E-field of an ununiformly-charged rod

[nahya]

ballistic particle and electric field

Throughout space there is a uniform electric field in the -y direction of strength E = 450 N/C. There is no gravity. At t = 0, a particle with mass m = 5 g and charge q = -11 µC is at the origin moving with a velocity v0 = 35 m/s at an angle q = 25° above the x-axis.
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i found out that the magnitude of the force acting on the particle is 0.00495 N.
the question is this: at t = 5 s, what are the x- and y-coordinatesof the position of the particle?
x: 158.60 m, but I'm having trouble finding out what the y-coordinate would be, because of the electric field present in the -y direction.

i tried this:
the initial velocity in the y diction is 35 sin 25'.
then, v_y = 35sin25 - at, where a = F/m. I've already found F, and m is given. so, a = 0.99
then, v_y = 35sin25 - 0.99t. plug in t, i get 9.8416. but that's wrong.
what am i doing wrong?

thanks.

 

[Galileo]

The integral is correct. How'd you get 7.2e10? K is of order e10 and A of the order e-6 so you should get something of the order e4.

 

[nahya]

doh! i forgot that A is in µC.
so... 2 * 9e9 * 4e-6 = 72000, and it's still wrong...
the answer's supposed to be in N/C.

edit:

oh. -72000. hehe.
thanks!

 

[nahya]

i'm recycling the thread. old post was edited to a new question.