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Electric Charge Problem - Book error?

  1. Jun 4, 2007 #1
    Hit enter a bit too fast for the title.

    TITLE: Electric Charge Problem - Book Error?

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

    6. In Fig. 21-22, four particles form a square. The charges are [itex]q_{1} = q_{4} = Q[/itex] and [itex]q_{2} = q_{3} = q[/itex].
    (a) What is [itex]Q/q[/itex] if the net electrostatic force on particles 1 and 3 is zero?
    (b) Is there any value of [itex]q[/itex] that makes the net electrostatic force on each of the four particles zero? Explain.


    2. Relevant equations

    \displaystyle{\left|\vec{F}_{12}\right| = \frac{k \left|q_{1}\right|\left|q_{2}\right|}{r_{12}^2}}

    3. The attempt at a solution

    I've actually worked through this already however, I am in conflict with how the problem is stated.

    Essentially all one has to do for part (a) is find Q/q, and the problem already states that the net forces on particles 1 and 3 are zero. Therefore, from the picture several conclusions can be made. In order for the net forces on particles 1 and 3 to be zero, charges Q and q must be unlike-sign.

    Next, because the problem states that the net force for particles 1 and 3 is zero, either particle can be used to find Q/q.

    Continuing on, therefore choosing particle three for example, we need to break the forces acting on three due to the other three particles in to components (Fx and Fy).

    From here the net force on three is zero, and therefore demands Fx and Fy be zero aswell.

    Now, either approach to summing up the components for net: Fx or Fy will lead to one of the summed components being zero.

    Choosing Fy, we arrive at N.III.L.

    \left|\vec{F}_{31}\right| = \frac{\left|\vec{F}_{32}\right|}{\sqrt{2}}}

    Now from here its simple algebra.

    \displaytype{\frac{\left|Q\right|}{\left|q\right|}} = \displaytype{\frac{1}{2\sqrt{2}}}

    and noting that Q and q must be unlike-sign, the above reduces to

    \displaytype{\frac{Q}{q}} = \displaytype{\frac{-1}{2\sqrt{2}}}

    However, the correct answer (from the solutions manual (SM)) is

    \displaytype{\frac{Q}{q}} = \displaytype{-2\sqrt{2}}

    The SM shows arrives at this solution through resolving the Fx components for the net force on particle 1.

    So here is the problem, I can get the same answer ([itex]\displaytype{\frac{Q}{q}} = \displaytype{-2\sqrt{2}}[/itex]) if I resolve either component (Fx or Fy) for particle 1.

    However, I do not get the same answer when I resolve the components (either Fx or Fy) for particle 3.

    I consistently get [itex]\displaytype{\frac{Q}{q}} = \displaytype{\frac{-1}{2\sqrt{2}}}[/itex] for particle 3.

    Also, I will try resolving the net Fx component on particle 3 and see what I get.

    SO, technically for this problem shouldn't I be able to get the same answer (Q/q), through all four ways; that is resolving the two components (Fx and Fy) for each particle (1 and 3)???

    Any help would be appreciated.
    Last edited: Jun 5, 2007
  2. jcsd
  3. Jun 4, 2007 #2


    User Avatar

    Staff: Mentor

    Welcome to the PF. You need to show some of your own work in order for us to help you. The relevant equation that you list is a good start. Now draw the square and start drawing the forces. Are you sure you've listed all the particles? Seems like it would be hard to get the forces to cancel with just 4 particles at the vertices of a square....
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