A pair of oppositely charged parallel plates are separated by a distance of 5.0

Question

A pair of oppositely charged parallel plates are separated by a distance of 5.0 cm with

the 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 from rest at the

negative plate. Neglect any interaction between the proton and the electron.

a. After what time will their paths cross?

b. How fast will each particle be going?

c. At what time will the electron reach the opposite plate?

d. At what time will the proton reach the opposite plate?

Please I need clear explanation of which formulas to use, Be very Clear, and i already have the answers, i just dont understand how to get to the answers.

Points go to the best explained answer. be decriptive but brief at the same time. I need to understand how to manipulate this type of question for the exam tomorrow. THANK YOU

Explanation / Answer

a)

V=ed
500=E(.05)
E=10000
E=Fe/Q
Fe=10000*1.602x10^-19
Fe=1.602x10^-15
sum of all forces=ma1
proton:
2x(1.602x10^-15)=(1.67x10^-27)a1
a1=1.92x10^12 m/s^2 for electron:
sum of all forces=ma
2(1.62x10^-15)=(9.12x10^-31)a2
a2=3.55x10^15 m/s^2
if they cross each other at a distance x from one of the plates, then
x = 1/2*a1*t^2
d-x = 1/2*a2*t^2.
Solve these equations to fine x and t.
c) for electron d=vit+1/2at^2
.05=0+.5(3.55x10^15)t^2
t=5.31x10^-9 sec

d) for proton D=vit+1/2at^2
.05=0+1/2(1.92x10^12)t^2
t=2.28x10^-7

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