An electron has an initial velocity of 1.00 ? 106 m/s in a uniform 9.00 ? 105 N/

Question

An electron has an initial velocity of 1.00 ? 106 m/s in a uniform 9.00 ? 105 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity.
(a) What is the direction of the electric field?

Correct: Your answer is correct.

(b) How far does the electron travel before coming to rest?
  
Incorrect: Your answer is incorrect.
m
(c) How long does it take the electron to come to rest?
  
Incorrect: Your answer is incorrect.
s
(d) What is the electron's speed when it returns to its starting point?
  
Correct: Your answer is correct.
m/s

Explanation / Answer

the field produces a retardation in electron = q x E / m = 1.6 x 10-19 x 9 x 105 / 9.1 x 10-31 = 1.58 x 1017

initial velocity = 106 m/s

use v2 = u2 + 2as to get 0 = 1012 - 2 x 1.58 x 1017 xs

solve to get s distance travelled = 3.1 x 106 m

for time use v = u + at to get

0 = 106 + 1.58 x 1017 x t

this gives t = 6.3 x 10-12 s

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