36. (a) The charge q3 in the figure is q3 = C3V = (4.00 F)(100 V) = 4.00 104 C . (b) V3 = V = 100 V.
1 (c) Using U i = 2 CVi 2 , we have U 3 = 1 C3V32 = 2.00 102 J . i 2
(d) From the figure, q1 = q2 = C1C2V (10.0 F)(5.00 F)(100 V) = = 3.33 104 C. C1 + C2
37. (a) They each store the same charge, so the maximum voltage is across the smallest capacitor. With 100 V across 10 F, then the voltage across the 20 F capacitor is 50 V and the voltage across the 25 F capacitor is 40 V. Therefore, the voltage across t
38. (a) We calculate the charged surface area of the cylindrical volume as follows: A = 2 rh + r 2 = 2 (0.20 m)(0.10 m) + (0.20 m) 2 = 0.25 m2 where we note from the figure that although the bottom is charged, the top is not. Therefore, the charge is q =
40. (a) We use C = 0A/d to solve for d: d=
0 A
C
=
(8.85 10
12
C2 /N m 2 ) (0.35 m 2 )
12
50 10
F
= 6.2 102 m.
(b) We use C . The new capacitance is C' = C(/air) = (50 pf)(5.6/1.0) = 2.8102 pF.
41. The capacitance with the dielectric in place is given by C = C0, where C0 is the capacitance before the dielectric is inserted. The energy stored is given by 1 1 U = 2 CV 2 = 2 C0V 2 , so
=
2U 2(7.4 106 J) = = 4.7. C0V 2 (7.4 1012 F)(652V) 2
According