placement_solutions

placement_solutions - Solution : Two resistors R 2 are in...

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PHYS309, Spring 2010 13 Jan. 2010 Placement test 1 Problem 1 A capacitor is connected to a battery with EMF E . The distance between the capacitor plates is d . An electron with charge e and mass m starts with zero initial velocity near one plate and moves towards the other. How long will take the electron to reach the other plate? ε Solution : The electric field in the capacitor gap is E = E /d . The force acting on the electron F = eE gives the electron an acceleration a = F/m = e E /md . The distance traveled by a particle with constant acceleration and zero initial velocity is given by x = at 2 / 2. To reach the other plate the electron has to cover a distance d , thus the time is found from an equation d = at 2 2 t = r 2 d a Substituting the expression for a we get t = r 2 md 2 e E 2 Problem 2 Consider the circuit shown in the Figure. Find the charge on the capacitor.
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Unformatted text preview: Solution : Two resistors R 2 are in parallel and can be replaced by one resistor R 2 / 2. The total current owing through the circuit is I = E R 1 + R 2 / 2 1 R C R R 2 2 1 The voltage on the parallel resistors and on the capacitor is V = I R 2 2 = R 2 2 R 1 + R 2 E This give the capacitor charge Q = CV = R 2 C E 2 R 1 + R 2 3 Problem 3 Consider a function f ( x ) = x 2 e x . Find the positions of the extremum points. Sketch the graph of the function. Solution : First, we have f ( x ) = 2 xe x + x 2 e x = x ( x + 2) e x The equation f ( x ) = 0 has two solutions x 1 = 0 and x 2 =-2. The graph of the function looks as-2 2...
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This note was uploaded on 12/13/2011 for the course PHYS 309 taught by Professor Staff during the Spring '10 term at South Carolina.

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placement_solutions - Solution : Two resistors R 2 are in...

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