A proton is projected n the positive x direction into a region of uniform electr

Question

A proton is projected n the positive x direction into a region of uniform electric field E = (-5.60 x 10^5) i N/C at t = 0. The proton travels 6.00 cm as It comes to rest. (a) Determine the acceleration of the proton. How do you find the acceleration of an object f you know the net force that acts on it? m/s^2 (b) Determine the initial speed of the proton. The electric field is constant, so the force is constant, which means the acceleration will be constant. m/s (c) Determine the time interval over which the proton comes to rest. You appear to have calculated the time correctly using your Incorrect results from parts (a) and (b). s

Explanation / Answer

here,

(a)

electric feild , E = -5.60 *10^5 i N/C

force exerted on proton , F = E * e

F = -5.60 *10^5 i * e N

accelration of proton, a = F / mp

a = -5.60 *10^5 i * 1.66 * 10^-19 / 1.672 * 10^-27

a = - 5.56 * 10^13 m/s^2

the accelration of the proton is 5.56 * 10^13 m/s^2 and is towards negative x direction

(b)

let the initial speed of proton be u

final speed, v = 0

distance travelled ,s = 0.06 m

using third equation of motion

v^2 - u^2 = 2 * a * s

- u^2 = - 2 * 5.56 * 10^13* 0.06

u = 2.57 * 10^6 m/s

the initial speed of proton is 2.57 * 10^6 m/s and towards the positive x-direction

(c)

let the time interval be t

using first equation of motion

v = u + at

0 = 2.57 * 10^6 - 5.56 * 10^13 * t

t = 4.62 * 10^-8 s

the time interval over which the proton comes to rest is 4.62 * 10^-8 s

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