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Consider the configuration consisting a +q charge...

  1. Sep 16, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    and two -q charges aligned along the x axis as follows...

    -q__________q__________-q <--- the space between them being d

    a) Suppose the +q charge is displaced perpendicularly by a small vertical distance dy. What is the total force F acting on it? Taylor expand your answer to leading order in dy



    b) Next suppose instead the +q charge is displaced along the x axis by a small dx distance. What is the force acting on it?

    --------------

    So for a) I'm kind of stuck at the Taylor expansion on the force.

    F = (q)(-q)/4piε0d2 yhat + (q)(-q)/4piε0d2

    where d= sqrt(x^2+dy^2)

    I'm at a point where I'm trying to simplify something but feel that I"m doing something wrong in the math

    Same for b)

    Thanks!
     
  2. jcsd
  3. Sep 16, 2016 #2

    vela

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    One obvious problem here is you're trying to add a vector to a scalar. A less obvious one is you're using the variable ##d## in two different ways. You essentially have the magnitude of the force between the pairs of charges right:
    $$F = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{x^2+y^2}$$ What should ##x## equal here?

    What are the x-component and y-component of the net force on the middle charge going to be?

    Once you have those expressions, then use the Taylor expansion about ##y=0##.

     
  4. Sep 16, 2016 #3
    The second expression should also have a yhat - I was posting in a bit of a hurry my bad.

    Net force in the x direction is going to be 0 due to symmetry. The restoring force would be pointing back down to the origin.
    I guess I'm stuck at getting that expression.

    I should have
    F = -2(q^2/4piε0(1/dy^2)yhat)


    EDIT: Pardon the bad formatting and all - trying to get used to how the forum works. Also, pardon the mad noob question - I guess I understand the logic of where the problem is going I just can't put it into math.

    Thanks again.
     
  5. Sep 16, 2016 #4

    vela

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    You were closer before. When the middle charge is collinear with the others, there should be no vertical force. Is your latest expression for the force consistent with that?
     
  6. Sep 17, 2016 #5
    I"m supposed to calculate the force once it's shifted in delta y.

    https://drive.google.com/open?id=0B76TBRMyBuffbkVPY3hjZU0xV3M [Broken]

    The picture above is where I always get to before not knowing what to do anymore.
     
    Last edited by a moderator: May 8, 2017
  7. Sep 17, 2016 #6

    gneill

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    From your image it doesn't look to me like you've added the vectors of the forces properly.
    upload_2016-9-17_8-29-36.png
    By symmetry only the y-components of the forces survive the vector summation, but you haven't shown how those y-components are determined from the force magnitudes. Your work seems to be taking the full force magnitudes and just calling them y-components. Unless there's something special about your ##\hat{y}## vectors?

    What do you need to multiply the force magnitudes by in order to "extract" the magnitude of their y-components?
     
  8. Sep 17, 2016 #7
    I should have called it something else but

    ##\hat{r}## = rvector/|r| = (0, y)/sqrt(x^2+y^2)
     
  9. Sep 17, 2016 #8

    vela

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    The numerator of what you wrote suggests ##\vec{r} = (0,y)## while the denominator suggests ##\vec{r} = (x,y)##.

    The vector ##\hat r = \frac{\vec{r}}{\|\vec{r}\|}## has to have a magnitude of 1. The vector ##\frac{(0,y)}{\sqrt{x^2+y^2}}## clearly doesn't, so the two can't be equal.

    Forget about vector notation for right now. By what do you need to multiply the magnitude of the force by to get just the y-component? Note that ##x## won't appear in the final expression because ##x=0## for the middle charge.
     
  10. Sep 17, 2016 #9
    r sin theta would give me that magnitude of y?

    I really appreciate the help guys - sorry that I"m being a bit dense at the moment.
     
  11. Sep 17, 2016 #10

    gneill

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    If r is the magnitude of the vector of interest and θ is the angle that r makes with the positive x-axis, then yes, r sin(θ) would give you the magnitude of its y-component.

    Now, see if you can "construct" sin(θ) from the given geometry using the triangle sides.
     
  12. Sep 17, 2016 #11
    I"m not quite sure what you mean by construct here, so I assumed this

    rsin theta = y

    sin theta = y/sqrt(d^2+y^2)
    So then I taylor expanded 1/sqrt(d^2+y^2) and since it's to the highest order of y, I ended up with

    sin theta approximately equals y/d


    Which when I put that back into the original force equation I come out with -2q^2y/4piepsilon0d^3
     
  13. Sep 17, 2016 #12

    vela

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    I don't think it matters in this case, but I'd wait to Taylor expand until after you get the expression for the force because the magnitude of the force also depends on ##y##.
     
  14. Sep 17, 2016 #13
    There was no other y in the expression so I figured it'd be safe to Taylor expand then.

    Glad I finally got this - thanks to you guys! My head was getting sore from beating it on the desk so much.

    Now onto part b
     
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