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Thread: Electrostatic Force

  1. #1
    Super Member Aryth's Avatar
    Feb 2007

    Electrostatic Force

    I know I typed this question before... But the answers on there turn out not to be the method I'm looking for. So I'm going to post what I have so far and maybe someone can help me out with second part of this:

    The question is:

    Four particles form a square. The charges are $\displaystyle q_1 = q_4 = Q$ and $\displaystyle q_2 = q_3 = q$.

    Just for your info, the particles are arranged as such:

    1 2
    3 4

    (a)What is $\displaystyle \frac Qq $ if the net electrostatic force on particles 1 and 4 is zero?

    Here's my solution for part a:

    (My notation for Force for particles is this, say I want to say the force of particle 2 on 1, I write $\displaystyle F_{21}$)

    $\displaystyle \sum F_1 = F_{21} + F_{31} + F_{41}$

    This says that the net force on 1 is the sum of the forces on 1 from 4, 3, and 2.

    $\displaystyle \sum F_1 = k\frac{qQ}{a^2} + k\frac{qQ}{a^2} + k\frac{Q^2}{(\sqrt{2}a)^2}$

    $\displaystyle = 2k\frac{qQ}{a^2} + k\frac{Q^2}{2a^2}$

    $\displaystyle 2k\frac{2qQ + Q^2}{2a^2} = k\frac{2qQ + Q^2}{a^2}$

    Turns out I get the same thing for $\displaystyle \sum F_4$ because the net force for Q is 0. So, now I get the net force on Q:

    $\displaystyle \sum F_Q = 2k\frac{2qQ + Q^2}{a^2} = 0$

    $\displaystyle \frac{2k}{a^2}(2qQ + Q^2) = 0$

    Since k and a are obviously not 0, then $\displaystyle (2qQ + Q^2)$ must be 0, so:

    $\displaystyle 2qQ + Q^2 = 0$

    $\displaystyle Q^2 = -2qQ$

    $\displaystyle Q = -2q$

    $\displaystyle \frac Qq = -2.0$

    Is that right? If so, here's part b.

    (b) Is there any value of q that makes the net electrostatic force on each of the four particles 0?
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  2. #2
    Senior Member vincisonfire's Avatar
    Oct 2008
    Here is a scheme of your problem. The axis are in black.
    The Qs are the red dot. The qs are the green dot.
    The force Q on Q is represented by the red arrow.
    The force of the qs on Q are represented by the green arrows.
    Their projection on each axis is represented by the yellow and blue arrows.
    You can see that the yellow arrow cancel each other.
    Two superposed blue arrows have to compensate for the long red arrow.
    The blue arrow have $\displaystyle cos(\frac{\pi}{4}) $ of the magnitude of the green arrows that represente the full force of the qs on Q.
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