2009S2UnitExam2BSolutions

2009S2UnitExam2BSolutions - ID: B UCSD Physics 2B Answer...

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ID: B 1 UCSD Physics 2B Unit Exam 2B Potential & Capacitors Answer Section MULTIPLE CHOICE 1. ANS: A If you remember the formula, just plug & play, otherwise derive it from first principles. First find the potential in cylindrical symmetry, and then apply the capacitor definition formula V = E d r a b = 2 k λ r dr a b = 2 k ln b a Ê Ë Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ C = Q V = L 2 k b a Ê Ë Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ = L 2 k b a Ê Ë Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ Now plug in the numbers: C = L 2 k b a Ê Ë Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ = 18 m 29 .0 × 10 9 Nm 2 / C 2 Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 1.0 × 3 m 0.90 × 3 m Ê Ë Á Á Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ = 9.49 × 9 F = 9.49 nF On the other hand you could get a rough answer by treating the coaxial cylinders like a parallel plates (the easiest case). Not exact, but the plates are fairly close together, so the field shouldn’t diverge too much. The surface area A = 2 π rL (you could use r = a or r = b or something in between) and the separation distance is d = b a . Plug in the formula: C = ε 0 A d = 0 2 a + b 2 Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ L b a = 0 2 0.95 () 0.10 = 9.50 Very close to the same answer. .. and a great way to check your work, even if you know the correct formula. Note that I used the mean distance, and since the units of millimeter cancel in numerator & denominator, I didn’t bother. TOP: CAP cylinder 2. ANS: D E =− V x , V y , V z Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ 33 V m , 22 V m 2 y ,66 V m 3 z 2 Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ V m i 8 + V m 2 y j 8 66 V m 3 z 2 k 8 TOP: E from V
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ID: B 2 3. ANS: B Since V = Ed
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2009S2UnitExam2BSolutions - ID: B UCSD Physics 2B Answer...

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