329lect09 - 9 Static elds in dielectric media Summarizing...

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9 Static felds in dielectric media Summarizing important results from last lecture: within a dielectric medium, displacement D = ± E = ± o E + P , and if the permittivity ± = ± r ± o is known, D and E can be calcu- lated from free surface charge ρ s or volume charge ρ in the region without resorting to P . we have, on surfaces separating perfect dielectrics, ˆ n · ( D + - D - )= 0 , while ˆ n · D = ρ s on a conductor-dielectric interface ( ˆ n pointing from the conductor toward the dielectric). ˆ n D + D - Gauss’s law ∇· D = ρ (and its integral counterpart) includes only the free charge density on its right side, which is typically zero in many practical problems. once D and E have been calculated (typically using the boundary condition equations), polarization P can be obtained as P = D - ± o E , if needed. These rules will be used in the examples in this section. 1

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z x E = 18ˆ x E = 18ˆ x E = 3ˆ x ± = ± o ± = ± o ± = ± r ± o Example 1: A perfect dielectric slab having a Fnite thickness W in the x direction is surrounded by free space and has a constant electric Feld E = 18ˆ x V/m in its exterior. Induced polarization of bound charges inside dielectric reduces the electric Feld strength inside the slab from 18ˆ x V/m to E = 3ˆ x V/m. What are the displacement Feld D and polarization P outside and inside the slab, and what are the dielectric constant ± r and electric susceptibility χ e of the slab? Solution: Displacement Feld outside the slab, where ± = ± o , must be D = ± o E x 18 ± o C m 2 .
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This note was uploaded on 07/19/2010 for the course ECE ECE329 taught by Professor Kudeki during the Summer '10 term at University of Illinois at Urbana–Champaign.

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329lect09 - 9 Static elds in dielectric media Summarizing...

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