exam_2a_09_sol

exam_2a_09_sol - Phys. 2306 Exam 2 F09 Form A October 16,...

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Phys. 2306 Exam 2 F09 Form A October 16, 2009 0
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Solutions Phys. 2306 Exam 2 F09 Form A In Thomson’s atomic model of Helium, electrons are considered point charges and are like chocolate chips in cookie dough. In this case the dough is the alpha particle, treated as a sphere of radius R with total charge 2e and uni- form volume charge density. The electrons can move freely inside the alpha. Place the origin at the center of the alpha and the two electrons, each of charge -e inside the alpha at (x,y,z)= ( ± d ,0,0), where d < R . 1) Consider the electron at x= d , what magnitude of force does it feel due to the alpha? a) e 2 R 2 π± 0 d 3 b) e 2 d 2 π± 0 R 3 c) e 2 2 π± 0 d 2 d) e 2 ( R - d ) 2 π± 0 d 3 e) none of these The electric field at radius r inside a sphere of radius R with uniform charge density ρ = 2 e/ 4 πR 3 3 C/m 3 depends only on the radial distance r and is directed radially outwards. So draw a Gauss sphere of radius r=d and get, E ( d )4 πd 2 = Q en ± 0 = ((2 e/ 4 πR 3 3 ) ( 0)) 4 πd 3 3 E ( d ) = 2 ed 4 π± 0 R 3 = ed 2 π± 0 R 3 F α = eE ( d ) = e 2 d 2 π± 0 R 3 , ( b ) . 2) What magnitude of force does it feel from the other electron? a) e 2 R 4 π± 0 d 3 b) e 2 16 π± 0 d 3 c) e 2 4 π± 0 d 2 d) e 2 R 16 π± 0 d 3 e) none of these This force is between two point charges so F e = e 2 4 π± 0 (2 d ) 2 = e 2 16 π± 0 d 2 , ( e ). 3) If the electrons are at rest in equilibrium, what is d? a) R/ 2 b) R/ 3 c) R/2 d) R/(2) 1 / 3 e) none of these 1
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F α,e is atractive,repulsive so they can add to zero for equilibrium. 0 = e 2 d 2 π± 0 R 3 - e 2 16 π± 0 d 2 0 = d R 3 - 1 8 d 2 d 3 = R 3 / 8 so d = R/ 2 , ( c ) . 4) Suppose there were no electrons, this is a doubly ionized atom, what is the potential at the surface of the alpha? The potential far away is zero. a) e 4 π± 0 R 2 b) e 2 π± 0 R 2 c) e 2 4 π± 0 R 2 d) e 4 π± 0 R e) none of these Outside the sphere of radius R, E ( r ) = 2 e 4 π± 0 r r 3 . V ( R ) - V ( ) = V ( R ) = - Z R E · d r = - e 2 π± 0 Z R dr/r 2 = e 2 π± 0 (1 /R - 1 / ) = e 2 π± 0 R , ( e ) . 5) Like the last problem, but now we want the ratio of the potential inside the sphere to the potential at R. This ratio r= V ( < R ) /V ( R ) is a) < 1 b) = 1 c) 1 < r 1 . 5 d) 1 . 5 < r < 2 e) none of these We now integrate from to D, where D < R. V
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exam_2a_09_sol - Phys. 2306 Exam 2 F09 Form A October 16,...

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