Charge/Field/Velocity Problem

  • #1
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Homework Statement


PHYSICS.png


Homework Equations


F=kqQ/D^2=Ma
E=kQ/D^2
D=vt?

The Attempt at a Solution


So far, I've been able to label the forces on each ship through superposition. But that is all I currently able to understand. I know that my end solution is a speed/velocity, but I am not sure how to get there with the given (Q, t, D, and M)

Any help and direction would be greatly appreciated!
 

Answers and Replies

  • #2
gneill
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Have you considered a conservation law approach?
 
  • #3
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I have not because we haven't discussed it. How might I use said approach?
 
  • #4
gneill
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I have not because we haven't discussed it. How might I use said approach?
Usually one considers the trade off between potential and kinetic energy. What form of potential energy is involved here?
 
  • #5
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Usually one considers the trade off between potential and kinetic energy. What form of potential energy is involved here?

There is definitely potential energy between each ship.
 
  • #6
gneill
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There is definitely potential energy between each ship.
Yes... what type and how much?
 
  • #7
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Yes... what type and how much?

Electrical Potential Energy
U12=kq1q2/D
U23=kq2q3/D
U13=kq1q3/D
 
  • #8
gneill
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What's the total PE of the system in terms of the variables that you were given?
 
  • #9
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U(total) = 3* kq^2/D, right?
 
  • #10
gneill
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U(total) = 3* kq^2/D, right?
Looks good.

So how will that change as the ships all tend towards infinite separation? Where will it go?
 
  • #11
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Looks good.

So how will that change as the ships all tend towards infinite separation? Where will it go?

All the potential will simply turn into kinetic energy right?
 
  • #12
gneill
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All the potential will simply turn into kinetic energy right?
That's right.
 
  • #13
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That's right.

Alright, so then 3* kq^2/D=1/2Mv^2 and just solve for v
 
  • #14
gneill
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Alright, so then 3* kq^2/D=1/2Mv^2 and just solve for v
Remember that the energy will be split three ways.
 
  • #15
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Remember that the energy will be split three ways.

So is the multiplying by 3 completely necessary if the charges and D are the same? Couldn't I just solve it as kq^2/D=1/2Mv^2
 
  • #16
gneill
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So is the multiplying by 3 completely necessary if the charges and D are the same? Couldn't I just solve it as kq^2/D=1/2Mv^2
Sure. In your presented solution be sure to state why you are justified in doing it.
 
  • #17
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Sure. In your presented solution be sure to state why you are justified in doing it.

Ah okay, I understand now! Thank you very much! :)
 

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