.

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chapter22--PHY131 JAB - 1 Chapter 22 Gauss’s law...

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