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Unformatted text preview: 9 Static fields 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 s or volume charge in the region without resorting to P . on surfaces separating perfect dielectrics, n ( D +- D- ) = 0 typ- ically, while n D + = s on a conductor-dielectric interface (with n pointing from the conductor toward the dielectric). n D + D- Gausss law D = (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 z x E = 18 x E = 18 x E = 3 x = o = o = r o Example 1: A perfect dielectric slab having a finite thickness W in the x direction is surrounded by free space and has a constant electric field E = 18 x V/m in its exterior. Induced polarization of bound charges inside dielectric reduces the electric field strength inside the slab from 18 x V/m to E = 3 x V/m. What are the displacement field D and polarization P outside and inside the slab, and what are the dielectric constant r and electric susceptibility e of the slab? Solution: Displacement field outside the slab, where = o , must be D = o E = x 18 o C m 2 . The outside polarization P is of course zero. Boundary conditions at the interface of the slab with free space require the continuity of normal component of D and tangential component of E both of these conditions would be satisfied if we were to take D = x 18 o C/m 2 also within the dielectric slab. Thus, with E = 3 x V/m inside the slab, the condition D = slab E within the slab requires that slab = 6 o ....
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- Spring '08