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Work Done, Uniformly Charged Ring

  1. Oct 28, 2009 #1
    Charge Q uniformly spread through the ring of radius a. Find the work done to bring point charge q from the center of ring to infinity along the axis through the center of ring.

    From a previous problem I calculated E=kQz/(z^2+a^2)^1.5

    I then tried to apply the formula W=eo/2[tex]\intE^2 dz[/tex] integrating from 0 to infinity.

    As a result I got -Q^2/(1024a^3).

    However the professor on my paper just put an x on it, indicating that it was wrong & then wrote down use the formula W=q(Vo-Vinfinity).

    So my question is, can I not apply the formula I used for the equation? If not, why?
     
  2. jcsd
  3. Oct 28, 2009 #2

    gabbagabbahey

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    The formula you are referring to is actually

    [tex]U=\frac{\epsilon_0}{2}\int_{\text{all space}}E^2 d^3\textbf{r}[/tex]

    It gives the amount of energy stored in the fields. You could use it to calculate;

    (1) the amount of energy stored in the fields when the point charge is at the center of the ring [itex]U_1[/itex]

    and

    (2) the amount of energy stored in the fields when the point charge is at infinity [itex]U_2[/itex]

    and then conservation of energy would mean that the work done by the fields was [itex]W=U_1-U_2[/itex]. But, in order to do this (since you integrate over all space), you would have to know the electric field due to both the ring and the point charge everywhere; not just on the axis of the ring.

    It can be done this way, but it is MUCH easier to just calculate the potential due to the ring along its axis, and then use the formula your professor suggested.
     
  4. Nov 1, 2009 #3
    Ok, I understand that you'd need to know the EF everywhere, but wouldn't it just be zero off the axis? Do to symmetry?
     
  5. Nov 1, 2009 #4

    gabbagabbahey

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    No, it would not be zero due to symmetry....off axis, there is no symmetry.
     
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