2009S2UnitExam1ASolutions

# 2009S2UnitExam1ASolutions - ID: A UCSD Physics 2B Answer...

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ID: A 1 UCSD Physics 2B Unit Exam 1A Charges, Force & Fields Answer Section MULTIPLE CHOICE 1. ANS: E This is one of the basic facts about charges - from grade school, I think. TOP: CHARGE CONCEPT 2. ANS: C 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 R so that E = 2 k r = 2 k 2 R r = 4 k R r = 49 .0 0 × 10 9 Nm 2 / C 2 Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 3.14 () 12.00 × 2 m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 14.0 × 9 C / m 2 Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 13.0 × 2 m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ = 1,460 N C = 1.46 kN C The much easier way is to apply Gauss's Law from scratch: Charged Cylinder Q = 2 RL Gaussian Cylinder around the Charged Cylinder Area A = 2 rL E = Q ε 0 A = 2 RL 2 rL 0 = R r 0 TOP: ELECTRIC FIELD CYLINDER 3. ANS: A By definition, E = F q = 6.9 × 2 N 3.3 × 3 C = 2.1 × 5 N C TOP: ELECTRIC FIELD

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ID: A 2 4. ANS: B 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 Note that the field inside due to the outer charged shell is zero. 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
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## This note was uploaded on 04/04/2010 for the course PHYSICS 2B taught by Professor Lever during the Summer '09 term at UCSD.

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2009S2UnitExam1ASolutions - ID: A UCSD Physics 2B Answer...

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