3dfe285p 7 g a76 767m b am65

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Unformatted text preview: 1.00 " 10!12 F EVALUATE:! The separation between the pennies is nearly a factor of 10 smaller than the diameter of a penny, so it is a reasonable approximation to treat them as infinite sheets. Q !A INCREASE:! C # . Vab # Ed . C # 0 . Vab d EXECUTE:! C # 24.8. SET UP:! We want E # 1.00 " 104 N/C when V # 100 V . V 1.00 " 102 V EXECUTE:! (a) d # ab # # 1.00 " 10!2 m # 1.00 cm . E 1.00 " 104 N/C A# Cd (5.00 " 10!12 F)(1.00 " 10!2 m) A # 4.24 " 10!2 m # 4.24 cm . # # 5.65 " 10!3 m 2 . A # , r 2 so r # !12 2 2 !0 8.854 " 10 C /(N + m ) , (b) Q # CVab # (5.00 " 10!12 F)(1.00 " 102 V) # 5.00 " 10!10 C # 500 pC !0 A . We could have a larger d, along with a larger A, and still achieve the required C without d exceeding the maximum allowed E. IDENTIFY:! Apply the results of Example 24.4. C # Q / V . EVALUATE:! C # 24.9. SET UP:! ra # 0.50 mm , rb # 5.00 mm EXECUTE:! (a) C # L 2, !0 (0.180 m)2, !0 # # 4.35 " 10!12 F . ln(rb ra ) ln(5.00 0.50) (b) V # Q / C # (10.0 " 10!12 C) /(4.35 " 10!12 F) # 2.30 V C # 24.2 pF . This value is similar to those in Example 24.4. The capacitance is determined entirely by L the dimensions of the cylinders. IDENTIFY:! Capacitance depends on the geometry of the object. 2, !0 L (a) SET UP:! The capacitance of a cylindrical capacitor is C # . Solving for rb gives rb # ra e 2, !0 L / C . ln . rb / ra / EVALUATE:! 24.10. EXECUTE:! Substituting in the numbers for the exponent gives 2, . 8.85 " 10!12 C2 /N + m 2 / (0.120 m) # 0.182 3.67 " 10!11 F Now use this value to calculate rb: rb = ra e0.182 = (0.250 cm)e0.182 = 0.300 cm (b) SET UP:! For any capacitor, C = Q/V and 8 = Q/L. Combining these equations and substituting the numbers gives 8 = Q/L = CV/L. EXECUTE:! Numerically we get !11 CV . 3.67 " 10 F / .125 V / # # 3.82 " 10!8 C/m = 38.2 nC/m 0.120 m L EVALUATE:! The distance between the surfaces of the two cylinders would be only 0.050 cm, which is just...
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