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Homework Help: Force due to a uniformly charged ring

  1. Jan 15, 2010 #1
    1. The problem statement, all variables and given/known data

    A uniformly charged ring has a radius a, lies in a horizontal plane, and has a negative charge given by -Q. A small particle of mass m has a positive charge given by q. The small particle is located on the axis of the ring.

    What is the minimum value of q/m such that the particle will be in equilibrium under the action of gravity and the electrostatic force? (Use the following variables as necessary: a, k, Q, and g.)?

    2. Relevant equations

    I let x = the distance the charge is from the center of the ring.

    [tex]E[/tex]=[tex]\frac{kQx}{\sqrt{a^2+x^2}^{3}}[/tex]

    3. The attempt at a solution

    I started by saying that "mg" must be equal to:

    [tex]\frac{kQxq}{\sqrt{a^2+x^2}^{3}}[/tex]

    However, I cannot eliminate the x variable from the problem. Am I missing something or do I have the wrong approach altogether?
     
  2. jcsd
  3. Jan 15, 2010 #2

    ehild

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    You do not need to eliminate x. Express q/m as function of x from the condition qE=mg, and find where this function has its minimum.

    ehild
     
  4. Jan 15, 2010 #3
    I can express q/m as a function of x however it is also a function of several other variables which have no numerical values. How do I go about finding a minimum for the function algebraically?
     
  5. Jan 15, 2010 #4

    ehild

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    "Use the following variables as necessary: a, k, Q, and g".

    You need to give a formula, not a number.

    ehild
     
  6. Jan 15, 2010 #5
    Okay. But how do I find a formula for the minimum using only the variables a, k, Q, and g? I tried to take a derivative (thinking I may be able to then find a local minimum) but it got really nasty.

    Am I missing something? - Another way to find the minimum?
     
  7. Jan 16, 2010 #6

    ehild

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    [tex]mg = \frac{kQxq}{\sqrt{a^2+x^2}^{3}} \rightarrow q/m= \frac{g}{kQ}\frac{(x^2+a^2)^{3/2}}{x}[/tex]



    Find the position of minimum of the function.

    [tex]f(x)=\frac{(x^2+a^2)^{3/2}}{x}[/tex]

    Replace back this x into the formula for q/m.

    ehild
     
  8. Jan 16, 2010 #7
    I got it! I had to grunt it out and it turns out the answer was not that pretty but it was correct nonetheless. Thanks for your help!
     
  9. Jan 16, 2010 #8

    ehild

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    Congratulation! It was not easy to get the result, but you did it!
     
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