106-20102-FIN - PROBLEM (1) (20 points) A solid insulating...

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Phys 106 Final Examination Page 1 Monday, 30-May, 2011 PROBLEM (1) (20 points) A solid insulating sphere of radius R has a volume charge density that varies with r according to the expression " = # r 2 where is a positive constant and r < R is measured from the center of the sphere. Concentric with the sphere is a conducting spherical shell with inner radius 2 R and outer radius 3 R , and the shell has a net positive charge Q . Express your answers in terms of the given quantities and fundamental constants as needed. R 2 R 3 R A B (a) (4 pts) Find the charge contained within the solid insulating sphere of radius R . Consider a charge element dq as a spherical shell of radius r and thickness dr inside the insulating sphere. The spherical shell has a volume dV = 4 r 2 dr . q = charge contained within solid insulating sphere q = dq 0 R " = dV 0 R " = $ r 2 (4 % r 2 dr ) 0 R " q = 4 "# r 4 dr 0 R $ = 4 r 5 5 % & ' ( ) * 0 R q = 4 R 5 5 (b) (6 pts) Using Gauss's law, find an expression for the electric field in the regions (i) R < r < 2 R , (ii) 2 R < r < 3 R , and (iii) r > 3 R . State the direction of the electric field in each region. (i) Consider a spherical gaussian surface of radius r such that R < r < 2 R , and concentric with the solid sphere. " E = ! E # $ d ! A = E cos dA = q in & 0 # E 4 r 2 = q in 0 = q 0 = 4 "$ R 5 5 0 E = R 5 5 0 r 2 (for R < r < 2 R , directed radially outward) (ii) = 0 (for 2 R < r < 3 R , inside a conductor) (iii) Following the steps in part (i): E 4 r 2 = q in 0 = q + Q 0 = 4 R 5 5 0 + Q 0 E = R 5 5 0 r 2 + Q 4 $# 0 r 2 (for r > 3 R ., directed radially outward) (c) (4 pts) Find the surface charge densities ! in and out on the inner and outer surfaces of the conducting spherical shell, respectively. Consider a spherical gaussian surface of radius r such that R < r < 3 R , and concentric with the solid sphere. " E = ! E # $ d ! A = E cos dA = q in 0 # E = 0 and hence q in = 0. So, q in = q + 4 (2 R ) 2 in = 0 in = # q 16 R 2 = # 1 16 R 2 4 $% R 5 5 ---> in = # R 3 20 Q = 4 (2 R ) 2 in + 4 (3 R ) 2 out = 16 R 2 in + 36 R 2 out out = Q 36 R 2 $ 16 R 2 in 36 R 2 = Q 36 R 2 $ 4 in 9 out = Q 36 R 2 $ 4 9 ( $ R 3 20 ) ---> out = Q 36 R 2 + R 3 45 Notice: There may be alternative solutions. (d) (6 pts) Find the potential difference V B ! V A between the points A and B , shown in the figure. V B " V A = " ! E # d ! s = A B $ " E R 3 R $ ds (cos 0 ° ) = " Edr R 3 R $ V B " V A = " Edr R 2 R # " Edr 2 R 3 R # = " Edr R 2 R # " 0 V B " V A = " R 5 5 0 r 2 dr R 2 R % = " R 5 5 0 [ " 1 r ] R 2 R = R 5 5 0 ( 1 2 R " 1 R ) V B " V A = " R 4 10 0
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Phys 106 Final Examination Page 2 Monday, 30-May, 2011 PROBLEM (2) (20 points) The switch S in the figure is open and the two capacitors C and 2 C are uncharged. Express your answers in terms of the given quantities and fundamental constants as needed.
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106-20102-FIN - PROBLEM (1) (20 points) A solid insulating...

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