2009S2UnitExam1BSolutions

2009S2UnitExam1BSolutions - ID: B UCSD Physics 2B Answer...

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ID: B 1 UCSD Physics 2B Unit Exam 1B Charges, Force & Fields Answer Section MULTIPLE CHOICE 1. ANS: B τ = p × E = pE sin θ = 35 × 10 9 C m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 2.2 × 3 N C Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ sin165 °=+ 2.0 × 11 N m TOP: ELECTRIC DIPOLE TORQUE 2. ANS: A To apply the line/cylinder formula, we need the linear charge density λ . Pick some length L for a sample section of cylinder. Then the lateral surface area of that section will be A = 2 π RL and the total charge is the area times the surface charge density Q = 2 RL σ . Now = Q L = 2 RL L = 2 R so that E = 2 k r = 2 k 2 R r = 4 k R r = 49 .0 0 × 9 Nm 2 / C 2 Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 3.14 () 19.00 × 2 m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 27.0 × 9 C / m 2 Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 31.0 × 2 m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ = 1.87 × 3 N = kN C The much easier way is to apply Gauss's Law from scratch: Charged Cylinder Q = 2 RL Gaussian Cylinder Area A = 2 rL E = Q ε 0 A = 2 RL 2 rL 0 = R r 0 = 19 31 27.0 × 9 8.85 × 12 Ê Ë Á Á Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ Note the extra advantage of not having to convert centimeters to meters since the units cancel TOP: ELECTRIC FIELD CYLINDER 3. ANS: A This is one of the basic facts about charges - from grade school, I think. TOP: CHARGE CONCEPT 4. ANS: A The electric field inside any ideal conductor, even if it’s hollow & empty inside, is always zero. TOP: CONDUCTOR FIELD CONCEPT
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ID: B 2 5. 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 ¯ ˜ ˜ ˜ ˜
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2009S2UnitExam1BSolutions - ID: B UCSD Physics 2B Answer...

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