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329fall08hw7sol - ECE-329 Fall 2008 Homework 7 Solution 1...

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ECE-329 Fall 2008 Homework 7 — Solution October 15, 2008 1. In HW6, we found that the charge density in a conductor with conductivity σ and permittivity o is ρ ( r , t ) = cos(100 x ) e - σ o t C / m 3 . a) Using the continuity equation ∇ · J + ∂ρ ∂t = 0 and noting that J must have only a component along x -direction (i.e., J = J x ˆ x ), we can write ∂J x ∂x = - ∂ρ ∂t . Solving this equation we obtain J = σ 100 o sin(100 x ) e - σ o t ˆ x A m 2 . The associated electric field is E = 1 σ J = 1 100 o sin(100 x ) e - σ o t ˆ x V m . b) The energy dissipation rate per volume in the conductor is simply J · E = σ 10 4 2 o sin 2 (100 x ) e - 2 σ o t W m 3 . 2. For the specified charge distribution, let us compute D , E , and P in all regions, as well as, the surface charge densities on each boundary. a) Region 1 ( r a ) is occupied by a conductor with σ = 10 6 S / m , therefore, we can directly write D 1 = 0 C m 2 , E 1 = 0 V m , P 1 = 0 C m 2 . In steady-state, charges can accumulate only on the surface of conducting materials. Since this material holds a net charge Q 1 = 1 C , the surface charge density on the sphere of radius r = a is ρ S | r = a = Q 1 Area = 1 4 πa 2 C m 2 . b) Region 2 ( a < r < b ) is occupied by a dielectric with 2 = 5 o . Applying Gauss’ law ¸ D · d S = Q enc and considering D = D r ˆ r , we find D 2 = 1 4 πr 2 ˆ r C m 2 for a < r < b. In addition, we get E 2 = 1 2 D 2 = 1 20 π o r 2 ˆ r V m , and P 2 = D 2 - o E 2 = 1 4 πr 2 - 1 20 πr 2 ˆ r = 1 5 πr 2 ˆ r C m 2 . 1
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ECE-329 Fall 2008 c) Region 3 ( b r c ) is also occupied by a conductor and therefore D 3 = 0 C m 2 , E 3 = 0 V m , P 3 = 0 C m 2 .
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