ass12 - SOLUTIONS, ASSIGNMENT 12 24.10. I DENTIFY :...

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Unformatted text preview: SOLUTIONS, ASSIGNMENT 12 24.10. I DENTIFY : Capacitance depends on the geometry of the object. (a) S ET U P : The capacitance of a cylindrical capacitor is ( 29 2 ln / b a L C r r = P . Solving for r b gives 2 / L C b a r r e = P . E XECUTE : Substituting in the numbers for the exponent gives ( 29 12 2 2 11 2 8.85 10 C /N m (0.120 m) 0.182 3.67 10 F -- = Now use this value to calculate r b : r b = r a e 0.182 = (0.250 cm) e 0.182 = 0.300 cm (b) S ET U P : For any capacitor, C = Q/V and = Q/L . Combining these equations and substituting the numbers gives = Q/L = CV/L. E XECUTE : Numerically we get ( 29 ( 29 11 8 3.67 10 F 125 V 3.82 10 C/m = 38.2 nC/m 0.120 m CV L -- = = = E VALUATE : The distance between the surfaces of the two cylinders would be only 0.050 cm, which is just 0.50 mm. These cylinders would have to be carefully constructed. 24.11. I DENTIFY and S ET U P : Use the expression for C / L derived in Example 24.4. Then use Eq.(24.1) to calculate Q. E XECUTE : (a) From Example 24.4, ( 29 2 ln / b a C L r r = P ( 29 ( 29 12 2 2 11 2 8.854 10 C / N m 6.57 10 F/m 66 pF/m ln 3.5 mm/1.5 mm C L -- = = = (b) ( 29 ( 29 11 10 6.57 10 F/m 2.8 m 1.84 10 F. C-- = = ( 29 ( 29 10 3 11 1.84 10 F 350 10 V 6.4 10 C 64 pC Q CV--- = = = = 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. E VALUATE : C depends only on the dimensions of the capacitor. Q and V are proportional....
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ass12 - SOLUTIONS, ASSIGNMENT 12 24.10. I DENTIFY :...

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