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notes13 2317

# notes13 2317 - Divergence Physical Concept ECE 2317 Applied...

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Prof. Filippo Capolino ECE Dept. Fall 2006 Notes 13 ECE 2317 ECE 2317 Applied Electricity and Magnetism Applied Electricity and Magnetism Notes prepared by the EM group, University of Houston . (used by Dr. Jackson, spring 2006) Divergence Divergence -- -- Physical Concept Physical Concept r < a : ± 23 0 2 0 4 4 3 C/m 3 encl S rv v r Dnd S Q rD r r D πρ π ρ ⋅= ⎛⎞ = ⎜⎟ ⎝⎠ ⎡⎤ = ⎣⎦ v x y z v = v 0 a r Start by considering a sphere of uniform volume charge density The electric field is calculated using Gauss's law: r > a : 0 3 2 0 2 4 4 3 C/m 3 v r a a D r = = Divergence Divergence -- -- Physical Concept (cont.) x y z v = v 0 a r 2 ˆ 4 r S S ψ =⋅ = 3 0 4 3 v a = ( r > a ) ( r < a ) Flux through a spherical surface: 3 0 4 3 v r = 2 0 C/m 3 v r r D = 3 2 0 2 C/m 3 v r a D r = Divergence Divergence -- -- Physical Concept (cont.) Physical Concept (cont.) ( r < a ) ( r > a )

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More flux lines are added as the radius increases (as long as we stay inside the charge). ± 0 S Dnd S ψ Δ Δ= ⋅ > Observation: Δ V S Divergence Divergence -- -- Physical Concept (cont.) Physical Concept (cont.) ± 0 S S Δ > The net flux out of a small volume Δ V inside the charge is not zero. Divergence is a mathematical way of describing this. ± 0 1 lim V S div D D n dS V Δ→ Δ ≡⋅ Δ v Δ V Definition of divergence: Gauss Gauss ’s Law s Law -- -- Differential Form Differential Form Note: the limit exists independent of the shape of the volume (proven later). ± 0 1 lim V S div D D n dS V Δ Δ v Δ V ρ v ( r ) ± () encl v S S Q r V Δ ⋅=≈ Δ v 0 1 lim v V v div D r r V V r Δ = Apply divergence definition to small volume inside a region of charge Gauss Gauss ’s Law s Law -- -- Differential Form Differential Form Gauss Gauss ’s Law s Law -- Differential Form (cont.) Differential Form (cont.) ( ) v div D r r = The electric Gauss law: this is one of Maxwell’s equations.
Example Example Choose Δ V to be small sphere of radius r: x y z ρ v = v 0 a r ± 0 1 lim V S div D D n dS V Δ→ Δ ≡⋅ Δ v 3 4 3 Vr π Δ= ± () 3 22 0 0 4 44 33 v rv S r r Dnd S D r r ππ Δ ⎛⎞ ⋅= = = ⎜⎟ ⎝⎠ v 3 0 00 3 4 1 lim lim 4 3 3 v vv VV r div D r ρπ ≡= = Verify that the differential form of Gauss’s law gives the correct result for the example of a sphere of uniform volume charge density.

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notes13 2317 - Divergence Physical Concept ECE 2317 Applied...

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