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A pair of oppositely charged, parallel plates are separated

  • Thread starter lilkidyea
  • Start date
  • #1
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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 interaction between the proton and the electron.
a. After what interval will their paths cross?
b. How fast will each particle be going when their paths cross?
c. At what time will the electron reach the opposite plate?
d. At what time will the proton reach the opposite plate?

Homework Equations



I seriously don't know there are soo many equations but these will probably do the job:
E=Fe/Q
E=k q/r2
Sum of all forces=ma
d=vit+1/2at2


The Attempt at a Solution



I don't know where to start. I honestly I tried using E=k q/r2 then find Fe with E=Fe/Q and use sum of all forces but then I took into account that both the positive plate and negative plate were acting on either proton or electron at the same time and from there I got completely lost
 
Last edited:

Answers and Replies

  • #2
rl.bhat
Homework Helper
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Hi lilkidyea, welcome to PF.
Proton and electron has the same charge but different mass. So they have different acceleration. Find their accelerations. Then using kinematic equations find the remaining results.
 
  • #3
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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 wrong
 
Last edited:
  • #4
rl.bhat
Homework Helper
4,433
7
E=Fe/Q
So Fe = .........?
 
  • #5
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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 ans or a or b
 
  • #6
rl.bhat
Homework Helper
4,433
7
2x(1.602x10^-15)=(1.67x10^-27)a
Why there is a factor 2 in the beginning?
For (a), 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.
 
  • #7
4
0
2x(1.602x10^-15)=(1.67x10^-27)a
Why there is a factor 2 in the beginning?
well I figured since the positive plate is pushing and the negative plate is pulling on the same proton I multiplied the magnitude by 2
 

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