18 16 chapter 18 b from figure 1866c from figure

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18-16 Chapter 18
(b) From Figure 18.66c, From Figure 18.66b, This one capacitor stores nearly half the total stored energy. 18.67. Set Up: For a parallel-plate capacitor, Solve: (a) U is proportional to so increase the charge by a factor of to double the stored energy. We didn’t use so this result applies to any capacitor. (b) Still connected to the battery means V stays constant and Q changes. means (c) goes to so Reflect: In parts (b) and (c) the change causes the charge on the capacitor to increase and this increases the electric field between the plates and therefore increases the stored energy. 18.68. Set Up: The field when there is air between the plates is The field when the dielectric is between the plates is Solve: 18.69. Set Up: for a parallel plate capacitor; this equation applies whether or not a dielectric is present. Solve: (a) per (b) Reflect: The dielectric material increases the capacitance and decreases the electric field that corresponds to a given potential difference. 18.70. Set Up: The capacitance with the dielectric present is The dielectric strength is the maximum allowed field between the plates. Solve: 18.71. Set Up: Solve: (a) With the dielectric, before: after: (b) The energy increased. Reflect: The power supply must put additional charge on the plates to maintain the same potential difference when the dielectric is inserted. so the stored energy increases. U 5 1 2 QV , D U 5 13.5 mJ 2 3.6 mJ 5 9.9 mJ. U 5 1 2 CV 2 5 1 2 1 46.9 3 10 2 6 F 21 24.0 V 2 2 5 13.5 mJ U 5 1 2 C 0 V 2 5 1 2 1 12.5 3 10 2 6 F 21 24.0 V 2 2 5 3.60 mJ C 5 1 3.75 21 12.5 m F 2 5 46.9 m F. U 5 1 2 CV 2 . C 5 KC 0 . A 5 Cd K P 0 5 1 1.50 3 10 2 9 F 21 2.00 3 10 2 4 m 2 1 3.20 21 8.854 3 10 2 12 C 2 / 1 N # m 2 22 5 0.0106 m 2 . d 5 V ab E 5 4.00 3 10 3 V 20.0 3 10 6 V / m 5 2.00 3 10 2 4 m. V ab 5 Ed . C 5 KC 0 5 K P 0 A d . E 5 V Kd 5 85 mV 1 10 21 7.5 3 10 2 9 m 2 5 1.13 3 10 6 V / m. cm 2 . C 5 1 10 2 1 8.85 3 10 2 12 F / m 21 1.0 3 10 2 4 m 2 2 7.5 3 10 2 9 m 5 1.18 m F V 5 Ed A 5 1.0 cm 2 5 1.0 3 10 2 4 m 2 . C 5 KC 0 5 K P 0 A d . K 5 E 0 E 5 2.03 3 10 6 V / m 0.330 3 10 6 V / m 5 6.15. E 0 5 Q P 0 A 5 0.180 3 10 2 6 C 1 8.854 3 10 2 12 C 2 / 1 N # m 2 221 100 3 10 2 4 m 2 2 5 2.03 3 10 6 V / m. E 5 E 0 K . E 0 5 Q P 0 A . U S 9 U 5 90 J. V S 3 V V 5 75 V. V 5 25 V U 5 1 2 CV 2 . U S 2 U 0 . d S d / 2 U 5 1 2 CV 2 5 1 2 1 P 0 A d 2 V 2 . C 5 P 0 A d " 2 Q 2 U 5 1 2 CV 2 5 1 2 C 1 Q / C 2 2 5 Q 2 / 1 2 C 2 . C 5 P 0 A d . C 5 Q V . U 5 1 2 CV 2 . U 4.8 5 1 2 CV 2 5 1 2 1 4.80 3 10 2 6 F 21 5.48 V 2 2 5 7.21 3 10 2 5 J 5 72.1 m J V 4.8 5 Q 4.8 C 4.8 5 2.63 3 10 2 5 C 4.80 3 10 2 6 F 5 5.48 V. Q tot 5 2.63 3 10 2 5 C. Q tot 5 C eq V 5 1 2.19 3 10 2 6 F 21 12.0 V 2 5 2.63 3 10 2 5 C. Electric Potential and Capacitance 18-17
18.72. Set Up: The capacitance with the dielectric present is The dielectric strength is the maximum allowed field between the plates. Solve: (a) (b) 18.73. Set Up: Solve: 18.74. Set Up: The potential of a point charge is Figure 18.74 Solve: (a) The positions of the two charges are shown in Figure 18.74. (b) Any point on the x axis is the same distance from each charge, so The potential is zero at all points on the x axis.

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