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Charges on an equilateral triangle

  1. Apr 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Charges Q, Q, and q lie on the corners of an equilateral triangle with sides of length a. Charge q lies on the top corner with Q and Q on the left and right corners.

    (a) What is the force on the charge q?

    (b) What must q be for E to be zero half-way up the altitudeat P?


    2. Relevant equations

    F=(1/4πε) * (q1q2/r2)

    E = (1/4πε) * (q/r2)

    3. The attempt at a solution

    (a) F = qQ / (4πεa2) iy

    (b) Eq = q / (4πε (3/16 a2)) iy

    and EQ = Q / (4πε (7/16 a2)) iy

    I just wanted to know if the attempts are in proportion to the required?

    Thanks alot in advance.
     
  2. jcsd
  3. Apr 6, 2012 #2

    collinsmark

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    No, that's not quite right.

    You are correct that the net force is completely in the [itex] \hat \imath_y [/itex] direction. That is because the [itex] \hat \imath_x [/itex] components of each force from the Q charges cancel.

    But the [itex] \hat \imath_y [/itex] force components do not lead to your answer.
    Those electric field calculations are not quite right either.
     
  4. Apr 6, 2012 #3
    I tried to do it several times based on may understanding but for some reason I am so not able to reach to the solution. Any hint?
     
  5. Apr 6, 2012 #4

    gneill

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    Did you begin with a diagram of the triangle, labeling all sides in cluding the altitude?
    attachment.php?attachmentid=45937&stc=1&d=1333748959.gif
    If you calculate the force that one Q exerts upon q along a side a, then what is the y-component of that force? Hint: you can determine the trigonometric functions (sin, cosine) of the angle using an appropriate ratio of sides.
     

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  6. Apr 6, 2012 #5

    collinsmark

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    Let's start with the first part (a).

    For each charge Q at the bottom, you can find the magnitude of the force from that particular charge on charge q using

    [tex] \vec F = \frac{1}{4 \pi \varepsilon_0} \frac{qQ}{r_2} \hat \imath_r. [/tex]

    where [itex] \hat \imath_r [/itex] points in the direction from the particular Q to q.

    But the [itex] \hat \imath_r [/itex] for one charge is different from the [itex] \hat \imath_r [/itex] of the other charge, because they point in different directions. So you can't just add them simply. For each case, you need to break up [itex] \hat \imath_r [/itex] into its [itex] \hat \imath_x [/itex] and [itex] \hat \imath_y [/itex] components.

    That might involve something like

    [tex] F_x = F\cos \theta [/tex]
    [tex] F_y = F \sin \theta [/tex]
    (This relationship is just for example purposes, the particular relationship may depend on the particular problem/particular pair of charges).

    Or if you don't want to use trigonometric functions, you can do it using the Pthagorean theorem (which works with this problem). If a is the side of an equilateral triangle, and you vertically bisect the triangle down the middle, forming two isosceles triangles, then the base of each isosceles triangles is a/2. What does Pthagorean theorem tell you about the height of the triangle relative to a?
     
  7. Apr 7, 2012 #6
    I tried and I finally reached the solution. Thanks for all your support!!
     
  8. Feb 20, 2014 #7
    I am working on the second part of this question and don't understand why the (7/16) is to the power of 3/2. Am I just missing something obvious?
     
    Last edited: Feb 20, 2014
  9. Feb 20, 2014 #8

    gneill

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    Staff: Mentor

    In order to be able to comment we'll have to see your work.
     
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