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) = aÂ·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)atÂ² + vÂ·t Here, s = .1 meters, a = 0, and one solves for v. 

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