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Electric Charge Question

  1. Jan 18, 2008 #1
    The problem statement, all variables and given/known data

    [​IMG]

    The attempt at a solution

    Kq1q3/(L+x)2 = Kq2q3/(x)2
    q1/(L+x)2 = q2/(x)2

    I came up with the question above. But when I plug in the numbers I get 2 different answers. For x, I got -7.363450837 and -21.73332335. I tired both answers in the system and I got it wrong. I am not sure why this is happening.

    Do I have to add +11 to the x to get the real answer? If so then to which one do I add 11 to?
     
  2. jcsd
  3. Jan 18, 2008 #2

    Doc Al

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    Staff: Mentor

    Hint: First figure out in what region along the x-axis must q3 be placed? (x < 0, x > L, or 0 < x < L).

    (If you can't figure out which region, try them all!)

    Also: Be careful with your signs. (You can't just use magnitudes.)
     
  4. Jan 18, 2008 #3
    I can't figure out which region it is in. I have only have 1 more attempt left to get the answer correct.
     
  5. Jan 18, 2008 #4

    Doc Al

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    Then try each region (one at a time) and set up the equation such that the net field is zero. Of the three regions, only one will give a sensible answer. Try it!

    Note: The direction of the field counts.
     
  6. Jan 18, 2008 #5
    I think its on the left of the q1. But then I am not sure about the formual.

    If it is to the left of q1, that means that x is the distance between q1 and q3. And L+x is the distance between q2 and q3. I am guessing I would use this question

    q1/(x)2 = q2/(L+x)2

    Am I right? Also since q2 is a negitive, when I try to solve in get complex number which doesn't let me solve for x.
     
  7. Jan 18, 2008 #6

    Doc Al

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    You are almost there.

    Let q1 and q2 stand for just the magnitudes of the charges. (Put in the signs by hand.) The field from q1 (which is the positive charge) points to the left, which we'll call negative. So: E1 = -q1/(x)^2. Similarly, E2 = +q2/(L+x)^2.

    Find x when E1 + E2 = 0.
     
  8. Jan 18, 2008 #7
    I am so lost now. I don't get it at all. Can you please give me the answer and I will be sure to study the soulution when it is posted.
     
  9. Jan 18, 2008 #8

    Doc Al

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    Don't stop now. You are practically done.

    That's the exact equation you need. Just use the magnitudes of the charges.
     
  10. Jan 18, 2008 #9

    Doc Al

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    Note that -q1/(x)^2 + q2/(L+x)^2 = 0 is the same as:
    q1/(x)^2 = q2/(L+x)^2

    (q1 and q2 are both positive numbers.)
     
  11. Jan 18, 2008 #10
    Okay I took the absoulute value of the charges and used the question. I got two values for X. I got -.0363654916 and .1073332336.

    I tried putting in the answer -.0363654916 and I got it wrong. I have one attempt left. What do I do?
     
  12. Jan 18, 2008 #11

    hage567

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    Homework Helper

    Are your answers expressed in meters? It looks like it. I think the homework program is asking for the answer in cm. Check your units.
     
    Last edited: Jan 18, 2008
  13. Jan 19, 2008 #12

    Doc Al

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    I thought we had agreed that the charge must be placed to the left of q1. And we used "x" to represent the distance to the left of q1. So "x" must be positive. (A negative value of "x" means to the right of q1--which means it's not a solution that satisfies our initial conditions.)

    It probably would have been a bit smarter if we called the distance to the left of q1 by the letter "d", so we wouldn't confuse it with the x-coordinates. Oh well! :uhh: (This is probably the source of your confusion!)

    So you solved for "x" and chose the only answer that makes sense (the positive value). Now you must translate our "x" to the coordinate along the x-axis. A positive value for "x" means a negative x-coordinate. (Since q1 is at the origin.)

    I hope this makes sense.

    Also: As hage567 advised, make sure your answer is in cm, not meters.
     
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