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 March 28th, 2008, 12:56 PM #1 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Electric Field 1. An electron has a mass of 9.11 x 10^(-31) kg. Its charge is -1.6 x 10^(-19) C. And, suppose it is given an initial horizontal velocity of 8.0 x 10^7 m/s. It then travels through a vertical (upward pointing) electric field with a constant strength of 2.0 x 10^4 N/C. In a time t, it has horizontal displacement of 10 cm. (a) How much force acts on the electron? In which direction is the force? (I got 3.2 x 10^(-15) N, downward.) (b) What is the acceleration of the electron? (c) What will be the horizontal velocity of the electron at the ned of time t? (d) What will be the vertical velocity of the electron at the end of time t? (e) At the end of time t, what angle will the electron's velocity make with its initial velocity? (f) What will be the vertical displacement of the electron during time t? Any help would be appreciated. Thanks, j. March 29th, 2008, 09:05 AM #2 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 (a) Since E = F_q / q, this can arrange to F_q = Eq = (2.0 x 10^4 N/C, upward)(-1.6 x 10^(-19)) = 3.2 x 10^(-15) N, downward. (b) Since v(t) = 8.0 x 10^7 m/s, v'(t) = a(t) = 0 m/s^2. (c) Since x'(t) = v(t) = v_0 + at, v(T) = 8.0 x 10^7 m/s, right + (0 m/s^2)T = 8.0 x 10^7 m/s, right. (d) Since no velocity is given in the problem, 0 m/s. Any help would be appreciated. Thanks, J. May 1st, 2008, 01:48 PM #3 Senior Member   Joined: Apr 2008 Posts: 435 Thanks: 0 Although it is true that acceleration can usually be denoted by v'(t), the v(t) that we are given is the horizontal component and the acceleration is the vertical component. Thus the acceleration is not zero (which we expect, as an unbalanced force is acting on it). As we know F=ma ||Thanks Isaac|| and we know both the force and the mass, we can find the vertical acceleration. As no force acts horizontally, the horizontal velocity does not change. The vertical velocity does change. We know that v(t) = a·t (as the vertical velocity starts at zero and there is constant acceleration). Thus at time t, the electron will be moving horizontally and down. One can imagine the two values forming a triangle. the angle between them would be the atan(v_y/v_x) where v_y is the vertical component, v_x is the horizontal component. Finally, to actually find t, one recalls that s= (1/2)at² + v·t Here, s = .1 meters, a = 0, and one solves for v. Tags electric, field Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ibungo Calculus 3 March 4th, 2013 07:16 PM watson Abstract Algebra 1 September 14th, 2012 09:07 PM Keroro Probability and Statistics 2 June 5th, 2012 12:41 PM r-soy Physics 1 March 7th, 2011 09:14 AM supermikong Calculus 0 August 15th, 2010 11:40 PM

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