A proton accelerates from rest in a uniform electric field of 590 N/C. At one la

Question

A proton accelerates from rest in a uniform electric field of 590 N/C. At one later moment, its speed is 1.40 Mm/s (nonrelativistic because v is much less than the speed of light). Find the acceleration of the proton. _____ m/s^2 Over what time interval does the proton reach this speed? _____ s How far does it move in this time interval? ____ m What is its kinetic energy at the end of this interval? _____ J A proton is projected in the positive x direction into a region of uniform electric Field E^rightarrow = (-6.80 times 10^5) i^cap N/C at t = 0. The proton travels 6.90 cm as it comes to rest. Determine the acceleration of the proton. magnitude ____ m/s^2 direction Determine the initial speed of the proton. magnitude ____ m/s direction

Explanation / Answer

1)

Magnitude of electric field is , E = 590 N/C

The final speed is , v = 1.4 *10^6 m/s

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) * 590N/C / (1.67 *10^-27 kg)

                         = 5.65 *10^10 m/s^2

b) The time taken to reach this speed is

       t = v/a     [since v = at]

        t = (1.4 *10^6 m/s) /5.65 *10^10 m/s^2

        t = 2.4 *10^-5 s

c) Distance moved in t is given by

   s = 1/2 at^2

   s = 1/2 * 5.65 *10^10 m/s^2 *(2.4*10^-5 s)^2

     = 17.3 m

d)The kinetic energy = 1/2 mv^2

                                 = 0.5 * (1.67 *10^-27 kg) *( 1.4 *10^6 m/s)^2

                                 = 1.66 *10^-15 J

2)

Given

Magnitude of electric field is , E = (-6.80*10^5) i N/C

Distance travelled by the porton is , s = 6.9 *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) * (-6.80*10^5) i N/C / (1.67 *10^-27 kg)

                         = -6.51 *10^15 m/s^2 (i)

Magnitude of the acceleration is 6.15 *10^15 m/s^2

The direction of the acceleration is towards - x axis

b) The initial speed of the proton is

         v^2 -vi^2 = 2as

         0 - vi^2 = 2 *( 6.15 *10^15 m/s^2 (-i)) *(6.9 *10^-2 m)

                vi = 2.9 *10^7 m/s (i)

The magnitude of the initial speed is 2.9 *10^7 m/s

Direction of the initial speed is towards + x axis

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