An electron moving parallel to the x axis has an initial speed of 3.70 x 106 m/s

Question

An electron moving parallel to the x axis has an initial speed of 3.70 x 106 m/s at the origin. Its speed is reduced to 1.40 x 105 m/s at the point x = 2.00 cm. (a) Calculate the electric potential difference between the origin and that point. (b) Which point is at higher potential? (c) If instead a proton had been traveling at the origin with the same velocity as the electron, how fast would it be moving when it reaches the point x = 2.00 cm?

I know the answers to parts A and B from this solution manual:

http://www.chegg.com/homework-help/principles-of-physics-4th-edition-chapter-20-problem-5p-solution-9780534491437

But, I can not figure out part C. Please, write out the solution clearly with all equations and the steps to find the solution. Thank you

Explanation / Answer

a) q(V2-V1) = 0.5 m(v2^2 - v1^2)

so V2-V1 = - 0.5 x 9.11E-31 x ( 1367.04E10 ) / 1.602E-19

or V1-V2 = 38.869 V

b) origin is at higher potential

c)now q x (v2-V1) = 0.5 m ( v2^2 - v1^2 )

so 1.602E-19 x 39.869 = 0.5 x 1.67E-27 x ( v1^2 - v2^2)

so ( v1^2 - v2^2) = 76.491 E8

so v2^2 = (3.7E6 )^2 - 76.491E8

so v2 = 369.89E4 or 3.69 x 10^6 m/s

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