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

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

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