A proton traveling at 3E6 m/s enters a region where the electric field has a magnitude of 3E5 N/C. The electric field is uniform and slows the proton's motions.(adsbygoogle = window.adsbygoogle || []).push({});

a)calculate the distamce the proton will travel before coming to a momentary halt?

B)Calculate the decel. of the proton and the time for it to trqavel this distance

F=qE = 1.6E-19 * 3E5 = 4.8E-14 N

a = F/m 4.8E-14 / 1.67E-31 a = 2.87E13 m/s

D = Vf^2 - Vi^2 / 2A = 3E6^2 / 2 * 2.87E13 = .157 m

I am not sure how to do part B or what formula to use

thanks joe

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# Homework Help: Answer check / little help

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