Nevermind! I got it :) I was using horizontal displacement and velocities instead of vertical.
It slipped my mind that that viy= 0 m/s.
Anyways, thank you! I figured it out this time :biggrin:
So this is what I ended up doing:
d= vit + ½at2
a= 2(d-vit) / t2
= [2(0.02-(7x106)(2.86x10-9))] / ((2.86x10-9)2)
= -4.89 x 1012 m/s/s
^ My first question is whether this should be a negative acceleration.. which I don't quite understand
E= ma/q
E= [(9.11x10-31)(-4.89x1012)]/(1.6x10-19)=...
My work is as shown below:
Mass of Electron: 9.11 x 10^-31 kg
Charge of Electron: 1.6 x 10^-19 C
E= F/q
F= ma
v= d/t
t= d/v= 0.02/(7x10^6)= 2.86 x 10^-9 seconds
a= v/t= (7x10^6)/(2.86x10^-9)= 2.45 x 10^15 m/s/s
F= ma= (9.11x10^-31)*(2.45x10^15)= 2.23 x 10^-15 N
E= F/q =...
Homework Statement
[PLAIN]http://img214.imageshack.us/img214/6328/physics2.jpg
An electron enters the lower left side of a parallel plate capacitor and exists precisely at the upper right side (just clearing the the upper plate). The initial velocity of the electron is 7x10^6 m/s parallel to...