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Electrostatics and velocity question

  1. Jun 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Electric field given by the vector [itex]\vec{E}=x\hat{i}+y\hat{j}[/itex] is present in xy plane. A small ring carrying charge +Q, which can freely slide on a smooth non conducting rod, is projected along the rod from point (0,L) such that it can reach the other end of the rod. What minimum velocity should be given to the rod? (Assume zero gravity).
    ziws28.jpg



    2. Relevant equations



    3. The attempt at a solution
    I tried finding the change in potential, finding the work done and equating the work done to kinetic energy. Let's assume that it moves a very small displacement dr as shown in the figure.
    15d9p54.jpg
    It will have components dx[itex]\hat{i}[/itex] and -dy[itex]\hat{j}[/itex].
    ∴[itex]dr=dx\hat{i}-dy\hat{j}[/itex].
    Change in potential, ΔV
    [tex]ΔV=-\int\vec{E}\cdot\vec{dr}[/tex]
    [tex]=-\int(x\hat{i}+y\hat{j})\cdot(dx\hat{i}-dy\hat{j})[/tex]
    [tex]=-\int_{0}^{L}xdx+\int_L^0ydy[/tex]
    [tex]=-L^2[/tex]

    [tex]QΔV=\frac{1}{2}mu^2[/tex]
    where u is the velocity with which ring is projected.
    Solving this i get,
    [tex]u=\sqrt{\frac{2QL^2}{m}}[/tex]

    But this is wrong, the answer is
    [tex]u=\sqrt{\frac{QL^2}{2m}}[/tex]

    I don't understand where i am doing wrong.
    Any help is appreciated. :smile:
     
  2. jcsd
  3. Jun 2, 2012 #2

    ehild

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    dr =dxi+dyj regardless of the direction of the of the path.

    ehild
     
  4. Jun 2, 2012 #3
    But then i get ΔV=0. :confused:
     
  5. Jun 2, 2012 #4

    ehild

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    Yes. The change of potential is zero between the initial and final points, but it has a maximum somewhere in between. The kinetic energy has to overcome that maximum potential.
    Think: You stand at a fence and you want to throw a ball over the other side. . How much must be the kinetic energy of the ball? You can not say that it is PE(f)-PE(i)=0...

    ehild
     
  6. Jun 2, 2012 #5
    Is that maximum potential point at (L/2,L/2) ?
     
  7. Jun 2, 2012 #6
    The field is radial so the potential difference between (0,L) and (L,0) will be 0. However. along the path, the ring will gain potential energy until it is at it's nearest to the origin and then lose it. So take the potential difference between (0,L) and where the ring is nearest the origin in a similar approach as you did.
     
  8. Jun 2, 2012 #7
    The nearest point would be then (L/2,L/2).

    Okay so i do the integration again.
    [tex]ΔV=-(\int_{0}^{\frac{L}{2}}xdx+\int_L^{\frac{L}{2}}ydy)[/tex]

    Solving this i get:
    [tex]ΔV=\frac{3L^2}{4}[/tex]

    But i don't think this is correct. It doesn't lead to the right answer.
     
  9. Jun 2, 2012 #8

    ehild

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    NO. Find it. Write up the potential along the path as function of the distance from the initial point, and find the local maximum.

    ehild
     
    Last edited: Jun 2, 2012
  10. Jun 2, 2012 #9
    Local maximum? How should i go on finding it?
    I don't have any equation to perform differentiation. Just a shot in the dark, do i have to use the following equation and set the derivative to zero?

    [tex]ΔV=-(\frac{x^2}{2}+\frac{y^2}{2})[/tex]
     
  11. Jun 2, 2012 #10
    Never mind my previous post ehild, i just read your post carefully.
    How can i write the potential as a function of distance from initial point? Should i assume that the ring reaches a point (x,y), find the distance?

    (I double posted just to let ehild know that there's activity in this thread.)
     
  12. Jun 2, 2012 #11

    ehild

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    You need the maximum along the straight line from (0,L) to (L,0). If s is the distance from the initial point, x=s/√2 and y=L-s/√2. Write the potential energy in terms of s.

    ehild
     
  13. Jun 2, 2012 #12
    I am lost here, i have no idea on this. I hope i am not annoying you. :uhh:
     
  14. Jun 2, 2012 #13

    ehild

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    The point charge moves along the blue line. The equation of that line is y=L-x. If the point has started from the upper end, and travelled s distance, it is in point x,y: x=s/√2 (as the line encloses 45°angle with the x axis) and y=L-s/√2.
    You derived already that the potential is -1/2(x^2+y^2) with respect to the origin as reference point. In terms of s , it is
    V=(-1/2) [(s/√2)2+(L-s/√2)2].

    The maximum is when s=L/√2, and this corresponds to the point (L/2;L/2) as you wrote before. The potential is -L^2/4, but you need the change of potential from point (0,L), where it is -L^2/2. The difference is ΔV=L^2/4. You get the same result with integrating along the line from (0,L) to (L/2;L/2); you made simply a sign error in post #7.

    ehild
     

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    Last edited: Jun 2, 2012
  15. Jun 2, 2012 #14
    Thanks for the help! :smile:
    Simplifying the expression you gave me,
    [tex]V=\frac{-1}{2}(L^2+s^2-\sqrt{2}Ls)[/tex]
    To find the maximum, we need to set the derivative equal to zero, am i right?
    [tex]\frac{dV}{ds}=\frac{-1}{2}(2s-\sqrt{2}L)[/tex]
    [tex]0=2s-\sqrt{2}L[/tex]
    [tex]s=\frac{L}{\sqrt{2}}[/tex]

    Therefore x=L/2 and y=L/2. I am afraid that i have done something wrong because these are the same points i stated before.

    EDIT:Oops, you edited your post.
     
  16. Jun 2, 2012 #15

    ehild

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    Yes, and (L/2;L/2 was correct, I made some mistakes, too, sorrrrrrry....

    Have you got the correct result at last?

    ehild
     
  17. Jun 2, 2012 #16
    Yes you are right, i did a mistake while integrating. Thanks for all the help ehild! :smile:
    I have got the right answer.
    It was really fun.
     
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