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Current due to a single charge

  1. Jan 31, 2013 #1
    I'm currently in an antennas class, and while my EM is fairly strong, it's been a long time since I've done very basic electromagnetics like this question asks, so I can't clearly remember the specific details and physics properties. Any help is appreciated.

    1. The problem statement, all variables and given/known data

    We have considered in class how an e/m disturbance is created by a positive charge
    that is oscillating along the z axis. Let the position of a particle with a charge of + 1 nC
    from the origin be described by

    z(t) = Asin(ωt)

    where ω is the angular frequency (related to the period of oscillation).

    a. Find expressions for the velocity and acceleration of the charge (Note: these are vector
    quantities)

    b. Find an expression for the current, I(t), associated with the motion of the charge.

    There are actually a few more parts, but I don't know if I need help on them just yet as such.


    2. Relevant equations

    z(t) = Asin(ωt)
    I = dQ/dt (?)

    The charge as given in class is q+.

    3. The attempt at a solution

    a)

    The first part, a), isn't too difficult, basic physics says that velocity is the time-derivative of position and likewise acceleration is the time-derivative of velocity, which leds me to:

    v(t) = z'(t) = ωAsin(ωt)
    a(t) = v'(t) = z''(t) = -ω2Acos(ωt)

    b)

    This is the part I'm stuck on. I'm not sure how to pull current out of this even though I know that current is just charge passing through a given surface per time, or I = dQ/dt. But I'm not sure how to relate it to the results of a), if it is even related to that.

    The best I've come up with is that IdS = vdq, but I'm not wholly sure on what to integrate nor what dS is seeing as it's a point charge on a one-dimensional line. Perhaps it's as simple as I = q*v? So would it just be I = qωAsin(ωt)?
     
  2. jcsd
  3. Jan 31, 2013 #2

    TSny

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    I don't think it's possible to write an expression for the current of a moving point charge. I = q v would not have the correct units for current.

    However, it is possible to write an expression for the current density if you use the Dirac delta function: j(r,t) = q v(t) δ(r-ro(t)) where r is an arbitrary point of space and ro(t) is the position of the particle at time t.

    Not sure if this will help.
     
  4. Jan 31, 2013 #3
    Perhaps Biot-Savart? Would that be applicable to help in some way?

    Alternatively, since v is in units of m/s, and q is in C, qv would produce units of Cm/s.

    Given that A is the total amplitude of its motion, and therefore in m, would something like I = qv/A [= qωsin(ωt)] be closer or reasonable?
     
  5. Jan 31, 2013 #4

    TSny

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    I don't think so. If you consider the definition of current, it's the rate at which charge is passing through a certain specified area. Usually that area is a particular cross-sectional area of a wire. Anyway, the current I is always in regard to a certain area. The relation between I and j is I = j dS if the area dS is perpendicular to j. In the case of a single point charge, I don't see a natural choice for dS. If you took an arbitrarily chosen area element dS perpendicular to the z-axis and try to define I for that area, then the current would be fairly singular. Most of the time for that area, the current would be zero. At the instant the charge passes through the area, the current would be divergent.

    So, I don't know. I have occasionally seen cases where one defines a "current element" for a small length of distance dz. The current element is the product I dz. Then people "argue" that for a point charge q you can write (sloppily, at best) I dz = (dq/dt) dz = (q/dt) dz = q (dz/dt) = q v. So, I dz = q v. I don't like that argument. The quantity q v is certainly independent of dz. So, I dz must then be independent of dz. That would make I inversely proportional to dz. Odd.
     
  6. Jan 31, 2013 #5
    Alternatively, what if you consider the area dS to be δ(x)δ(y)?

    i.e., you could treat the path the charge is on as an infinitely thin wire? Then the charge would be oscillating in it, which is what AC current is anyway. I mean realistically you could do that to a wire anyway, choose a differential section of it (in this case dz) so small that it only has a single oscillating charge carrier in it?

    Edit: Sorry if I'm spamming, this may be relevant: http://www.physics.sfsu.edu/~lea/courses/ugrad/460notes5.PDF
     
    Last edited: Jan 31, 2013
  7. Jan 31, 2013 #6

    TSny

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    δ(x) has dimension of 1/x. So, δ(x)δ(y) would not have the dimension of area it would have dimension of 1/area.

    In the current density expression, note that δ(r-ro) = δ(x-xo)δ(y-yo)δ(z-zo) so the product of δ functions is already there.
     
  8. Jan 31, 2013 #7
    This question doesn't seem to be worth much and it's an EE class versus physics so I figure the answer has to be much simpler. Regardless, I did find something of note:

    http://www.physics.sfsu.edu/~lea/courses/ugrad/460notes4.PDF

    On page 7 it talks about moving point charges, and the other link I provided up there

    http://www.physics.sfsu.edu/~lea/courses/ugrad/460notes5.PDF

    Gives an expression for I on page 2 or 3 as I = -ωqsin(ωt).

    This is done somewhat cleverly by realizing the parallel between either an oscillating point charge or having a varying charge. If some light could be shed on this explanation it'd be most appreciated.

    Just purely from a physical standpoint, a charge oscillating as such could be realized "physically" by having a wire that is thin enough and short enough such that only one electron is oscillating within it (per AC electricity). Thus there's got to be some way to figure this out, just blimey if I know it.
     
  9. Jan 31, 2013 #8
    Could you explain the charge/current density thing to me, namely where the delta-dirac functions come in?

    Nevermind, I finally understand this part, thank you. What about above with the finding for current?
     
    Last edited: Jan 31, 2013
  10. Jan 31, 2013 #9
    The prof just sent an email that a value of current may not be sensibly calculated.

    Thus, thank you kindly for your answer which appears to be correct!

    Thanks for helping out an ignorant EE :)
     
  11. Jan 31, 2013 #10

    TSny

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    Well, here they are no longer talking about a single point charge oscillating up and down. They use a model from Griffith's text to represent an oscillating dipole. Here I will quote directly from Griffiths since his description is more complete than the paper you linked to:

    "Picture two tiny metal spheres separated by a distance d and connected by a fine wire; at time t the charge on the upper sphere is q(t), and the charge on the lower sphere is -q(t). Suppose that we drive the charge back and forth through the wire, from one end to the other, at an angular frequency ω: q(t) = qocos(ωt). The result is an oscillating electric dipole: p(t) = pocos(ωt) [itex]\hat{z}[/itex] where po = qod "

    Then the current in the connecting wire would be I = dq/dt.
     
  12. Jan 31, 2013 #11

    TSny

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    OK, good.
     
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