solutions_27 - PHY2061 R. D. Field Chapter 27 Solutions...

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PHY2061 R. D. Field Solutions Chapter 27 Page 1 of 8 Chapter 27 Solutions Problem 1: Consider a spherical conducting shell S 1 of radius R on which charge +Q is placed. Without touching or disturbing it, this shell is now surrounded concentrically by a similar shell S 2 of radius 2R on which charge -Q is placed (see Figure). What is the magnitude of the electric field in the region between the two shells ( R <r < 2R )? Answer: KQ/r 2 Solution: Gauss’ Law tells us that Φ E Surface enclosed EdA Q =⋅ = rr ε 0 . If we choose our Gaussian surface to be a sphere with radius r such that R < r < 2R then we get E(r)4 π r 2 = Q/ ε 0 and solving for E gives E = KQ/r 2 . Problem 2: In the previous problem, what is the electric field inside shell S 1 (r < R)? Answer: zero Solution: Gauss’ Law tells us that Φ E Surface enclosed Q = 0 . If we choose our Gaussian surface to be a sphere with radius r such that r < R then we get E(r)4 π r 2 = 0 , since Q enclosed = 0 , and solving for E gives E = 0 . Problem 3: A solid insulating sphere of radius R has charge distributed uniformly throughout its volume. What fraction of the sphere&s total charge is located within the region r < R/2 ? Answer: 1/8 Solution: The amount of charge, Q(r) , within the radius r < R is given by R 2R Shell S 1 has charge +Q Shell S 2 has charge -Q
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PHY2061 R. D. Field Solutions Chapter 27 Page 2 of 8 ∫∫ = = = rr r dr r dV r Q 00 3 2 3 4 4 ) ( ρ π , where ρ is the volume charge density and dV = 4 π r 2 dr . Thus, 3 ) ( = R r Q r Q , where Q = 4 π R 3 ρ /3 is the sphere&s total charge. For r = R/2 , (r/R) 3 = 1/8 . Problem 4: A solid insulating sphere of radius R has a non-uniform volume charge distribution given by ρ (r) = ar , where a is a constant. What is the total charge Q of the insulating sphere? Answer: π aR 4 Solution: The amount of charge, Q , within the sphere is given by = = = = RR R aR dr r a dr r ar dV r r Q 4 0 3 2 4 4 ) ( ) ( ) ( . Problem 5: In a certain region of space within a distribution of charge the electric field is given by r ar r E ± ) ( = r . It points radially away from the origin and has a magnitude E(r) = ar , where a = 150N/(Cm) . How much electric charge ( in nanoC ) is located inside a shell with an inner radius of 0.5 meters and an outer radius of 1.0 meters ? Answer: 14.6 Solution: Gauss’ Law tells us that nC C Nm m m Cm N r r K a r E r r E r r E r r E r dA E Q enclosed 6 . 14 / 10 99 . 8 ) ) 5 . 0 ( ) 1 ))(( /( 150 ( ) ( )) ( ) ( ( 4 )) ( 4 ) ( 4 ( 2 2 9 3 3 3 1 3 2 1 2 1 2 2 2 0 1 2 1 2 2 2 0 0 = × = = = = = πε ε
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PHY2061 R. D. Field Solutions Chapter 27 Page 3 of 8 Problem 6:
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This note was uploaded on 05/31/2011 for the course PHY 2061 taught by Professor Fry during the Spring '08 term at University of Florida.

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solutions_27 - PHY2061 R. D. Field Chapter 27 Solutions...

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