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Charged balls connected by spring

  1. Sep 29, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider two small balls with charges +q and -q, connected by a spring with spring constant k. A homogeneous electric field E has been turned on. E is collinear with the line that connects -q to +q. By how much will the string stretch?

    2. Relevant equations
    Coulomb's Law
    Hooke's Law
    3. The attempt at a solution
    I figured that we could use F=-kx and substitute Coulomb's Law in for F. Doing this and solving for x, I obtained (1/(-4πεk))(q^2/r^2), where ε is the permittivity constant and k is the spring constant. Was this the correct way to approach this problem? Any input would be appreciated. Thanks!
     
  2. jcsd
  3. Sep 29, 2015 #2

    ehild

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    I do not see that you took the applied electric field E into account.
     
  4. Sep 29, 2015 #3
    How would one go about doing that? I know that F=qE, so that's how I figure the electric field could be taken into account, but I'm not sure how to go about representing that mathematically. Wouldn't it look very similar to Coulomb's Law?
     
  5. Sep 30, 2015 #4

    ehild

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    No, E is homogeneous, it does not depend on the position of the charges.The force due to E on charge q is F=qE.
    Draw the FBD at the positive charge. What forces act on it?
     
  6. Sep 30, 2015 #5
    The forces acting on the ball would be the coulomb force and the homogeneous electric field. So, you'd represent the force on the left side of the equation as (coulomb force)+qE?
     
  7. Sep 30, 2015 #6

    ehild

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    Left side of what equation?? Yes, and you have the spring force, too.
     
  8. Sep 30, 2015 #7
    the left side of Hooke's law, ##F=-kx##.
     
  9. Sep 30, 2015 #8

    ehild

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    OK, then write out in detail. The relaxed length of the spring was not given?
     
  10. Sep 30, 2015 #9
    He told us to assume initial length as ##l_{0}##. So, therefore, I'm thinking we can represent this system as ##\frac{1}{-4\pi\varepsilon _{0}} \frac{q^{2}}{r^{2}}+\mathbf{E}=-kx##. Does this seem reasonable?
     
  11. Sep 30, 2015 #10
    Could x be replaced by ##(l_{0}-r)##?
     
  12. Sep 30, 2015 #11

    ehild

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    It would be useful to make a drawing, and show the direction of the forces.
    I do not know, what you call x and what direction you assumed for E.
     
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