A proton initially moves left to right long the x axis at a speed of 2.00 x 10³m/s.It moves into an electric field,which points in the negative x direction,and travels a distance of 0.200 m before coming to rest.If the proton's mass and charge are 1.67 x 10^-27kg and 1.60 x 10^-19C respectively,what is the magnitude of the electric field?

Question

Quizplus · Accepted Answer

We can use the formula for the work done by an electric field on a charged particle to solve for the magnitude of the electric field:

W = qEd

where W is the work done, q is the charge of the particle, E is the electric field strength, and d is the distance traveled in the electric field.

Since the proton comes to rest, all of its initial kinetic energy is converted into potential energy due to the electric field:

KEi = PEf

(1/2)mv^2 = qEd

We can solve for E:

E = (1/2)mv^2 / qd

Plugging in the given values:

E = (1/2)(1.67 x 10^-27 kg)(2.00 x 10^13 m/s)^2 / (1.60 x 10^-19 C)(0.200 m)

E = 0.104 N/C

Therefore, the correct answer is (C).

A Proton Initially Moves Left to Right Long the X