A proton is projected in the positive x direction into a region of uniform elect

Question

A proton is projected in the positive x direction into a region of uniform electric field E = -5.50 105 i N/C at t = 0. The proton travels 6.10 cm as it comes to rest. (a) Determine the acceleration of the proton. . Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. --Direction-- -i +i +j -j +k -k m/s2 (b) Determine the initial speed of the proton. --Direction-- +i -i +j -j +k -k m/s (c) Determine the time interval over which the proton comes to rest.

Explanation / Answer

PY212 with Bochinski?? Use this formula for the first part. sin2(theta)=(R*E*e)/(m*(v^2)) R=range travelled by protons E=electric field e=charge of proton (1.6e-19) m=mass of proton (1.67e-27) v=initial velocity Just use inverse sin and divide by two. The answer you get is the small angle. The large angle is 180 - that last answer. Second part...Find the velocity for the x component. Vx=V*cos(theta). Do that for both angles. Then divide your range by both x-component velocities and you got your times.

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