Physics. Urgent. Please Help!!!?

Question

Two horizontal metal plates, each 100 mm square, are aligned 10.0 mm apart, with one above the other. They are given equal-magnitude charges of opposite sign so that a uniform downward electric field of 2500 N/C exists in the region between them. A particle of mass 2.50 10-16 kg and with a positive charge of 1.20 10-6 C leaves the center of the bottom negative plate with an initial speed of 1.00 105 m/s at an angle of 37.0° above the horizontal. Where does the particle strike relative to a starting point?

Answer 1

This is similar to projectile motion.

The downward force on the particle from the E-field is 0.003 N. compare this to the downward force from gravity on the mass=

(2.5E-16)(9.8)=2.5E-15 N << than the E-field force we can ignore it.

In the x-direction

vx=cos(37)*1E5=79.9E3 m/s

the equations of motion in the y direction are

vy=sin(37)*1E5=60.2E3
ay=-0.003/2.5E-16=-1.2E13 m/s^2

so we have
y=-1.2E13/2 t^2+60.2E3t

Let's check to see if the particle hits the other plate. The maximum high occurs at t=5.02E-9
this height is 0.00015<<.01 so the particle doesn't get high enough to touch the other plate.

It must land some distance from the center of ejection.

This distance is when s=0 again. so looking at the projectile motion equations in the y-direction, the particle hits the bottom plate in

60.2E3/1.2E13=5.02E-9 seconds

In this time the particle will travel in the x-direction
(79.9E3 m/s)(5.02E-9)=4.01E-4 m or 0.4 mm