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notes13 2317 - ECE 2317 ECE Applied Electricity and...

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Prof. David R. Jackson ECE Dept. Spring 2011 Notes 13 ECE 2317 ECE 2317 Applied Electricity and Magnetism Applied Electricity and Magnetism
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Divergence -- Physical Concept Divergence -- Physical Concept r < a : 2 2 3 0 0 C/m ˆ 4 4 3 3 encl S r v v r D n dS Q r D r r D π ρ π ρ = = = Ñ Start by considering a sphere of uniform volume charge density The electric field is calculated using Gauss's law: x y z ρ v = ρ v 0 a r
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r > a : 2 2 3 0 3 0 2 C/m 4 4 3 3 r v v r r D a a D r π ρ π ρ = = Divergence -- Physical Concept (cont.) Divergence -- Physical Concept (cont.) x y z ρ v = ρ v 0 a r
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2 ˆ 4 r S D ndS r D ψ π = = 3 0 4 3 v a ψ π ρ = ( r > a ) ( r < a ) Recall that 3 0 4 3 v r ψ π ρ = 2 0 C/m 3 v r r D ρ = 2 3 0 2 C/m 3 v r a D r ρ = Divergence -- Physical Concept (cont.) Divergence -- Physical Concept (cont.) ( r < a ) ( r > a ) Flux through a spherical surface: Hence
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More flux lines are added as the radius increases (as long as we stay inside the charge). ˆ 0 S D n dS ψ = Observation: V S Divergence -- Physical Concept (cont.) Divergence -- Physical Concept (cont.) ˆ 0 S D n dS ψ = The net flux out of a small volume V inside the charge is not zero. Divergence is a mathematical way of describing this.
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0 1 ˆ lim V S div D D n dS V Ñ V Definition of divergence: Gauss’s Law -- Differential Form Gauss’s Law -- Differential Form Note: The limit exists independent of the shape of the volume (proven later). A point with a positive divergence acts as a “source.” A point with a negative divergence acts as a “sink.”
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0 1 ˆ lim V S div D D n dS V Ñ V ρ v ( r ) ( 29 ˆ encl v S D n dS Q r V ρ = Ñ ( 29 ( 29 ( 29 ( 29 0 1 lim v V v div D r r V V r ρ ρ = = Apply divergence definition to a small volume inside a region of charge: Gauss’s Law -- Differential Form Gauss’s Law -- Differential Form Gauss's law: Hence
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Gauss’s Law -- Differential Form (cont.) Gauss’s Law -- Differential Form (cont.) ( 29 ( 29 v div D r r ρ = The electric Gauss law in point form: This is one of Maxwell’s equations .
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Example Example Choose V to be small sphere of radius r : x y z ρ v = ρ v 0 a r 0 1 ˆ lim V S div D D n dS V Ñ 3 4 3 V r π = ( 29 ( 29 3 2 2 0 0 4 ˆ 4 4 3 3 v r v S r r D n dS D r r ρ π π π ρ = = = Ñ 3 0 0 0 0 0 3 4 1 lim lim 4 3 3 v v v V V r div D r
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