# displacement of particle

• January 15th 2008, 07:14 PM
rcmango
displacement of particle
A particle of charge +3 µC and mass 2.20 10-5 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.
• January 16th 2008, 11:17 PM
calculus_jy
should be right
Q= +3 µC
m= 2.20 10-5 kg
$\vec{E}$= +390 N/C.
t =2.90 x 10^-2 s?
$\vec{u}$=0
use $\frac{\vec{F}}{Q}=\vec{E}$
to find F then use $\vec{F}=m\vec{a}$ to find a
the use $\vec{s}=\vec{u}t+1/2 \vec{a}t^2$ 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