Homework #2 - Solutions

Homework #2 - Solutions - Physics 107 Homework#2 Solutions...

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Physics 107 – Homework #2 Solutions 19.52 E = σ ε 0 = 0 E = 8.85 × 10 12 C 2 N m 2 7.0 × 10 5 N C ⎟ = 6.2 × 10 6 C/m 2 19.55 The Gaussian surface is chosen to be a cylinder with radius r and length L , and its axis is along the wire. Φ= EA = E (2 π rL ) is assumed to be parallel to the end caps, so only the area of the curved surface is considered. The total charge enclosed by the Gaussian surface is q r E = λ L . Apply Gauss’s law. q 0 = L 0 = E rL ) . Solve for E . E = 2 πε 0 r 20.13 13. (a) W = q Δ V = e Δ V = (1.60 × 10 19 C)(0.070 V) = 1.1 × 10 20 J (b) The work required to “pump” a Na ion out of the cell is independent of the thickness of the cell wall, so the answer to part (a) + 20.15 (a) Δ K =−Δ U =− ( U C U A ) = K A U A U C = K A U C = U A K A Δ K ( U C U B ) = 2 K A U B U C = 2 K A U C = U B 2 K A So, U B 2 K A = U A K A . Find K A . U B 2 K A = U A K A U B U A = K A Find V . C U C = U A K A = U A ( U B U A ) = 2 U A U B V C = 2 V A V B = 2(352 V) 129 V = 575 V Also accept V C = 2(332V) – 149 V = 515 V (b) K A = U B U A eV B ( eV A ) = e ( V A V B ) = (1.60 × 10 19 C)(352 V 129 V) = 3.57 × 10 17 J Also accept K A = (1.6 x 10 –19 C ) (332 V – 149 V) = 2.93x 10 –17 J
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20.17 1 2 mv 2 = e Δ V v = 2 e Δ V m = 2(1.60 × 10 19 C)(25,000 V) 9.11 × 10 31 kg = 9.4 × 10
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Homework #2 - Solutions - Physics 107 Homework#2 Solutions...

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