A proton is acted on by an uniform electric field of magnitude 283 N/C pointing

Question

A proton is acted on by an uniform electric field of magnitude 283 N/C pointing in the positive x direction. The particle is initially at rest.

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

(b) Determine the work done by the electric field when the particle has moved through a distance of 2.55 cm from its initial position.
answer: 1.15464e-18 J Correct

(c) Determine the change in electric potential energy of the charged particle.
answer: ___________ J

(d) Determine the speed of the charged particle.
answer: ___________ m/s

Explanation / Answer

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

The electric field is given E = 283 Newton/Coulomb in the y direction.

The force on the particle is F = EQ

F = 283 N/C x 1.6 x 10^-19 C

F = 4.528 x10^-17 N, 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 0.0255 m,

W = 4.528 x 10^-17 N x 0.0255 m = 1.1546 x 10^-18 J

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 = 2.7065 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)
v = sqrt(2 x 2.7065 x 10^10 x 0.0255)

v = 3.7153 x 10^4 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.