Question

A proton is acted on by an uniform electric field of magnitude 443 N/C pointing in the negative x direction. The particle is initially at rest.

(a) In what direction will the charge move?
---Select--- +x direction ?x direction +y direction ?y direction +z direction ?z direction

(b) Determine the work done by the electric field when the particle has moved through a distance of 3.15 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

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

The electric field is given E = 443 Newton/Coulomb in the -x direction.

The force on the particle is F=EQ
F = 7.09706 x10^-17 Newtons, in the -x direction.
Since the force is in this direction, the acceleration will also be in this direction. Thus, the proton will begin moving in the -x direction.

The work done is W = integral F dx. Since F is constant here, W = F x. So if the proton moves 0.0315m, the work is 2.2355x10^-18 Joules.

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

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 = 4.242 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 0.0315 m, then its velocity is
51695.8 m/s