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FWLec10

# FWLec10 - A F Peterson Notes on Electromagnetic Fields...

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A. F. Peterson: Notes on Electromagnetic Fields & Waves 10/04 Fields & Waves Note #10 Capacitance and Energy Objectives: Introduce the notion of capacitance, which is the amount of charge a device can store per unit of applied voltage. Illustrate its calculation by solving Laplace’s equation for several examples. Introduce an expression for the energy stored in the electric field associated with a charge distribution and use it to calculate the energy stored in capacitors. Capacitance A capacitor is an energy-storage device. Consider the “black box” device in Figure 1. Suppose that a voltage is applied between the terminals, resulting in a current flow into the device, and further suppose that some charge is stored in the device as a result. The capacitance of the device is the amount of charge stored per unit of applied voltage, or C Q V = (10.1) The units of capacitance are Farads (F). The external system performed work in order to place the charge into the device, and that work is stored in the device in the form of potential energy. The energy stored in the capacitor is W QV CV e = = 1 2 1 2 2 (10.2) The stored energy can be removed from the capacitor and used for some purpose. In this Note, we illustrate the operation of a capacitor from the electromagnetic field perspective. In addition, the capacitance per unit length of parallel strip and coaxial transmission lines will be determined. Finally, the energy stored in a capacitor will be related to the energy density of the associated electric fields. The ideal parallel-plate capacitor Figure 2 depicts the cross section of a parallel-plate capacitor with plates located at x = 0 and x = d . For simplicity, we initially consider an “ideal” capacitor, for which the fields are only functions of x . Equivalently, we neglect any fringing of the fields near the edges of the plates. Suppose that the two plates are identical in shape and have area A . The problem can be posed mathematically as that of determining the voltage between the plates, given that a voltage difference of V 0 is maintained between the plates. The appropriate equation to solve is Laplace’s equation = 2 0 V (10.3)

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A. F. Peterson: Notes on Electromagnetic Fields & Waves 10/04 subject to the boundary conditions V x x ( ) = = 0 0 (10.4) V x V x d ( ) = = 0 (10.5) The Laplacian simplifies in this situation to = + + = 2 2 2 2 2 2 2 2 2 V V x V y V z d V dx (10.6) and the resulting differential equation is d V dx 2 2 0 = (10.7) Since this equation only involves one variable, the solution may be found by integration. The initial step d V dx dx dx 2 2 0 Ú Ú = (10.8) produces dV dx C = 1 (10.9) where C 1 is a constant of integration. A second integration dV dx dx C dx = Ú Ú 1 (10.10) yields the general solution V x C x C ( ) = + 1 2 (10.11) The constants C 1 and C 2 are determined from the boundary conditions. Imposing (10.4) results in V x C C x ( ) ( ) = = + = 0 1 2 0 0 (10.12)

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A. F. Peterson: Notes on Electromagnetic Fields & Waves 10/04 implying that C 2 0 = (10.13) Imposing (10.5) yields V x
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