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A proton, initially traveling in the +x-direction with a speed of 5.15×105 m/s , enters a...

A proton, initially traveling in the +x-direction with a speed of 5.15×105 m/s , enters a uniform electric field directed vertically upward. After traveling in this field for 3.96×10−7 s , the proton’s velocity is directed 45 ∘ above the +x-axis. Part A What is the strength of the electric field?

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Answer #1

acceleration of proton

a=V/t =(5.15*105)/(3.96*10-7) =1.3*1012 m/s2

Since

F=ma =qE

The strength of the electric field is

E=ma/q =(1.6727*10-27)(1.3*1012)/(1.602*10-19)

E=1.358*104 N/C

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