Question

At1=0, a proton is projected in the positive x-direction into a region of a uniform electric field of E=-6x1051. The proton travels 7.00 cm as it comes to rest. Determine Its initial speed.
in the simplest steps

Answer 1

To determine the initial speed of the proton, we can use the equation of motion for an object in a uniform acceleration:

vf^2 = vi^2 + 2ax

Where vf is the final velocity (0 m/s since the proton comes to rest), vi is the initial velocity (unknown), a is the acceleration (caused by the electric field), and x is the displacement (7.00 cm or 0.07 m).

Given that the electric field E = -6x10^51 N/C, we can calculate the acceleration of the proton using Newton's second law:

F = ma

Where F is the force acting on the proton and m is the mass of the proton. The force acting on the proton is given by:

F = qE

Where q is the charge of the proton. The charge of the proton is the elementary charge e:

q = e = 1.6x10^-19 C

Substituting this into the equation:

Fe = ma

qE = ma

Solving for the acceleration:

a = qE/m = (1.6x10^-19 C)(-6x10^51 N/C)/(1.67x10^-27 kg) = -9.6x10^-7 m/s^2

Substituting the known values into the equation of motion:

(0 m/s)^2 = vi^2 + 2(-9.6x10^-7 m/s^2)(0.07 m)

0 = vi^2 - 1.344x10^-8 m^2/s^2

Solving for vi:

vi^2 = 1.344x10^-8 m^2/s^2

vi = sqrt(1.344x10^-8 m^2/s^2) ≈ 3.6667x10^-4 m/s

Therefore, the initial speed of the proton is approximately 3.67x10^-4 m/s.