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Homework Help: Electric Force, Equilibrium Config of Charges

  1. May 14, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the charge Q that should be placed at the centre of the square of side 8.50E+0 cm, at the corners of which four identical charges +q = 3 C are placed so that the whole system is in equilibrium.

    2. Relevant equations
    F=(k*q1*q2) / r^2

    3. The attempt at a solution
    I know the solution is something like this:

    Let r = side of square. That's 8.5 cm, right?
    Without the center charge, F1 (on each charge, away from center) = kq^2/(sqrt(2)*r)^2 (diagonal charge) + 2*sqrt(0.5)*kq^2/r^2 (2 adjacent charges) =
    k(q/r)^2*(0.5+sqrt(2)) = 2.1430705E13 N
    F1 = -F (center charge Q to each corner charge q) = kqQ/(r*sqrt(0.5))^2 = 2kqQ/r^2
    ==> Q = -(F1)r^2/(2kq) = -(0.5+sqrt(2))q/2 = -2.87132 C

    my only question is this:
    why does the F of the 2 adjacent charges = 2*sqrt(0.5)*kq^2/r^2
    where does the 2*sqrt(1/2) come from? why isnt it just (2*k*q^2)/r^2
  2. jcsd
  3. May 15, 2012 #2
    Forces are vectors. You can't add them like normal scalars. Draw a diagram and split components, and you'll see why :smile:
  4. May 16, 2012 #3
    i actually dont get this whole thing at all lol ...

    so when you break it down, F1 = F12 + F13 + F14
    F12 and F14 are the adjacent charges. aren't those just kq^2/r^2 for each?
    and the F13 is the diagonal charge .. isnt that just kq^2/r^2 but in the x dir it has a cos(theta) attached to it and for the y dir it has a sin(theta) attached ..
  5. May 16, 2012 #4
    it doesnt make sense at all. F12 should just be (k*q^2)/r^2 and same with F14 .. it doesn't make sense to me why F12+F14 = (2*sqrt(0.5)*k*q^2)/r^2
  6. May 16, 2012 #5
    Do you know addition of vectors? Can you tell me what the sum of the two vectors in the attachment would be? Assume the magnitude of both the forces is F, and they are perpendicular.

    Attached Files:

  7. May 16, 2012 #6
    adding those 2 vectors equals the 'hypotenuse' .. i dont know what you want for an answer
  8. May 16, 2012 #7
    Yes, its the hypotenuse. What is the magnitude of that resultant sum?
  9. May 16, 2012 #8
  10. May 16, 2012 #9
    Yep. Now what you have in your original question is,

    F12 = F14 = (kq2)/r2 = F

    And they are perpendicular to each other....Apply the same concept you used for my question...
  11. May 16, 2012 #10
    then F12 + F14 = sqrt(2)*(kq^2)/r^2 + sqrt(2)*(kq^2)/r^2 = 2*sqrt(2) (kq^2)/r^2 ...
  12. May 16, 2012 #11
    How did you get the second term from!?

    When they're perpendicular,

    F + F = sqrt(2) F, thats it.
  13. May 16, 2012 #12
    oh yea right. so just sqrt(2)*(kq^2)/r^2 so whats that mean now
  14. May 16, 2012 #13
    Well, there you go :biggrin:
  15. May 17, 2012 #14
    well you didnt answer my question ... in the problem's answer i posted in my original first post, which gave me the right answer, it says that F12+F14 = 2*sqrt(0.5)*kq^2/r^2

    2*sqrt(1/2) does not equal sqrt(2) ..
  16. May 17, 2012 #15

    Are you really sure....?
  17. May 17, 2012 #16
    maybe they made a typo since it still works out *shrugs* thats all the reason why i was confused ... thanks though :)

    ohh yea duhh i dont know what i was thinking for that part .. i think i got the radical on the bottom mixed up
  18. May 17, 2012 #17
    ohhh yea duh idk what i was thinking for that. mixed up the radical on the bottom
  19. May 17, 2012 #18
    [itex]2\frac{1}{\sqrt{2}} = \sqrt{2}[/itex]

  20. May 17, 2012 #19
    haha yea like i said mixed up the radical. wasnt thinking to take out the radical on the bottom lol
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