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 dielectricvacuum 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 APDF Split DEMO : Purchase from www.APDF.com to remove the watermark...
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 Fall '11
 Liu
 Electric charge, Permittivity, Dielectric, Electric displacement field, Lmped Electromechanical Elements

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