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Electromechanical Dynamics (Part 1).0053

Electromechanical Dynamics (Part 1).0053 - the upper plate...

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Lmped Electromechanical Elements Volume for relating fields in S"vacuum and in dielectric-" Dielectric of _" Conducting plates permittivity e of area A Fig. 2.1.7 A parallel-plate capacitor. Example 2.1.4. Consider the simple parallel-plate capacitor of Fig. 2.1.7. It consists of two rectangular, parallel highly conducting plates of area A. Between the plates is a rectangular slab of dielectric material with constant permittivity e, D = EE. The lower plate and the dielectric are fixed and the upper plate can move and has the instantaneous position x with respect to the top of the dielectric. The transverse dimensions are large compared with the plate separation. Thus fringing fields can be neglected. The terminal voltage is constrained by the source v which is specified as a function of time. We wish to calculate the instantaneous charge on the upper plate and
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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...
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