8.022pset4solution

8.022pset4solution - Massachusetts Institute of Technology...

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Massachusetts Institute of Technology Department of Physics 8.022 Spring 2004 Solution 4:Capacitance, currents, resistance. TA: Yi Mao, [email protected] 1. Purcell 3.17: Spherical capacitor. b a + + + + + + + + - - - - - - - - σ σ out inn - Figure 1: Spherical vacuum capacitor. We frst consider the capacitance C o± a spherical vacuum capacitor. Assume a total charge Q is uni±ormly distributed on the sur±ace o± inner sphere and -Q on the outer sur- ±ace. The electric feld between the spheres are E ( r ) = Q/r 2 and hence δV = Q ( 1 b - 1 a ). So C = Q/δV = 1 / ( 1 b - 1 a ). The electric feld on the sur±ace o± inner sphere is E inn = 4 πσ inn E 0 . The energy stored in this capacitor is W = 1 2 Q 2 C = 1 2 ( σ inn 4 πb 2 ) 2 C 1 2 E 2 0 b 4 /C. Put everything together, we get W max ( b ) = 1 2 E 2 0 ( b 3 - b 4 a ) , (1)

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which is regarded as a function of b. Maximize W max ( b ) by W 0 max ( b ) = 0. We get b = 3 a/ 4 (2) W max = (27 / 512) a 3 E 2 0 . (3) 2. Purcell 3.23: Cylindrical capacitor. a b + - + + + + + + + + + + + - - - - - - - - - - - L Figure 2: Cylindrial vacuum capacitor (solid lines). The dashed lines are a Gaussian surface to calculate the electric ±eld in between. (a) Assume a total charge Q is uniformly distributed on the surface of inner cylinder of radius a, and -Q on that of outer cylinder of radius b. Consider a Gaussian surface that is a coaxial cylinder of radius r ( a < r < b ) and length L plus horizontal cross-sections on the top and bottom. Gauss’s law reads E ( r ) × 2 πrL = 4 πQ, or E ( r ) = 2 Q/ ( rL ) for a < r < b . So the potential di²erence V = Z b a E ( r ) dr = 2 Q L ln b a . C = Q V = L/ (2 ln b a ) . (4) As s = b - a ¿ a , ln b a = ln (1 + s a ) s a . C aL 2 s = A 4 πs , (5)
where A = 2 πaL is the surface area. This is exactly the capacitance of a parallel-plate capacitor of area A and seperation s. (b) Let the overlapping length of the cylinders be L, now a variable. The energy stored in this capacitor is W ( L ) = 1 / 2 CV 2 12 where V 12 is the potential diFerence between cylinders. W

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This homework help was uploaded on 04/11/2008 for the course PHYSICS 8.022 taught by Professor Hughes during the Spring '08 term at MIT.

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8.022pset4solution - Massachusetts Institute of Technology...

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