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Unformatted text preview: 0.50 mm. These cylinders would have to be carefully constructed. 8# Capacitance and Dielectrics 24.11. 24-3 IDENTIFY and SET UP:! Use the expression for C/L derived in Example 24.4. Then use Eq.(24.1) to calculate Q. C 2, !0 EXECUTE:! (a) From Example 24.4, # L ln . rb / ra / !12 2 2 C 2, . 8.854 " 10 C / N + m / # # 6.57 " 10!11 F/m # 66 pF/m ln . 3.5 mm/1.5 mm / L (b) C # . 6.57 " 10!11 F/m / . 2.8 m / # 1.84 " 10!10 F. Q # CV # .1.84 " 10!10 F / . 350 " 10!3 V / # 6.4 " 10!11 C # 64 pC 24.12. The conductor at higher potential has the positive charge, so there is +64 pC on the inner conductor and !64 pC on the outer conductor. EVALUATE:! C depends only on the dimensions of the capacitor. Q and V are proportional. IDENTIFY:! Apply the results of Example 24.3. C # Q / V . SET UP:! ra # 15.0 cm . Solve for rb . 1% r r & EXECUTE:! (a) For two concentric spherical shells, the capacitance is C # ' a b ( . kCrb ! kCra # ra rb and k ) rb ! ra * !12 kCra k (116 " 10 F)(0.150 m) rb # # # 0.175 m . kC ! ra k (116 " 10!12 F) ! 0.150 m (b) V # 220 V and Q # CV # (116 " 10!12 F)(220 V) # 2.55 " 10!8 C . EVALUATE:! A parallel-plate capacitor with A # 4, ra rb # 0.33 m 2 and d # rb ! ra # 2.5 " 10!2 m has !0 A # 117 pF , in excellent agreement with the value of C for the spherical capacitor. d IDENTIFY:! We can use the definition of capacitance to find the capacitance of the capacitor, and then relate the capacitance to geometry to find the inner radius. (a) SET UP:! By the definition of capacitance, C = Q/V. C# 24.13. EXECUTE:! C # Q 3.30 " 10!9 C # # 1.50 " 10!11 F = 15.0 pF V 2.20 " 102 V (b) SET UP:! The capacitance of a spherical capacitor is C # 4, !0 ra rb . rb ! ra EXECUTE:! Solve for ra and evaluate using C = 15.0 pF and rb = 4.00 cm, giving ra = 3.09 cm. (c) SET UP:! We can treat the inner sphere as a point-charge located at its center and use Coulomb’s law, 1q E# . 4, !0 r 2 . 9.00 "...
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## This document was uploaded on 03/11/2014 for the course PHYSICS 240 at University of Michigan.

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