Question

A proton is acted on by a uniform electric field of magnitude 313 N/C pointing in the negative z-direction. The particle is initially at rest.

(a) In what direction will the charge move?

(b) Determine the work done by the electric field when the particle has moved through a distance of 3.75 cm from its initial position.

____________J

(c) Determine the change in electric potential energy of the charged particle.

___________J

(d) Determine the speed of the charged particle.

_______m/s

Answer 1

a.)

The charge of a proton is
Q = 1.602 x 10^−19 Coulombs

The electric field is given E = 313 Newton/Coulomb in the -Z direction.

The force on the particle is F=EQ
F = 5.0706 x10^-17 Newtons, in the -z direction.
Since the force is in this direction, the acceleration will also be in this direction. Thus, the proton will begin moving in the -Z direction.

Answer: -ve Z- direction

b.)

The work done is W = integral F dx. Since F is constant here, W = F x. So if the proton moves 0.0375m, the work is 1.9x10^-18 Joules.

Answer: w= 1.90*10^-18 J

C.)

This is the same as the decrease in the proton's electric potential energy but in opposite direction.

W = -1. 90*10^-18 J.. Answer.

d.)

The acceleration of the proton can be found from
F = ma;
a = F/m
The mass of a proton is
m = 1.673 x 10^−27 kg

a = 3.0308 x 10^10 m/s^2

Now we can use the equations of motion:
v = at
x = a/2 t^2

We can relate v to x:
v = sqrt(2ax)

So if the proton has moved 0.0375 m, then its velocity is

V = √(2*3.0308*10^10*0.0375)
= 33713 m/s....Answer.

Hope this helps you.

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