29. (a) At any instant the displacement current idin the gap between the plates equals the conduction current iin the wires. Thus imax= idmax= 7.60 µA. (b) Since id= ε0(dΦE/dt), ddtiEdΦFHGIKJ==××=×⋅−−max...06125760 10859 10A8.85 10F mVms (c) According to Problem 32-13, the displacement current is iCdVdtAddVdtd0. Now the potential difference across the capacitor is the same in magnitude as the emf of the generator, so V= msin ωtand dV/dt= ωεmcos t. Thus, 0m(/)cosdiAdt=and max0m/.dd=This means ()( )( )( )21236max8.85 10F m0.180 m130rad s 220 V3.39 10 m,7.60 10 AdAdiεωε−−−×π=××where A = πR2was used. (d) We use the Ampere-Maxwell law in the form GGBdsId⋅=z0, where the path of integration is a circle of radius rbetween the plates and parallel to them. Idis the displacement current through the area bounded by the path of integration. Since the
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.