UnitExam1Bsolutions

UnitExam1Bsolutions - ID: B UCSD Physics 2B Answer Section...

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ID: B 1 UCSD Physics 2B Unit Exam 1B Charges, Force & Fields Answer Section MULTIPLE CHOICE 1. ANS: B Make a ratio of the two forces using the fact that the only change is that the new distance is r new = 1.65 * r F new F = k q 1 q 2 r new Ê Ë Á ˆ ¯ ˜ 2 k q 1 q 2 r () 2 = r r new Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ 2 F = F r r new Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ 2 = 22 N 1 1.65 Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ 2 = 8.08 N TOP: COULOMB FORCE CONCEPT 2. ANS: D The electric field from a point charge is E = kQ r 2 . Outside a charged spherical shell, the field looks the same as if the whole charge were located in a point at the center. Halfway between the shells, the distance from their center is 3 2 R . Hence the field is E = r 2 = 3 2 R Ê Ë Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ 2 = 4 9 R The other way to do this is to construct a Gaussian Surface at r = 3 R /2. The enclosed charge is + Q and the surface area A = 4 π r 2 = 4 3 2 R Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ 2 = 9 R 2 . The electric field is E = Q enc ε 0 A = + Q 4 πε 0 9 4 R 2 Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ = 4 9 R 2 where we used k = 1 4 0 TOP: ELECTRIC FIELD SPHERICAL 3. ANS: D Fundamental properties of conductors : Electric field is zero inside Electric field is perpendicular at surface Excess charge must reside at surface only TOP: CONDUCTOR FIELD CONCEPT
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ID: B 2 4. ANS: A For a sphere with radial symmetry, we can take the volume element to be the product or the surface of a spherical shell times its thickness dV = A surface × dr = 4 π r 2 . We integrate the charge elements dq = ρ r ()
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UnitExam1Bsolutions - ID: B UCSD Physics 2B Answer Section...

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