ch25-p069 - 69. (a) Since the field is constant and the...

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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, =×−× = × −− σ 177 10 4 60 10 2 83 10 66 6 .. . . c h c h Cm 2 69. (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 A is equal to the field in B : G E V d == × 200 10 5 Vm (b) 5 | | 2.00 10 V m . E G See the note in part (a). (c) For the air-filled capacitor, Eq. 25-4 leads to σε = × q A E 0 62 177 10 G Cm (d) For the dielectric-filled capacitor, we use Eq. 25-29:
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