Calculate final velocity of electric particle

[beeteep]

Homework Statement

A particle of mass 7.3 × 10⁻⁵ g and charge 24 mC moves in a region of space where the electric field is uniform and is 6.3 N/C in the x direction and zero in the y and z direction. If the initial velocity of the particle is given by v_y = 4.1 × 10⁵ m/s, v_x = v_z = 0, what is the speed of the particle at 0.4 s? Answer in units of m/s

Homework Equations

F=ma
F=qE
v=v_i + at

The Attempt at a Solution

m=7.3x10^-8 kg
q=.024 C
E=6.3 N/C
F = .024(6.3) = .1512 N

.1512 = 7.3x10^-8a
a = 2071232.877 m/s²x = doesn't matter
x_i = doesn't matter
v = ?
v_i = 4.1x10^5
a = 2071232.877
t = .4

v = v_i + at
v = 4.1x10⁵ + 2071232.877(.4) = 1238493.151 m/s

Final Answer = 1238493.151

But the system says it's wrong...
Any help is appreciated! Thanks!

 

[Bystander]

Ten significant figures?

 

[beeteep]

The system is supposed to count your answer correct if you're within +/- 1% of the correct answer, so I err on the side of caution and put in every digit I have since it did not specify how many significant figures it wanted.

 

[haruspex]

charge 24 mC

Isn't that rather large? Should it be μC?

v = vi + at

Yes, but not v_xi+a_yt.
Correction, I meant:
Yes, but not v_yi+a_xt.

 

[beeteep]

Most of the problems I have worked have been μC but I just double checked and this problem definitely says mC.

 

[haruspex]

Most of the problems I have worked have been μC but I just double checked and this problem definitely says mC.

OK, but please see the edit to my previous post.

 

[beeteep]

OK. I think I see what you're saying. According to the problem statement, since the v_yi is given and there is no force acting in the y-direction, v_yf=v_yi. Then I'd calculate the v_xf with v_xi=0 and include the acceleration for .4s which would give me two components to a vector, right? With that, I could take the square-root of (v_x² + v_y²) and that would give me the speed?

 

[haruspex]

OK. I think I see what you're saying. According to the problem statement, since the v_yi is given and there is no force acting in the y-direction, v_yf=v_yi. Then I'd calculate the v_xf with v_xi=0 and include the acceleration for .4s which would give me two components to a vector, right? With that, I could take the square-root of (v_x² + v_y²) and that would give me the speed?

Yes.

 

[beeteep]

Some heroes don't wear capes.

Thank you so much for your help!