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Fa06Soln7

# Fa06Soln7 - Physics 7B Fall 2006 Homework 7 Solutions...

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Physics 7B Fall, 2006 Homework 7 Solutions 22.12 ( a ) The charge on a conducting sphere must be on the surface. If we construct a spherical Gaussian surface inside the metal sphere, there will be no enclosed charge and thus E a = 0 . ( b ) Because the point is still inside the metal sphere, we have E b = 0 . ( c ) Because the point is outside the metal sphere, the field is radial, given by E c = Q /4 π Å 0 r c 2 = (– 3.50 × 10 –6 C)/4 π (8.85 × 10 –12 C 2 /N · m 2 )(3.10 m) 2 = – 3.28 × 10 3 N/C (toward the sphere). ( d ) Because the point is outside the metal sphere, the field is radial, given by E d = Q /4 π Å 0 r d 2 = (– 3.50 × 10 –6 C)/4 π (8.85 × 10 –12 C 2 /N · m 2 )(6.00 m) 2 = – 8.75 × 10 2 N/C (toward the sphere). ( e ) Because the charge on the shell is the same as the charge on the surface of the metal sphere, all of the fields will be the same . ( f ) For points inside the nonconducting sphere, only the charge inside a spherical surface with a radius to the point will provide the field: E ( r r 0 ) = [( Q / 9 π r 0 3 )( 9 π r 3 )]/4 π Å 0 r 2 = Qr /4 π Å 0 r 0 3 . Thus we have E a = (– 3.50 × 10 –6 C)(0.15 m)/4 π (8.85 × 10 –12 C 2 /N · m 2 )(3.00 m) 3 = – 1.75 × 10 2 N/C (toward the center). E b = (– 3.50 × 10 –6 C)(2.90 m)/4 π (8.85 × 10 –12 C 2 /N · m 2 )(3.00 m) 3 = – 3.38 × 10 3 N/C (toward the center). At points outside the nonconducting sphere, the field will be the same as before; E c = – 3.28 × 10 3 N/C (toward the sphere); E d = – 8.75 × 10 2 N/C (toward the sphere). 22.22 [For reference for problem 22.23] From the symmetry of the charge distribution, we know that the electric field must be radial, with a magnitude independent of the direction. For a Gaussian surface we choose a sphere of radius r . On this surface, the field has a constant magnitude and E and d A are parallel, so we have E · d A = E d A . The charge density is ρ = Q /[ 9 π ( r 0 3 r 1 3 )]. ( a ) For the region where r < r 1 , there is no charge inside the Gaussian surface, so we have E = 0; r < r 1 . ( b ) For the region where r 1 < r < r 0 , we apply Gauss’s law: ı E · d A = EA = Q enclosed / ε 0 ; r r 1 r 0 Q

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E 4 π r 2 = ρ 9 π ( r 3 r 1 3 )/ Å 0 , which gives E = ρ 9 π ( r 3 r 1 3 )/4 π Å 0 r 2 = Q ( r 3 r 1 3 )/4 π Å 0 ( r 0 3 r 1 3 ) r 2 ; r 1 < r < r 0 . ( c ) For the region where r > r 0 , the electric field is that of a point charge; E = Q /4 π Å 0 r 2 ; r > r 0 . 22.23 From the symmetry of the charge distribution, we know that the electric field must be radial, with a magnitude independent of the direction. The charge density of the sphere is ρ = Q /[ 9 π ( r 0 3 r 1 3 )]. We can add the field of the point charge at the center to the fields found in Problem 22: ( a ) For the region where r < r 1 , the electric field is that of the point charge at the center: E = q /4 π Å 0 r 2 ; r < r 1 .
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