follows. Adapting Eq. 25-35 to this problem, we see that the difference in charge densities between parts (c) and (d) is due, in part, to the (negative) layer of charge at the top surface of the dielectric; consequently, ′=×−×=−×−−−σ177 104 60 102 83 10666....chchCm269. (a) Since the field is constant and the capacitors are in parallel (each with 600 V across them) with identical distances (d= 0.00300 m) between the plates, then the field in Ais equal to the field in B: GEVd==×200 105Vm (b) 5||2.00 10 V m .E=×GSee the note in part (a). (c) For the air-filled capacitor, Eq. 25-4 leads to σε=×−qAE062177 10GCm (d) For the dielectric-filled capacitor, we use Eq. 25-29:
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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.