hw_ch20 - Instructor Tomoyuki Nakayama Tuesday PHY 2005...

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Instructor: Tomoyuki Nakayama Tuesday, May 25, 2010 PHY 2005 Applied Physics 2 - Summer C 2010 Solutions for Suggested Homework Problems (Chapter 20) ________________________________________________________________________________ 2. The definition of capacitance ( C = Q/ Δ V ) yields Q = C Δ V = 2.64 × 10 -6 C. 6. The capacitance of a parallel plate capacitor is given by C = ε 0 A/d . We solve it for the area of the plate to obtain A = Cd/ ε 0 = 102 m 2 11. (a) If capacitors are connected in series, the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances. Therefore, we get 1/ C eq = 1/2 + 1/7 + 1/14 = 10/14 C eq = 1.4 μ F. (b) If capacitors are connected in parallel, the equivalent capacitance is the sum of the individual capacitances. We have C eq = 2+7+14 = 23 μ F. 13. (a) When multiple capacitors are connected in parallel across a battery, the potential drop across each capacitor is the same as the potential difference provided by the battery. We obtain Q 1 = C 1 Δ V = 24 C, Q 2 = C 2 Δ V = 72 C and Q 3 = C 3 Δ V = 144 C (b) The equivalent capacitance is 1/ C eq = 1/2 + 1/6 + 1/12 = 3/4 C eq = 1.33 μ F. Since the capacitors are connected in series, the charge stored in each
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This note was uploaded on 07/16/2011 for the course PHY 2005 taught by Professor Lee during the Summer '08 term at University of Florida.

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hw_ch20 - Instructor Tomoyuki Nakayama Tuesday PHY 2005...

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