Force exerted by an external electric field

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
13 replies · 4K views
lacar213
Messages
29
Reaction score
0

Homework Statement


A particle of mass 1.20 kg with a charge of +2.30 mC is shot into a region with an electric field of strength 18.0 N/C, having a direction opposite the particle's velocity. If the particle has an initial speed of 2.40 m/s, how long does it take to come to a complete (though momentary) standstill? Assume there are no other forces acting on the particle.


Homework Equations


F = qE
F= ma


The Attempt at a Solution


I'm not sure what equation to start with because of all the information already given. I couldn't find any equations dealing with time in the chapter this problem was in. It sounds like you may have to use kinematics but I'm not sure if that is the correct way to do it.
 
Physics news on Phys.org
0 = 2.40 + 9.8 (t)
t = 9.8 / 2.4 = 4.08333
 
I forgot to find the force using F=qe then you can plug it into f=ma

A = .000035

.000035t = 2.40 - 0
t = .000014

would the acceleration be negative also since the field of strength is opposite the particle
 
the problem states millicoulombs not micro - t = .014375
 
lacar213 said:
I forgot to find the force using F=qe then you can plug it into f=ma

A = .000035

.000035t = 2.40 - 0
You're good up until this point. However, your final answer is wrong.
lacar213 said:
t = .000014
You may want to check your arithmetric
lacar213 said:
would the acceleration be negative also since the field of strength is opposite the particle
Yes, but you took care of that (be it intentionally or otherwise) in your second line above, by multiplying through by -1.
 
lacar213 said:
the problem states millicoulombs not micro - t = .014375
Your final answer is still wrong. Recheck your basic arithmetic. Notice that you are dividing a number greater than one by a number smaller than one. How can the result be less than one?
 
2.4 / .0345 = 69.5652
 
GOT IT thanks for the help!