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Homework Help: Equilbrium charge between 2 charges

  1. Jan 16, 2005 #1
    So what I did was:

    Kq1/(1+r) + Kq2/r = 0, making r =the distance away from q2 outside of the system to be the equilibrium point. so q1 is 1m + r distance away from the point, and q2 = r distance away from the point.

    K's cancel out, you get:
    q1/(1+r) + q2/r = 0
    Some algebra, turns out r = -0.2m.

    Does that mean the point is 0.2m inwards from q2->q1 instead of outside of the system?
    aka: (q1) -----0.8m----(qe)--0.2--(q2), where qe = equilbirium point?

    Also, arent there two points which are in equliibrium of the system? We did a problem in class similar to this and had 2 pointsat equlibrium. Any help is greatly appreciated.
    Last edited: Jan 16, 2005
  2. jcsd
  3. Jan 16, 2005 #2


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    Are you sure you're supposed to be using the first power of r in your equations ? How does the inverse square law pertain here ? When should you instead use the first power as you did here (hint : potential).

    I get a "nice" answer between the two charges.
  4. Jan 16, 2005 #3
    isn't the potential of a charge: kq/r?

    well q(Vb - Va) = kq(q-initial) / ra - kq(q-initial) / rb
    but thats for moving a charge from one place to another, right?
  5. Jan 16, 2005 #4
    OHhhhh is it:

    Kq1 / r-from-q1 + kq2 / r-from-q2 = 0

    so the K's cancel out leaving:

    q1 / r1 + q2 / r2 = 0

    q1 and q2 are knowns but r1 and r2 are unknowns. how would that be solved?

    Would it be, kq1qe / r1^2 = kq2qe / r2^2
    where q1 = q1
    q2 = q2, and qe = equilbrium charge
    r1 = distance from q1 - qe,
    r2 = distance from q2 - qe

    from that you get kq1qe / r1^2 = kq2qe / r2^2
    becomes: q1 / r1^2 = q2 / r2^2
    where q1, and q2 are known.

    Then using system of equations, solve for r1 and r2?
    Last edited: Jan 16, 2005
  6. Jan 16, 2005 #5
    This makes no sense to me...

    equation 1) q1/r1 +q2/r2 = 0
    equation 2) r1 [distance between equilibrium charge -> q1] = 2(R2)

    plugging it in just gives me zero. help please, anyone?
  7. Jan 16, 2005 #6


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    Potential yes, electric field strength, no.

    What is the definition of electric field strength ? What is the electric field strength at distance r from point charge q ? Show some equations to prove understanding.
  8. Jan 16, 2005 #7
    Alright, well.

    The definition of electric field strength? The amount of force that is pulling or pushing on charges surrounding a charge?

    E=F/q, so the Electric Field strength would be the Force [kq1q2/r^2] divided by the charge, q.

    That means, E = kq1 / r^2.

    Oh, also - Eleectric Potential Energy = qV, where V = kq/r. So, EPotential Energy = kq^2/r.
    Last edited: Jan 16, 2005
  9. Jan 16, 2005 #8


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    This is correct. So you realise you were wrong to use the first power in your first attempt, correct ? Because there you were trying to find a zero potential point, which is NOT what the question was asking for.

    So rewrite your first attempt using the second power of r and get a result and I'll check to see if it's right.
  10. Jan 16, 2005 #9
    OHHH wow. Thanks a lot.

    Ok, so for the net electric field to be 0, kq1 / (1+r)^2 + kq2/(r^2) = 0

    Alright so:
    1) k's drop out, leaving q1 / (1+r)^2 + q2/(r^2) = 0
    2) (6.8x10^-6) / (1+2r+r^2) + (-1.7x10^-6) / (r^2) = 0
    - multiply through by (1+2r+r^2)(r^2)
    3) (6.8x10^-6)(r^2) + (-1.7x10^-6)(1+2r+r^2) = 0
    4) R becomes a quadratic formula giving +.66667 and -.66667??

    I feel so lost. Does that seem remotely correct?

    *For q2, should I be putting in the negative (-) in front of it since it's a negative charge??
    Last edited: Jan 16, 2005
  11. Jan 16, 2005 #10


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    I don't see why you're having so much trouble.

    Let q1 (the [itex]+6.8 \mu C [/itex] charge) lie at position r = 0 and the other q2 ([itex]-1.7 \mu C[/itex])lie at position r = R. Here R = 1 m.

    Then at a position r, the net electric field strength of the system is :

    [tex]+ \frac{kq_1}{r^2} - \frac{kq_2}{(R - r)^2}[/tex]

    isn't it ?

    Equate that expression to zero, and solve for r in terms of R. What do you get ?

    You should get [tex]r = R(1 + (\frac{q_2}{q_1})^{\frac{1}{2}})^{-1}[/tex]

    BTW, q1 and q2 refer to the absolute numerical values of the charges, since I've taken care of the signs in my formulation.
    Last edited: Jan 16, 2005
  12. Jan 16, 2005 #11
    Yep [tex]r = R(1 + (\frac{q_2}{q_1})^{\frac{1}{2}})^{-1}[/tex] equates to 0.66666. That's what I got too using what I said above. I think it's doing the same thing.

    Anyway, thanks a lot for the help.
  13. Jan 17, 2005 #12


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    Then you have basically the right idea, but you don't have to solve a quadratic to get the answer. Just do this :

    [tex]+ \frac{kq_1}{r^2} - \frac{kq_2}{(R - r)^2}[/tex]

    [tex]\frac{q_2}{q_1} = (\frac{R}{r} - 1)^2[/tex]

    [tex]\frac{R}{r} = 1 + (\frac{q_2}{q_1})^{\frac{1}{2}}[/tex]

    and you can manipulate that easily.

    But it's OK to do it via a quadratic, just more tedious. You do understand why you should ignore the negative root though, correct ?
    Last edited: Jan 17, 2005
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