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FWLec11

# FWLec11 - A F Peterson Notes on Electromagnetic Fields...

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A. F. Peterson: Notes on Electromagnetic Fields & Waves 10/04 Fields & Waves Note #11 Capacitors in Series and Parallel: Field Mapping Objectives: Continue the study of capacitance, by considering examples with multiple layers or regions of different permittivities. Expand the concept of series and parallel capacitance to describe the field mapping approach for general capacitor shapes. Example: A parallel-plate capacitor with two layers of dielectric Figure 1 depicts the cross section of a parallel-plate capacitor with plates located at x = 0 and x = d , as in Note #10. However, in this situation the region between the plates is divided into two materials, at location x = c , as illustrated. As in the previous examples, we assume that the fields are “ideal” in the sense that they are only functions of x , and neglect any fringing of the fields near the edges of the plates. As in the preceding examples, the problem can be posed mathematically as determining the voltage between the plates, using Laplace’s equation. However, we must separately treat the two regions between the plates. Therefore the differential equation to consider is separated into two equations, by region: = = < < 2 1 2 1 2 0 0 V d V dx x c , (11.1) = = < < 2 2 2 2 2 0 V d V dx c x d , (11.2) The solutions V 1 and V 2 are subject to the boundary conditions V x x 1 0 0 ( ) = = (11.3) V x V x d 2 0 ( ) = = (11.4) at the plates. In addition, the continuity of the voltage may be imposed at x = c : V x V x x c x c 1 2 ( ) ( ) = = = (11.5) A fourth condition is required, and must be obtained from the conditions that are satisfied by the electric field at the material interface. Since E V = -— , the only component of E present between the plates is E dV dx x = - (11.6) Because E x is the normal component at the interface, the appropriate condition (Note #9) is

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A. F. Peterson: Notes on Electromagnetic Fields & Waves 10/04 e e 1 1 2 2 E E x x c x x c = = = (11.7) By combining the relations in (11.6) and (11.7), we obtain the fourth boundary condition as e e 1 1 2 2 dV dx dV dx x c x c = = = (11.8) The general solutions to the differential equations in (11.1) and (11.2) are easily obtained as V x K K x x c 1 1 2 0 ( ) , = + < < (11.9) V x K K x c x d 2 3 4 ( ) , = + < < (11.10) There are four constants to be determined by the boundary conditions. Imposing (11.3) yields V x K K x 1 0 1 2 0 0 ( ) ( ) = = + = (11.11) or equivalently, K 1 0 = (11.12) Imposing (11.4) results in the equation V x K K d V x d 2 3 4 0 ( ) ( ) = = + = (11.13) The condition in (11.5) produces V x K c V x K K c x c x c 1 2 2 3 4 ( ) ( ) ( ) ( ) = = = = = + (11.14) Finally, (11.8) results in e e e e 1 1 1 2 2 2 2 4 dV dx K dV dx K x c x c = = = = = (11.15) In this situation, the application of the four boundary conditions results in one of the four constants being given directly ( K 1 = 0), and three equations that must be solved to determine the other constants: K dK V cK K cK K K 3 4 0 2 3 4 1 2 2 4 0 0 + = - - = - = e e (11.16)
A. F. Peterson: Notes on Electromagnetic Fields & Waves 10/04 From the last equation, we determine K K 2 2 1 4 = e e (11.17) Substituting this result into the second equation, we obtain K c K 3 2 1 4 1 = - Ê Ë Á ˆ ¯ ˜ e e (11.18) Finally, by combining (11.18) with the first equation of the three, we obtain the result K V c d 4 0 1 2 1 1 = - + e e e e ( ) (11.19) By substituting back into (11.17) and (11.18), we obtain K V

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