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Unformatted text preview: Solution for Practice Test 4 for Test 2 Solution to Practice Test Problem 4.1(Field Given Potential) Problem: In a region of space the electric potential is V ( y ) = γ y 3 , where γ is a constant. Calculate the electric field in the region. Select One of the Following: (a) vector E = γ 2 y 2 ˆ y (b-Answer) vector E = 3 γ y 4 ˆ y (c) vector E =- 3 γ y 4 ˆ y (d) vector E =- γ 2 y 2 ˆ y (e) Not enough information given to find field. Solution If the field depends on only one variable, then electric field is related to the potential by vector E =- dV dy ˆ y =- d dy γ y 3 ˆ y = 3 γ y 4 ˆ y Total Points for Problem: 3 Points Solution to Practice Test Problem 4.2(Coil Gun Feasibility) Problem: Paul and Brad are building a coil gun. This problem is to help them with the feasibility study. The capacitor we use to fire our electromagnetic coil launcher has capacitance 2000 μ F and we charge it to a voltage of 150V . We have found the lightest projectile we can get our hands on has a mass of about 4g = 0 . 004kg . If our coil gun is 100% efficient, that is all the energy stored in the capacitor is turned into kinetic energy of the projectile, how fast can the projectile be shot? Select One of the Following: (a) . 13 m s (b) 22 . 5 m s (c) 11 , 000 m s (d) 75 m s (e-Answer) 106 m s Solution 1 The energy stored in a capacitor is U = 1 2 C (Δ V ) 2 = 1 2 (2000 μ F)(150V) 2 = 22 . 5J If all this energy is converted into kinetic energy, then U = 1 2 mv 2 so the velocity is v = radicalbigg 2 U m = radicalBigg 2(22 . 5J) . 004kg = 106 m s = 240mph Total Points for Problem: 3 Points Solution to Practice Test Problem 4.3(Power Dissipated in Graphite) Problem: Graphite has resistivity 1 . 4 × 10- 5 Ωm (it is a lousy conductor, but still a conductor). A pencil lead is made of graphite (one of the molecular forms of elemental carbon) and is 15cm in length and has radius . 05cm . If the pencil lead carries a current of 100A , how much heat is produced per second? Select One of the Following: (a-Answer) 26700J (b) 267J (c) 7 . 3J (d) . 73J Solution The resistance of the graphite rod is given by R = ρℓ A = ρℓ πr 2 = (1 . 4 × 10- 5 Ωm)(0 . 15m) π (0 . 0005m) 2 = 2 . 67Ω The power dissipated in the resistor is P = I Δ V , substitute Ohm’s Law, P = I 2 R = (100A) 2 (2 . 67Ω) = 26700W or 26700J per second. Total Points for Problem: 3 Points Solution to Practice Test Problem 4.4(Bulbs in Parallel Problem) Problem: The figure below shows two light bulbs, B A and B B in parallel connected across a battery. When the switch is closed a third bulb, is connected in parallel with the first two bulbs. Which expression below correctly expresses the brightness of the three bulbs after the switch is closed and the relation between the current drawn from the battery before the switch is closed, I before to the current drawn from the battery after the switch is closed, I after . Assume the battery is perfect....
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