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Question with simple current and charge of an electric dipole.

  1. Jul 8, 2011 #1
    This is from Field and Waves Electromagnetic by Cheng page 602. THis is regarding to elemental dipoles. Given:


    [Quote Book]
    [tex] i(t)=\Re e [Ie^{j\omega t}]=I\; cos\;\omega t\;\;\hbox { (11.7)}[/tex]

    Since the current vanishes at the ends of the wire, charge must be deposited there.

    [tex] i(t)=^+_-\frac {d q(t)}{dt} \;\;\hbox { (11.8)}[/tex]

    In phasor notation, [itex]q(t)=\Re e[Qe^{j\omega t}][/itex], we have

    [tex]I=^+_-j\omega Q\;\;\hbox{ (11.9)}[/tex]

    or [tex]Q = ^+ _- \frac {I} {j \omega}\;\;\hbox {(11.10)}[/tex]

    [End Book]




    This is what my calculation of q(t):

    [tex] i(t)=\frac {d [q(t)]}{dt} = \Re e \left [ Q \frac {d (e^{j\omega t})}{dt}\right ] =\Re e [ Q j\omega e^{j\omega t}] = -\omega \;Q \;sin\;\omega t [/tex]

    You see [itex] i(t)[/itex] not equal to [itex]I\; cos\;\omega t [/itex]

    I just don't see how this work!!! Please help.
     
    Last edited: Jul 9, 2011
  2. jcsd
  3. Jul 9, 2011 #2

    Born2bwire

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    But you defined I to be a complex number and then you relate the real valued i(t) with I\cos \omega t which is complex.
     
  4. Jul 9, 2011 #3
    I did not define anything. The first part I copy from the book. Looks to me both I and Q are a constant which stand for the peak value of the current and charge resp. as you see the [itex]cos\;\omega\;t[/itex] that make the current/charge harmonically time varying.

    I double check the differentiation in phasor form and showed the result of differentiating q(t) resp to t and it don't show [itex] I\;cos\;\omega t[/itex]

    Thanks
     
  5. Jul 9, 2011 #4
    The book did something wrong then. As Born2bwire mentioned, you have this formula:

    [tex]I=\pm j \omega Q [/tex]
    If Q is real, then I is complex.

    If I is complex, then this formula does not hold:

    i(t)=ℜe[Iejωt]=Icosωt

    If Q is not real, then this formula does not hold:

    q(t)=ℜe[Qejωt]=Qcosωt

    So there are some inconsistencies in your equations.
     
  6. Jul 9, 2011 #5
    Thanks for the help. I edit the original post. I actually type the paragraph word to word straight from the book and then at the bottom I show my evaluation that show the inconsistency.

    I know, something seems wrong with the book and I cannot make sense out of it.

    Given

    [tex] i(t)=\Re e [Ie^{j\omega t}]=I\; cos\;\omega t\;\;\hbox { (11.7) , }q(t)=\Re e[Qe^{j\omega t}]\;\;\;\hbox { and }\;\;I=^+_-j\omega Q\;\;\hbox{ (11.9)}[/tex]

    You really lock it in. I hope by typing word for word, you might make some sense out of it.

    Thanks

    Alan
     
  7. Jul 9, 2011 #6
    You derived this:

    i(t)=d[q(t)]/dt=ℜe[Qd(ejωt)/dt]=ℜe[Qjωejωt]=−ωQsinωt

    which is incorrect.

    Q is imaginary, so you can't just take it out of the Re[]. What you do is plug in Q=+- I/(jw), and then take the real part. Then you should get something like: I(t)=Icos(wt)
     
  8. Jul 9, 2011 #7
    This is the hertzian dipole, I don't get why Q is imaginary. Convensionally, Q is just the amplitude and q(t) is just Q cos wt.
     
  9. Jul 9, 2011 #8
    Your equation (11.9) says that one of the quantities must be imaginary. So it's either Q or I.

    A Hertzian dipole behaves like a capacitor, where the current leads the voltage. Now for a capacatitor Q follows V. So you expect the current to lead the charge. Since a hertzian dipole behaves like a pure capacitance, the lead is 90 degrees.

    So if your current behaves like cos(wt), then the charge would lag behind by 90, or cos(wt-90)=sin(wt).

    So both the charge and voltage can't both vary as cos(wt). One of them has to vary as sin(wt).

    So notice Q=Real[Qe(iwt)]=Real[I/(jw)e(iwt)]=(I/w)sin(wt)

    So Q varies by sin(wt), and I by cos(wt).
     
    Last edited: Jul 9, 2011
  10. Jul 9, 2011 #9
    Thanks, I think I got it.

    Good night.
     
  11. Jul 9, 2011 #10

    Please see the question in Blue above.
     
  12. Jul 9, 2011 #11
    I meant voltage. For a capacitor, Q=CV, where C (the capacitance) is real. So whatever phase the voltage has, the charge Q has the same phase. A Hertzian dipole (i.e. a very tiny antenna) is pretty much an AC generator connected to a capacitor. If you want you can put two spheres at the end of the wires coming out of the generator, so it looks more like a capacitor. In a circuit that contains just a capacitor and a generator, if the current is I(t)=I cos(wt) where I is real, then the voltage is V(t)=I/(wC) sin(wt). The charge would then be C times this, or Q(t)=(I/w)sin(wt).

    This is the same result that you got. The charge varies as sin(wt), and the current as cos(wt), so the charge and current aren't in phase but are 90 degrees out of phase.

    This capacitor analogy only works if the dipole is very tiny (compared to the wavelength of the frequency that it's being driven at). As the dipole gets bigger the behavior of the antenna changes, sometimes acting as pure inductors, or series and parallel resonant circuits.

    So basically I was trying to give another picture of what's going on in the antenna. You can model it as an AC generator connected to a pure capacitor, and solve for the charge and current as you would a normal circuit.
     
  13. Jul 9, 2011 #12
    I mis-understood. When I saw "Q follow V", I thought you meant Q is behind V............V first and then Q like Q follow ( lagging behind) I in phase. I understand Q is in phase with V where Q=CV.

    Thanks

    Alan
     
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