A proton accelerates from rest in a uniform electric field of 650 N/C. At one later moment, its speed is 1.40 Mm/s (nonrelativistic because v is much less than the speed of light).(c) How far does it move in this time interval?(d) What is its kinetic energy at the end of this interval?

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A proton accelerates from rest in a uniform electric field of 650 N/C. At one later moment, its speed is 1.40 Mm/s (nonrelativistic because v is much less than the speed of light).
(c) How far does it move in this time interval?
(d) What is its kinetic energy at the end of this interval?

Expert Solution

Step 1

$G i v e n : E = 650 N / C, v = 1.2 \times 10^{6} m / s$

The acceleration of the proton due to the electric field is found by :

$F = q E m a = q E H e r e, m a n d q a r e m a s s a n d c h a r g e o f p r o t o n r e s p e c t i v e l y . a = \frac{q E}{m} = \frac{1.6 \times 10^{- 19} \times 650}{1.67 \times 10^{- 27}} a = 6.22 \times 10^{10} m / s^{2}$

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A proton accelerates from rest in a uniform electric field of 650 N/c. At one later moment, its speed is 1.40 Mm/s (nonrelativistic because v is much less than the speed of light). (c) How far does it move in this time interval? (d) What is its kinetic energy at the end of this interval?