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Electrostatic equilibrium

  1. May 10, 2006 #1
    "Two charges, q1, and q2, are located at the origin and at (0.50m, 0), respectively. Where on the x axis must a third charge, q3, of arbitrary sign be placed to be in electrostatic equilibrium if (a) q1 and q2 are like charges of equal magnitude, (b) q1 and q2 are unlike charges of equal magnitude, and (c) q1/sub] = +3.0 microCoulombs and q2 = -7.0 microCoulombs?"

    for (a), I thought since they are like charges, q3 can only be in electrostatic equilibrium if it's between the two charges.
    let r = distance from (0,0)
    F31 = F32
    (kq3q1)/r2 = (kq3q1)/(0.50-r)2
    r2 = (0.50-r)2
    r = 0.25 m
    but the answer is -0.25m. Is that a typo?

    for (b), I thought since they are opposite charges, q3 can only be in equil. if it's not between the two charges.
    let r = distance from (0,0) or (0.50, 0)
    F31 =- F32 (is it correct to put the negative sign here?)
    (kq3q1)/r2 = -(kq3q1)/(0.50+r)2
    r2 = (0.50+r)2
    0 = 0.50
    but this doesn't make sense, so there is no location for q3 to be in equilibrium. (This answer is correct, but I think I could've solved it by just thinking about it... but I'm not sure how)

    for (c), I thought the only place for q3 (regardless of its sign) is to the left of (0,0)
    F31 =- F32
    (kq3q1)/r2 = -(kq3q1)/(r+0.50)2
    q1(r+0.50)2 = -q2r2
    (q1 + q2)r2 + q1r +0.25q1 = 0
    I use the quadratic formula, and take only the negative root, since r < 0
    r = -0.20m. So q3 would be at (-0.20,0).
    but the answer here is -0.94m
     
  2. jcsd
  3. May 10, 2006 #2

    nrqed

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    Yes, that is a typo
    One way to see it quickly is this: You know that it cannot be between the two charges because of the signs of the charges (the two forces must be in opposite directions for them to cancel out). Now, you must also impose that the *magnitudes* of the two forces must be the same for them to cancel out. But the charges q1 and q2 have the same magnitude..therefore the *distance* q1 to q3 and q2 to q3 must be the same...But wait! That's impossible if we are not between the charges. QED
    I will tell you in a minute
     
  4. May 10, 2006 #3

    nrqed

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    No, because the meaning of "r" in this equation is that it is a *distance*!
    (this is why you use (r+0.50)). So when you solve, you *must* have a positive value for r. Once you know the distance from the origin, then you get the *x coordinate* by saying x=-r.
     
  5. May 10, 2006 #4
    Thanks a lot nrged! I get it now :smile:
     
  6. Jan 16, 2009 #5
    Two charges, q1, and q2, are located at the origin and at (0.50m, 0), respectively. Where on the x axis must a third charge, q3, of arbitrary sign be placed to be in electrostatic equilibrium if (a) q1 and q2 are like charges of equal magnitude, (b) q1 and q2 are unlike charges of equal magnitude, and (c) q1/sub] = +3.0 microCoulombs and q2 = -7.0 microCoulombs?"

    for (a), I thought since they are like charges, q3 can only be in electrostatic equilibrium if it's between the two charges.
    let r = distance from (0,0)
    F31 = F32
    (kq3q1)/r2 = (kq3q1)/(0.50-r)2
    r2 = (0.50-r)2
    r = 0.25 m

    I have almost the same exact problem in my homework. All except c is the same. I am comfused on making the relationship between F31 and F32. If they are equal, how can we have /(0.50-r)2? My question is, how do we know it is .5-r? not .5+r? Either way, I don't understand how one know to write any number -r or +r. I need some conceptual help or something.
     
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