ps05soln - Physics 217 - Homework 5 Solutions Ruxandra...

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Unformatted text preview: Physics 217 - Homework 5 Solutions Ruxandra Bondarescu March 4, 2005 Problem 1. Purcell 3.20 The potential outside a sphere of charge Q is ( r ) = Q/r . The capacitance a conductor in the form of a sphere is C = Q/ ( R ) = R where R is the radius of the sphere. The Capacitance for a prolate spheroid is given to be C = 2 aepsilon1 ln parenleftBig 1+ epsilon1 1- epsilon1 parenrightBig (1) where epsilon1 = radicalbig 1- b 2 /a 2 is the eccentricity. In the limit that b=a, epsilon1 0. As epsilon1 0 in Eq. (1) we have an inderminate of the form . So, we apply LHopitals rule. The equation becomes lim epsilon1 C = lim epsilon1 2 a 2 / (1- epsilon1 2 ) = a (2) So, the in this limit the capacitance C = a , where a = b is the radius of the sphere and it agrees with the formula we derived above. As we go from a sphere to a very prolate spheroid ( b << a ) epsilon1 = sqrt 1- b 2 a 2 = 1- b 2 2 a 2 (3) So as a becomes very large C = 2 a ln parenleftBig 2 b 2 / (2 a 2 ) parenrightBig a- ln ( b/a ) (4) Now, ab 2 = constant V and thus b radicalbig V/a C a 3 / 2 lna lim a a lna lim a 1 1 /a (5) The total charge remains constant. So, the energy of the spheroid U = Q 2 / 2 C 0 as a becomes very large. The energy of the sphere remains constant and positive and thus it is larger than the energy of the spheroid. Since states tend to go to the lowest energy configuration, it is energeticallyenergy of the spheroid....
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This note was uploaded on 08/07/2008 for the course PHYS 2217 taught by Professor Leclair, a during the Spring '06 term at Cornell University (Engineering School).

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ps05soln - Physics 217 - Homework 5 Solutions Ruxandra...

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