2009S2UnitExam2ASolutions

# 2009S2UnitExam2ASolutions - ID A UCSD Physics 2B Answer...

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ID: A 1 UCSD Physics 2B Unit Exam 2A Potential & Capacitors Answer Section MULTIPLE CHOICE 1. ANS: B The field is uniform between parallel plates, so V ||=− E dL 0 d | | | | | | | | = Ed = 150 × 10 3 V / m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 3.77 × 2 m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ = 5.66 × 3 V = 5.66 kV TOP: V from E 2. ANS: D Just apply the formula & plug & play W = 1 2 CV 2 = 1 2 55 × 3 F Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 790 V () 2 = 1.72 × 4 J TOP: CAP Energy 3. ANS: A We first have to reduce the circuit before we can calculate anything at all. C 1 and C 2 are in parallel so the capacitances ADD: C 12 = C 1 + C 2 = 9.0 μ F . The network C12 is in series with C3, so 1 C series = 1 C + 1 C 3 = 1 9.0 F + 1 3.0 F = 4 9.0 F Hence the equivalent capacitance C = 2.25 F . One way to proceed is to use U 3 = Q 3 2 2 C 3 and the fact that Q 3 must be the same as Q TOT for the whole network. Thus Q TOT = C TOT V TOT = 2.25 × 6 F Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ V = 2.7 × 5 C so that U 3 = Q 3 2 2 C 3 = 2.7 × 5 C Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 2 23 .0 × 6 F Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ = 1.22 × 4 J But the fastest way is to use Q = C 3 V 3 = C V which splits the voltages inversely to the ratio of the capacitances. V 3 V = C C 3 = 9 F 3 F = 3 1 . Since the two voltages must add to 12V total, we V = 3 V and V 3 = 9 V and we can immediately use the equivalent energy formula: U 3 = 1 2 C 3 V 3 2 = 1 2 3.0 F Ê Ë Á ˆ ¯ ˜ 9.0 V 2 = 1.22 × 4 J = 122 J TOP: CAP network Energy

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ID: A 2 4. ANS: B We solve this by equating the Energy gain through Work done on the electron by the field W = q Δ V with the final kinetic energy of the moving electron KE = 1 2 mv 2 W = KE v = 2 q Δ V m = 21 .6 0 × 10 19 C Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 24.0 × 3 V Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 9.11 × 31 kg Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ = 9.18 × 7 m s TOP:
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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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2009S2UnitExam2ASolutions - ID A UCSD Physics 2B Answer...

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