P26_044 - 44. (a) The electric field E1 in the free space...

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Unformatted text preview: 44. (a) The electric field E1 in the free space between the two plates is E1 = q/ε0 A while that inside the slab is E2 = E1 /κ = q/κε0 A. Thus, V0 = E1 (d − b) + E2 b = q ε0 A d−b+ b κ , and the capacitance is C = = = q ε0 Aκ = V0 κ(d − b) + b 8.85 × 10−12 C2 N·m2 (115 × 10−4 m2 )(2.61) (2.61)(0.0124 m − 0.00780 m) + (0.00780 m) 13.4 pF . (b) q = CV = (13.4 × 10−12 F)(85.5 V) = 1.15 nC. (c) The magnitude of the electric field in the gap is E1 = q 1.15 × 10−9 C = = 1.13 × 104 N/C . 2 ε0 A 8.85 × 10−12 NC 2 (115 × 10−4 m2 ) ·m (d) Using Eq. 26-32, we obtain E2 = E1 1.13 × 104 N/C = = 4.33 × 103 N/C . κ 2.61 ...
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