46. (a) The electric field E1in the free space between the two plates is E1= q/ε0Awhile that inside the slab is E2= E1/κ= q/κε0A. Thus, VEdbE bqAdbb012=−+=FHGIKJ−+FHGIKJbgεκ0, and the capacitance is ()()()()()()()221242CN m008.85 10115 10m2.6113.4pF.2.610.0124m0.00780m0.00780mAqCVdbbεκκ−−⋅××====−+−+(b) q = CV= (13.4 ×10–12F)(85.5 V) = 1.15 nC. (c) The magnitude of the electric field in the gap is
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