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Finding the Delta Function of a Thin Ring

  1. Apr 5, 2016 #1
    1. The problem statement, all variables and given/known data

    A very thin plastic ring (radius R) has a constant linear charge density, and total charge Q. The ring spins at angular velocity [itex] \omega [/itex] about its center (which is the origin). What is the current I, in terms of given quantities? What is the volume current density J in cylindrical coordinates? (This may be a little tricky, since the ring is "very thin", there will be some [itex] \delta [/itex] functions. I suggest writing down a formula for [itex] \rho [/itex](s, [itex] \phi[/itex], z) first. And, remember that J should be a vector!)

    2. Relevant equations

    My problem is finding these, one might say it is "My real question".

    3. The attempt at a solution

    For the current I claimed that I=[itex]\omega \lambda[/itex] and [itex]\lambda [/itex]=Q/2[itex]\pi [/itex]R therefore I=[itex]\omega [/itex]Q/2[itex]\pi [/itex]R

    I feel pretty strong about that on the basis that current is due to a moving charge and that the ring itself is a charge distribution that is spinning. However, if someone could verify that I'd be very grateful.

    My real question is in regards to finding the delta function. My thoughts were that radially the ring is infinitely thin and it would also be going along the Z axis. However the length is just 2[itex]\pi[/itex]R so I just need to multiply these together as such my volume would become:

    V=[itex]\delta[/itex](s-R)[itex]\delta[/itex](z-z')2[itex]\pi[/itex]R

    I choose to use z' since the problem never actually tells us where along z the loop is situated.

    Since [itex]\rho[/itex] is just charge per unit volume then:

    [itex]\rho[/itex]=Q/[itex]\delta[/itex](s-R)[itex]\delta[/itex](z-z')2[itex]\pi[/itex]R

    Once I have this figured out I can move on to finding J.
     
  2. jcsd
  3. Apr 5, 2016 #2

    Orodruin

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    This is not correct, it even has the wrong units! Note that the delta functions have the units of the reciprocal of their arguments. The volume is also not a function of s and z.

    Hint: You want the density to be a function such that when you integrate it over all of space you get the total charge.
     
  4. Apr 5, 2016 #3
    First, thank you for your help. I recall now about the delta's having reciprocal units of their arguments. I'm confused as to why isn't the volume of the wire a function of s and z? Since I'm working in cylindrical coordinates doesn't any volume in this space have to be a function of the three coordinates?

    I had found a similarly worked example for spherical coordinates, where after some work with the delta function they they tested it by integrating to find the total charge Q. So, would it be correct to think that your hint is a way of testing that I've gotten the correct form, once I have a form to test? Thanks again in advance. :D
     
  5. Apr 5, 2016 #4

    Orodruin

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    Volume is not a function of coordinates, it is a property of a region.

    Right.
     
  6. Apr 5, 2016 #5
    This is what I've been able to come up with thus far:

    It would not matter where z' is, the range will still simple be a point. As such I'll call z'=0. The radial part has a point at s'=R. And there is no dependence on [itex]\phi[/itex] so omit it.

    Now in general [itex]\delta[/itex](r-r')=1/s*[itex]\delta[/itex](s-s')[itex]\delta[/itex]([itex]\phi-\phi')\delta[/itex](z-z')

    But given our conditions and that we need a resulting charge, add a to be determined constant A and it becomes:

    [itex]\delta[/itex](r-r')=1/s*[itex]\delta[/itex](s-R)[itex]\delta[/itex](z-0)

    So

    Q=[itex]\iiint [/itex] A/s [itex] \delta (s-R) \delta [/itex] (z) s dsd [itex]\phi [/itex]dz

    Which evaluates to
    Q=A*1*2[itex]\pi[/itex]*1=A2[itex]\pi[/itex]

    So

    A=Q/2[itex]\pi[/itex]

    Then;

    [itex]\rho [/itex]=[itex] Q \delta (s-R) \delta (z) \over 2 \pi s [/itex]

    Is this correct? If so, is it correct for the right reasons? :rolleyes:o_O:rolleyes:
     
  7. Apr 5, 2016 #6

    Orodruin

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    Yes, it is correct.
     
  8. Apr 5, 2016 #7

    vanhees71

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    I think there's something missing in the question. Is the ring rotating around the axis through its center perpendicular to its plane or in this plane. This makes a difference!
     
  9. Apr 5, 2016 #8

    Orodruin

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    From the attempt, I would have taken it as rotation around the axis perpendicular to the circle's plane and this is also how I would probably have interpreted the question if unspecified. But in general, yes, it makes a huge difference.

    Apart from that, OP has still not given a final expression for the current density.
     
  10. Apr 5, 2016 #9

    vanhees71

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    Sure, but if the rtation axis is really perpedicular to the circle's plane, it's now no more problem. Hint: how to get ##\vec{j}## from ##\rho##? You need a vector, and a current density is the number of charges flowing through a test area per area and time!
     
  11. Apr 5, 2016 #10

    vela

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    The problem statement says it's rotating about its center, not about an axis passing through the center that lies in the plane of the ring, so I'd interpret it the way the OP did.
     
  12. Apr 6, 2016 #11
    Ha ha, yeah the interpretation of how the ring was rotating was ambiguous. Happened a few times. From the example problems in the Griffith's text I assumed what he meant is the ring rotates about the z-axis and in the xy plane. Turns out it's what he wanted. :D
     
  13. Apr 6, 2016 #12
    Thank you for helping me out with this! It helps even just knowing "yes, that's right" or "no, try again." I liked your tips too.

    I turned in the assignment and turns out I got everything else right as well.
     
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