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Unformatted text preview: Physics 1B: Electricity & Magnetism Summer Session I, 2010 Quiz #2, July 8, 2010 This is version A ! Useful coefficients: Mass of proton, m p = 1 . 67 × 10 − 27 kg; Mass of electron, m e = 9 . 11 × 10 − 31 kg Unit of elementary charge e = 1 . 6 × 10 − 19 C Coulomb’s constant, k e = 8 . 99 × 10 9 N m 2 C − 2 Permittivity of free space ǫ = 1 4 πk e = 8 . 85 × 10 − 12 C 2 N − 1 m − 2 1 eV = 1.6 × 10 − 19 J g = 9 . 8 m s − 2 Questions 1 & 2: You are given 7 identical capacitors, each with capacitance 14 mF. They are all connected IN SERIES with a 9V battery. 1: Calculate the equivalent capacitance. a. 0.14 mF b. 0.5 mF c. 2 mF d. 49 mF e. 14 mF 2: While the capacitors are all kept connected to the battery, a dielectric material with a dielectric constant κ = 2 . 0 is inserted into the gaps of each capacitor. What is the ra tio of the equivalent capacitance now to that derived in problem 1, i.e., what is C eq , 2 / C eq , 1 ? a. 1/ κ b. 7/ κ c. κ d. κ/ 7 e. 7 κ Solution for Number 1: Let C denote the capacitance on each individual capacitor. 1 C eq, 1 = 1 C + 1 C + 1 C + 1 C + 1 C + 1 C + 1 C = 7 C 1 C eq, 1 = 1 14 mF + 1 14 mF + 1 14 mF + 1 14 mF + 1 14 mF + 1 14 mF + 1 14 mF = 7 14 mF = 1 2 mF C eq , 1 = C 7 = 2 mF. 1 Solution for Number 2: Let C denote the original capacitance, before the dielectrics are inserted. When the dielectrics are inserted, we have C → κ C on each capacitor. 1 C eq, 2 = 1 κC + 1 κC + 1 κC + 1 κC + 1 κC + 1 κC + 1 κC = 1 κ × 7 C Taking the reciprocal of both sides, we get C eq, 2 = κ C 7 = κC eq, 1 So C eq , 2 / C eq , 1 = κ 3. Consider a circuit consisting of five identical resistors (the resistance R in each resistor is identical) all connected in series to a 9volt battery. A current of 0.18 Amps flows through the circuit. What is the resistance of EACH resistor?the circuit....
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 Summer '09
 Berman
 Charge, Magnetism, Energy, Mass, TA, Dielectric, a. b. c., b. c. d.

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