49. (a) According to Chapter 26, the capacitance is C = ºµ0A/d. In our case º =4.5, A =(0.50 µm)2, and d =0.20 µm, so ºµ0A (4.5)(8.85 × 10-12 F/m)(0.50 µm)2 C = = =5.0 × 10-17 F . d 0.20 µm (b) Let the number of elementary charges in question be N. Then, the total amount of charges that appear in the gate is q = Ne. Thus, q = Ne = CV , which gives CV (5.0 × 10-17 F)(1.0V) N = = =3.1 × 102 . e 1.6 × 10-19 C