Question

A proton accelerates from rest in a uniform electric field of 664 N/C. At some later time, its speed is 1.46 106 m/s. (a) Find the magnitude of the acceleration of the proton. m/s2 (b) How long does it take the proton to reach this speed? µs (c) How far has it moved in that interval? m (d) What is its kinetic energy at the later time? J

Answer 1

Answer:

Explanation:

In an electric field E force on charge q

F = Eq , acceleration a = Eq / m

a = 664 x 1.6 x 10⁻¹⁹ / 1.67 x 10⁻²⁷

= 636.16 x 10⁸ m /s²

b )

initial velocity u = 0

final velocity v = 1.46 x 10⁶ m/s

v = u + at

1.46 x 10⁶ = 0 + 636.16 x 10⁸ x t

t = 2.29 x 10⁻⁵ s

c )

s = ut + 1/2 a t²

= 0 + .5 x 636.16 x 10⁸ x (  2.29 x 10⁻⁵ )²

= 1668 x 10⁻²

= 16.68 m

d )

Kinetic energy = 1/2 m v²

= .5 x 1.67 x 10⁻²⁷ x ( 1.46 x 10⁶ )²

= 1.78 x 10⁻¹⁵ J .