An electron moving parallel to the x axis has an initial speed of 4.14 at the or

Question

An electron moving parallel to the x axis has an initial speed of 4.14 at the origin. Its speed is reduced to 1.30 105 m/s at the point x = 2.00 cm. Calculate the electric potential difference between the origin and that point. Which point is at the higher potential? the point x = 2.00 cm the origin both have the same potential

Explanation / Answer

a) q V = 0.5 m v^2 ==> V = 0.5 m v^2/q = 0.5*9.109e-31*(4.14e6*4.14e6-1.3e5*1.3e5)/1.6e-19 = 48.7 V ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- b) the origin

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