A proton, initially traveling in the +x-direction with a speed of 5.65×105 m/s ,

Question

A proton, initially traveling in the +x-direction with a speed of 5.65×105 m/s , enters a uniform electric field directed vertically upward. After traveling in this field for 4.12×10?7 s , the proton’s velocity is directed 45 ? above the +x-axis.

Problem 15.82 A proton, initially traveling in the +z-direction with a speed of 5.65x105 m/s, enters a uniform electric field directed vertically upward. After traveling in this field for 4.12x10-7 s the proton's velocity is directed 45 above the +z-axis. Part A What is the strength of the electric field? Express your answer with the appropriate units. Value nits Submit My Answers Give Up Provide Feedback Continue

Explanation / Answer

The proton's 45 degree direction can only be achieved if the vertical velocity equals to the horizontal one, that is (5.65 * 105 m/s)

So the necessary strengh of the electric field is: F=m*a =q*E....E=m*a/q

a= v/t=(5.65 * 105 m/s)/(4.12x10-7s)=1.37 X 1012 m/s2

m=1.67 x 10-27 kg(proton's mass)

q=1.60x1-19 C

E = m*a/q = (1.67 * 10-27 kg)*(1.37 * 1012 m/s2)/(1.60 * 10-19 C) = 1.4299 X104 N/C

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