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Oppositely Charged Parallel Plates

  • Thread starter mattwild
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  • #1
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Homework Statement



Two oppositely-charged parallel plates are separated by a distance of 0.03m.

a. What is the strength of the electric field between the two plates if it takes 2.4x10^-19 J of energy to move an electron from one plate to the other.
b. What is the acceleration of that electron as it moves toward the positively-charged plate?
c. If the electron starts at rest, how long does it take to go from one plate to the other?
d. A Proton is launched in between the plates, moving parallel to them. Sketch the path this proton will take.

Homework Equations





The Attempt at a Solution



I'm really lost and not sure even what formulas to use except electric field strength but I have joules and I don't know what to do with that.
 

Answers and Replies

  • #2
gneill
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You're given the energy required to move the electron form one plate to the other. Sounds like work to me...

What do you know about work?
 
  • #3
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Ahhh that was the missing key. THANKS!
 
  • #4
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For b. I get the answer 8.7278 x 10^12 m/s2 after doing F=qE, and the F=ma using the electron mass and found force. Does that make sense?
 
  • #5
gneill
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20,792
2,770
For b. I get the answer 8.7278 x 10^12 m/s2 after doing F=qE, and the F=ma using the electron mass and found force. Does that make sense?
That's the right ballpark. The value looks a bit off, but that might be due to the constants you're using. Can you post more of the work to show how you arrived at that acceleration?

By the way, keep in mind that it's always better to keep more digits through intermediate calculations and to not round anything until you get to the end. That prevents rounding errors from creeping into significant figures.
 
  • #6
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That's the right ballpark. The value looks a bit off, but that might be due to the constants you're using. Can you post more of the work to show how you arrived at that acceleration?

By the way, keep in mind that it's always better to keep more digits through intermediate calculations and to not round anything until you get to the end. That prevents rounding errors from creeping into significant figures.

W = Fd so F= W/d

E =Fe/q so E=W/dq

E= 2.4x10^-19 J/(0.03m)(1.6x10^-19)

E= 50 N/C

F=qE

F=1.6x10^-19 x 50

F = 8 x 10 ^ -18

F = ma

a = F/m

a= 8x10^-18/9.11x10^-31

a= 8.7819x10^12 m/s^2
 
  • #7
gneill
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Much better :smile:
 

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