This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: ECE-329 Fall 2008 Homework 10 — Solution October 30, 2008 1. Two infinite perfectly conducting plates at z = 0 and z = z o are kept at potentials Φ = 0 and Φ = Φ o , respectively. The region between the plates is filled with two slabs of perfect dielectric materials having permittivities 1 for < z < d (region 1) and 2 for d < z < z o (region 2), as shown in the following figure. z z = z o z = d z = 0 ǫ 2 ǫ 1 Φ = Φ o Φ = 0 a) Because there are zero volumetric charge densities inside the dielectrics, Laplace’s equation can be used to find the electric potential Φ in these two regions. In the case that Φ = Φ( z ) , the general form solution of Laplace’s equation is simply a line, and therefore, we can write that Φ( z ) = ( a 1 z + b 1 for < z < d a 2 z + b 2 for d < z < z o . This implies that the associated electric field is simply E ( z ) =-∇ Φ = (- a 1 ˆ z for < z < d- a 2 ˆ z for d < z < z o . Notice that Φ = 0 at z = 0 , and thus b 1 = 0 . In addition, given that Φ = Φ o at z = z o , we have that a 2 z o + b 2 = Φ o . We also know that Φ must be continuous at the interface x = d , and therefore, we have that a 2 d + b 2- a 1 d = 0 . Maxwell’s boundary conditions tell us that the normal component of the electric flux density D ( z ) must be continuous across the interface z = d. Because of this we can write that ˆ n · ( D 1- D 2 ) = 0 1 a 1- 2 a 2 = 0 ....
View Full Document
- Spring '08
- Electrostatics, zo, surface charge density