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Electrostatic, 3 point charges

  1. Nov 18, 2009 #1
    This is from Sadiku's Elements of electromagnetics. I have come to a result but it's different from that of the book.

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

    Three identical small spheres of mass m are suspended by threads of negligible
    masses
    and equal length l from a common point. A charge Q is divided equally
    between the spheres
    and they come to equilibrium at the corners of a horizontal equilateral
    triangle
    whose sides are d. Show that

    [tex]Q^2 = 12\pi\epsilon_0mgd^3 \begin{bmatrix} l^2 - \frac{d^2}{3}\end{bmatrix}^{-\frac{1}{2}}[/tex]


    2. Relevant equations

    [tex]F = \frac{q_1q_2}{4\pi\epsilon_0R^2}\mathbf{a_r}[/tex]

    This is all given by the book.

    3. The attempt at a solution

    The way I see this problem is as follows:

    http://img256.imageshack.us/img256/6063/sadikuex4.png [Broken]

    P being the common point, Fe being the force exerted by the other 2 speheres, mg being the force exerted by gravity.
    The equilateral triangle would be the top view, while the right triangle is a side view of one of the spheres.

    The length of w in the right triangle is:

    [tex]d\sqrt{3}[/tex]

    from the inscribed circle formula in an equilateral triangle.

    If we define alpha as the P angle on the right-angle triangle, and T as the tension on each thread caused by both the electrostatic force and weight of a sphere, then:

    [tex]T\sin \alpha = F_e[/tex]

    [tex]T\cos \alpha = mg [/tex]

    [tex]\frac{\sin \alpha}{\cos \alpha} = \frac{F_e}{mg} (1)[/tex]

    But, by the superposition principle:

    [tex]F_e = \frac{1}{4\pi\epsilon_0} \frac{q^2}{d^2} + \frac{1}{4\pi\epsilon_0} \frac{q^2}{d^2} = \frac{1}{2\pi\epsilon_0} \frac{q^2}{d^2} (2) [/tex]

    where

    [tex]q = \frac{Q}{3} (3)[/tex]

    is the charge of an individual sphere.

    Now,

    [tex]\tan \alpha = \frac{w}{h}[/tex]

    [tex]h = \frac{w}{\tan \alpha}[/tex]

    [tex]l^2 = \frac{w^2}{\tan^2 \alpha} + w^2 = \frac{3d^2}{\tan^2 \alpha} + 3d^2[/tex]

    [tex]\tan^2 \alpha = \frac{3d^2}{l^2 - 3d^2}[/tex]

    [tex]\tan \alpha = \sqrt{3}d( l^2 - 3d^2 )^{-\frac{1}{2}} (4)[/tex]

    Substituting (2), (3) and (4) in (1) and solving for Q yields:

    [tex]Q^2 = 18\sqrt{3}\pi\epsilon_0mgd^3( l^2 - 3d^2 )^{-\frac{1}{2}}[/tex]

    What's wrong?

    Thanks and best regards.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 18, 2009 #2
    No time to read it all, but you should have [itex] d = \sqrt{3} w [/itex] and not [itex] w = \sqrt{3} d [/itex]
     
  4. Mar 12, 2011 #3
    not understand how you get d=(3)^(1/2)
     
    Last edited by a moderator: May 5, 2017
  5. Mar 12, 2011 #4
    w=d(3)^(1/2)
     
  6. Mar 12, 2011 #5
    w is not d*(3^1/2)
    its d/(3^1/2)

    in the picture youre looking at the pyramid from the top..
     

    Attached Files:

  7. Mar 12, 2011 #6
    got the solution....thanks
     
    Last edited: Mar 12, 2011
  8. Mar 12, 2011 #7
    Again looking at the pyramid from the top we see that the horizontal component of the force due to the two charges gets cancelled out...
     

    Attached Files:

  9. Mar 12, 2011 #8
    in the end use the followin equation

    tsin a/tcos a= Fe/mg
     
  10. Mar 12, 2011 #9
    gr8 work qureshi......
     
  11. Mar 13, 2011 #10
    thank you inti
     
  12. Nov 13, 2011 #11
    Thanks, I got my answer..
     
  13. Dec 30, 2011 #12
    Suggestion, its better to use (I) and (IV) in equation labelling than (1) (4) for it looks like it is a scalar multiplication
     
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