3j70d363g0re1b303 00145

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Unformatted text preview: Q decreases and U decreases. Q decreases and E # Q / !0 A so E decreases. We come to the same conclusion from E # V / d . (c) E # 24.52. 24.53. 24.54. 24.55. IDENTIFY:! C # KC0 # K !0 A . V # Ed for a parallel plate capacitor; this equation applies whether or not a dielectric d is present. SET UP:! A # 1.0 cm 2 # 1.0 " 10!4 m 2 . (8.85 " 10!12 F/m)(1.0 " 10!4 m2 ) EXECUTE:! (a) C # (10) # 1.18 $ F per cm2. 7.5 " 10!9 m V 85 mV (b) E # # # 1.13 " 107 V/m . K 7.5 " 10!9 m EVALUATE:! The dielectric material increases the capacitance. If the dielectric were not present, the same charge density on the faces of the membrane would produce a larger potential difference across the membrane. IDENTIFY:! P # E/t , where E is the total light energy output. The energy stored in the capacitor is U # 1 CV 2 . 2 SET UP:! E # 0.95U EXECUTE:! (a) The power output is 600 W, and 95% of the original energy is converted, so E # Pt # (2.70 " 105 W)(1.48 " 10!3 s) # 400 J . E0 # 400 J # 421 J . 0.95 2U 2(421 J) 2 1 # 0.054 F . (b) U # 2 CV so C # 2 # V (125 V) 2 EVALUATE:! For a given V, the stored energy increases linearly with C. !A IDENTIFY:! C # 0 d SET UP:! A # 4.2 " 10!5 m 2 . The original separation between the plates is d # 0.700 " 10 !3 m . d - is the separation between the plates at the new value of C. A! (4.20 " 10!5 m 2 )!0 # 5.31" 10!13 F . The new value of C is C # C0 0 0.25 pF # 7.81 " 10!13 F . EXECUTE:! C0 # 0 # d 7.00 " 10!4 m A! (4.20 " 10!5 m 2 )!0 A! But C # 0 , so d - # 0 # # 4.76 " 10!4 m . Therefore the key must be depressed by a distance of C 7.81 " 10!13 F d7.00 " 10!4 m ! 4.76 " 10!4 m # 0.224 mm . EVALUATE:! When the key is depressed, d decreases and C increases. 2, !0 L IDENTIFY:! Example 24.4 shows that C # for a cylindrical capacitor. ln(rb / ra ) SET UP:! ln(1 0 x ) C x when x is small. The area of each conductor is approximately A # 2, ra L . 2, !0 L 2, !0 L 2, !0...
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