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