# Electric field and electric forces

by Olgapoollamas
Tags: electric, field, forces
 P: 1 1. The problem statement, all variables and given/known data An electron is projected with an initial speed v0 = 2.00×10^6 m/s into the uniform field between the parallel plates, as shown in figure. Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates. 2. Relevant equations a). If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field. b)Suppose that in the figure the electron is replaced by a proton with the same initial speed v0. Would the proton hit one of the plates? c)If the proton would not hit one of the plates, what would be the magnitude of its vertical displacement as it exits the region between the plates? 3. The attempt at a solution Part a) ok so i found the acceleration by using basic mechanics x=v0 *t then, y=(1/2)*a*t^2 where t=x/v0, so y= (1/2)*a*(x/v0)^2 then solved for a: a=(2*y*v0^2)/x^2 where y=.5 cm x=2 cm, and v0=2.0*10^6 m/s since F=ma and F=Eq solved for E E=ma/q Then plugged in a E=(m/q)((2*y*v0^2)/x^2) The mass and the charge of the electron is known: Mass of e=9.019*10^-31 Charge of e=1.602*10^-19 I keep getting the same answer when i plug numbers back into the equation and the computer is telling me that is wrong. I don't know if there's anything wrong with my calculations or the equation. Please let me know if I'm on the right track. Part b) no it will not (correct answer). Same force with more mass, the proton will go past. Part c) F=qE a=F/m(mass of a proton) where acceleration we already know from part a. So, (2*y*v0^2)/(2 cm)^2 = F/m then solved for y, and I still can't get the answer. Anyone sees a mistake? Please let me know. Thank you so much.

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