This preview shows page 1. Sign up to view the full content.
Unformatted text preview: the upper plate. To solve this problem we need the given relation between D and E, (1.1.24) and (1.1.25), and the definition of the voltage of point a with respect to point b v = -jE dl With the neglect of fringing fields, the field quantities D and E will have only vertical components. We take them both as being positive upward. In the vacuum space D, = EOE, and in the dielectric Da = EEd. We assume that the dielectric has no free charge; consequently, we use (1.1.25) with a rectangular box enclosing the dielectric-vacuum interface as illustrated in Fig. 2.1.7 to obtain .oE, = eE . We now use the expression for the voltage to write v= = Ed' -'f+dE, da'. Integration of these expressions yields the vacuum electric field intensity V E , + (Eo/e)d A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
View Full Document
This note was uploaded on 02/10/2012 for the course MECE 4371 taught by Professor Liu during the Fall '11 term at University of Houston.
- Fall '11