So with that I did D=vit+1/2at^2
.05=0+1/2(1.92x10^12)t^2
t=2.28x10^-7 for d. Is this right?
for electron:
sum of all forces=ma
2(1.62x10^-15)=(9.12x10^-31)a
a=3.55x10^15
d=vit+1/2at^2
.05=0+.5(3.55x10^15)t^2
t=5.31x10^-9 for c. Is this right?
but with this I still don't know how to find the...
So uh for a this is what I've done:
V=ed
500=E(.05)
E=10000
E=Fe/Q
Fe=10000*1.602x10^-19
Fe=1.602x10^-15
sum of all forces=ma
proton:
2x(1.602x10^-15)=(1.67x10^-27)a
a=1.92x10^12 m/s^2
at this point I just went completely Wahh?! because my acceleration is just HUGE so I must have done something...
Homework Statement
A pair of oppositely charged, parallel plates are separated by a distance of 5.0 cm with a potential difference of 500 V between the plates. A proton is released from rest at the positive plate, and at the same time an electron is released at the negative plate. Neglect any...