Question

A proton is projected in the positive x direction into a region of uniform electric field = (-5.60 10⁵) N/C at t = 0. The proton travels 6.30 cm as it comes to rest.

(a) Determine the acceleration of the proton.

(b) Determine the initial speed of the proton.

(c) Determine the time interval over which the proton comes to rest.
s

magnitude	m/s2
direction	---Select--- +x −x +y −y +z −z

magnitude	m/s
direction	---Select--- +x −x +y −y +z −z

Answer 1

Given

Magnitude of electric field is , E = (-560*10^5) i N/C

Distance travelled by the porton is , s = 6.30 *10^-2 m

Mass of proton , m= 1.67 *10^-27 kg

a)The force on the proton is F = qE

                                        ma = qE

Acceleration, a = (1.6*10^-19 C) * (-560*10^5) i N/C / (1.67 *10^-27 kg)

                         = -5.36 *10^15 m/s^2 (i)

Magnitude of the acceleration is  5.36 *10^15 m/s^2

The direction of the acceleration is towards - x axis

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b) The initial speed of the proton is

         v^2 -vi^2 = 2as

         0 - vi^2 = 2 *( 5.36 *10^15 m/s^2 (-i)) *(6.30 *10^-2 m)

                vi = 2.6 *10^7 m/s (i)

The magnitude of the initial speed is 2.6 *10^7 m/s

Direction of the initial speed is towards + x axis

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c) From the relation v = vi + at

                         2.6 *10^7 m/s (i) = - ( -5.36 *10^15 m/s^2 (i)) t

                                         t = 4.85 *10^-9 s

The time interval over which the proton comes to rest is 4.85 *10^-9 s