
displacement of particle
A particle of charge +3 µC and mass 2.20 105 kg is released from rest in a region where there is a constant electric field of +390 N/C.
What is the displacement of the particle after a time of 2.90 x 10^2 s?
distance = in meters
couldn't this equation be manipulated to find x?
http://img88.imageshack.us/img88/7949/83637160mr5.png
I just have missing variables.

should be right
Q= +3 µC
m= 2.20 105 kg
$\displaystyle \vec{E}$= +390 N/C.
t =2.90 x 10^2 s?
$\displaystyle \vec{u}$=0
use $\displaystyle \frac{\vec{F}}{Q}=\vec{E}$
to find F then use $\displaystyle \vec{F}=m\vec{a}$ to find a
the use $\displaystyle \vec{s}=\vec{u}t+1/2 \vec{a}t^2$ to find s
and this is physics, think and dont just apply formulas! make sure you understand the theory behind each formula!
ps i do not mean to yell at u