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: Electrostatics 1st year engineering

  1. Mar 9, 2009 #1
    at each corner of a rhobus , made of rods, each with a lenght of B, there are electric charges of "Q" and "q" so that each q charge is opposite a q charge and each Q charge opposite a Q charge. what does the ratio of Q/q need to be so that the angle (alpha) is

    a)90 degrees
    b)60 degrees
    c)45 degrees
    (take the tension of the rods into account)

    i started up by taking one of the corners and analyzing the forces on it, since the shape is static, Ftotal=0.
    i took the top Q charge and said, what are the forces acting on it,

    the Coulumb force F(Q-Q) which pushes it upwards
    the Coulumb force F(Q-q) which (depends on the charge)
    the tension os the rods T

    my problem is that i do not know which of the charges are negative(if any) or positive, so i cannot decide on the directions of the forces, the only one i do know is F(Q-Q) since Q must repell Q,

    diagram below

    diagram below
    http://picasaweb.google.com/devanlevin/DropBox?authkey=Gv1sRgCL_4l4PpvP_YsQE#5311131135994735922 [Broken]
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 9, 2009 #2
    what i did so far is:

    since i dont know the sign of q, the coulomb force can either repell or attract Q

    F(QQ)=2Tcos(α/2) ± 2F(Qq)cos(α/2)

    q will always repell q so T is the force which balance that
    2Tsin(α/2) = F(qq)

    F(QQ)=F(qq)cot(α/2) ± 2F(Qq)cos(α/2)

    K[tex]\frac{QQ}{(2b*cos(alpha/2))^2}[/tex] = K[tex]\frac{qq}{(2b*sin(alpha/2))^2}[/tex]*cot(alpha/2) ± 2K[tex]\frac{Qq}{b^2}[/tex]cos(alpha/2)

    (Q/q) * [tex]\frac{Q}{cos^2(alpha/2)}[/tex] = [tex]\frac{q}{sin^2(alpha/2)}[/tex]*cot(alpha/2) ± 8Qcos(alpha/2)

    1st of all, is this alright?
    2nd of all, how do i find Q/q from this?? how do i get the q's and Q's out of this equation?
    Last edited: Mar 9, 2009
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook