Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Static equilibrium for three point charges

  1. Jan 22, 2012 #1
    The result I got seemed sort of messy and I'm not sure I've gotten the basic idea correctly, so I thought I'd make it sure. I think this was the easiest problem we had to solve for charge distributions and point charges and whatnot, so I know for sure I'll have to practice more if I made any big mistakes. I'm especially uncertain of my reasoning.

    1. The problem statement, all variables and given/known data

    A charge of 2q is placed at the origin and a charge of 3q is apart from it by a distance L. Where should a third charge be placed, and how large would it be, so the system would be in a static equilibrium.

    2. Relevant equations

    Electrostatic force F=(1/4πε_0)*((q_1*q_2)/r^2)*û_r

    3. The attempt at a solution

    The force on each charge has to be zero in order for the system to be in a static equilibrium. It is assumed that the charges are point charges and there are no external forces acting on them. Let the magnitude of the unknown charge be Q and the distance between 2q and Q be r.

    The forces on the particle with a charge of 2q:

    F = F_3q+F_Q = 0
    k*(2q*3q)/L^2)*û_L + k*(2q*Q)/r^2)*û_r = 0
    2qk* [ (3q/L^2)*û_L + (Q/r^2)*û_r ] = 0
    (3q/L^2) * û_L = (Q/r^2) * û_r

    Which is possible only if û_L || û_r, as 3q/L^2 and Q/r^2 are both scalars (I didn't make a mistake with this one, right?). So the unit vectors are parallel and three charges have to be on the same line.

    Static equilibrium is only possible when the charge Q is negative, and placed inbetween the two other charges (as there must be forces to two directions on each particle).

    Solving the magnitudes of Q and r, as the direction of r is no longer needed. Q is |Q| for now on, and I've noted the directions of the forces, choosing the direction from 2q to 3q to be positive. Two equations are enough for the two variables Q and r.

    The forces on the particle with a charge of 2q:

    F = F_Q - F_3q = 0
    <=> k*(2q*Q)/r^2 - k*(2q*3q)/L^2 = 0
    <=> 2q*k*( Q/r^2 - 3q/L^2 ) = 0
    <=> Q/r^2 = 3q/L^2
    <=> Q = 3q *(r/L)^2

    The forces on the particle with a charge of 3q:

    F = F_2q - F_Q = 0
    <=> k*(3q*2q)/L^2 - k*(Q*3q)/(L-r)^2 = 0
    <=> 3q*k*( 2q/L^2 - Q/(L-r)^2 ) = 0
    <=> 2q/L^2 = Q/(L-r)^2 // The Q has been solved for r
    <=> 2q/L^2 = (3q *(r/L)^2) /(L-r)^2
    <=> 2q = 3q * (r^2*L^2)/(L^2*(L-r^2))
    <=> 2/3 = r^2/(L-r)^2
    <=> 2(L-r)^2 = 3r^2
    <=> 2L^2 - 4Lr + 2r^2 = 3r^2
    <=> r^2 + 4Lr - 2L^2 = 0

    Solving for r eventually gives

    r = (√6-2)L (~= 0.449L, makes sense to me as Q should be closer to 2q than to 3q)

    Solving Q with r solved

    <=> Q = 3q * ((sqrt6-2)L/L)^2
    <=> Q = 3q * (6 - 4√6 - 4)
    <=> Q = 6q * (5 - 2√6) (~= 0.606q, no idea whether this makes any sense or not. And the actual Q is obviously ~= -0.606q, as this was |Q|.)

    Answer: r = (√6-2)L, Q = -6q*(5 - 2√6)

    I'm not a native English speaker, so I apologize for any mistakes. (I also apologize for the lack of Latex, I've never gotten around learning it.)
    Last edited: Jan 22, 2012
  2. jcsd
  3. Jan 22, 2012 #2
    I assume the charges 2q and 3q are fixed. Consider the electric field left of, right of, and in between the fixed charges. At those locations place a positive charge, which way does it go, or where can the electric field sum to zero? See,

    Attached Files:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook