Question

11. A proton accelerates from rest in a uniform electric field
of 650 N/C. At one later moment, its speed is 1.10 Mm/s
(nonrelativistic because *v* is much less than the speed of
light).

(a) Find the acceleration of the proton.

m/s^{2}

(b) Over what time interval does the proton reach this speed?

s

(c) How far does it move in this time interval?

m

(d) What is its kinetic energy at the end of this interval?

J

Answer #1

here,

electric feild , E = 650 N/C

final speed , v = 1.1 * 10^6 m/s

a)

the accelration , a = e * E /mp

a = ( 1.6 * 10^-19 * 650 )/(1.67 * 10^-27)

a = 6.23 * 10^10 m/s^2

b)

let the time taken be t

v = u + a * t

1.1 * 10^6 = 0 + 6.23 * 10^10 * t

solving for t

t = 1.76 * 10^-5 s

c)

the distance travelled , s = v^2 / ( 2 a)

s = (1.1 * 10^6)^2 /( 2 * 6.23 * 10^10) m

s = 9.7 m

d)

the kinetic energy at the end of this interval , KE = 0.5 m * v^2

KE = 0.5 * 1.67 * 10^-27 * ( 1.1 * 10^6)^2 J

KE = 1.01 * 10^-15 J

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