Charged particle accelerates in an electric field?

  • #1

Homework Statement

A positively charged particle initially at rest on the ground accelerates upward to 200m/s in 2.60s. The particle has a charge-to-mass ratio of 0.100 C/kg and the electric field in this region is constant and uniform.
What are the magnitude and direction of the electric field?

Homework Equations

a=delta v/ delta t
Electric Fields E=F/q
Coulomb's Law F=kq/r^2 Not sure I need this one.

The Attempt at a Solution

E=(10 kg/C)(200/2.6 m/s^2) = 769.23 N/C = 7.7*10^2 N/C Direction upward.

Question seems pretty straightforward but it's telling me this answer is incorrect. If I have to incorporate Coulomb's law, I'm not sure how.
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  • #3
Arcone said:
Hint: Gravity ;)

Thank you! Knew I was missing something obvious. Let me see if it works if I factor in gravity.
  • #4
ma = Fe - Fg
ma = Eq - mg
a = E(q/m)-g

E = (a + g)/(q/m) = (76.923 + 9.81 m/s^2)/(0.100 C/kg) = 867.33 N/C = 8.7*10^2 N/C

This look right?
  • #5
Yes, that's how i would solve it.

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