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Homework Help: Four-potential of moving charge

  1. Jan 14, 2015 #1
    1. The problem statement, all variables and given/known data

    (a) Find the four-vector potential of a moving charge
    (b) Find source time and z-component of electric field
    (c) Find electric potential to first order of x and hence electric field


    8zl6jn.png

    2. Relevant equations


    3. The attempt at a solution

    Part(a)

    [tex]\phi = \frac{q}{4\pi \epsilon_0 r} [/tex]
    [tex] \vec A = 0 [/tex]

    Consider world line of aritrarily moving charge, where the vector potential is only dependent on what charge is doing at source event. The relevant distance is source-field distance ##r_{sf}##. We represent this in 4-vector displacement ## R = (ct, \vec r)## where ##t = \frac{r_{sf}}{c}##. Consider ##R \cdot U = \gamma(-rc + \vec r \cdot \vec v)##. It gives the right form in rest frame. Thus:

    [tex]A = \frac{q}{4\pi \epsilon_0} \frac{ \frac{U}{c} }{-R \cdot U} [/tex]


    Part (b)

    [tex] t_s = \frac{r_{sf}}{c} = \frac{\sqrt{z^2+a^2}}{c} [/tex]
    [tex]R \cdot U = \gamma (-ct + \vec r \cdot \vec v) = -\gamma r c = -\gamma c \sqrt{z^2 + a^2} [/tex]

    Thus the four-potential is
    [tex]A = \frac{q}{4\pi \epsilon_0 c^2 \sqrt{z^2 + a^2}} (c, \vec v_s) = (\frac{\phi}{c}, \vec A)[/tex]
    To find electric field,
    [tex]E_z = -\frac{\partial \phi}{\partial z} - \frac{\partial A_z}{\partial t}[/tex]
    z-component of ##\vec A## is zero, so
    [tex]E_z = -\frac{\partial \phi}{\partial z} = -\frac{qz}{4\pi \epsilon_0 \left( z^2 + a^2 \right)^{\frac{3}{2}}} \hat k[/tex]


    Part (c)

    [tex]r = \sqrt{ z^2 + \left[ a cos(\omega t) - \Delta x \right]^2 + a^2 sin^2(\omega t) } [/tex]
    [tex]r = \approx \sqrt{ z^2 + a^2 - 2a \Delta x cos(\omega t) } \approx \sqrt{z^2 + a^2} \left[ 1 - \frac{a \Delta x cos(\omega t)}{z^2 + a^2} \right][/tex]

    For ##R \cdot U##, we have

    [tex]R \cdot U = \gamma(-rc + \vec r \cdot \vec v) = \gamma \left[ -cr + a\omega \Delta x sin (\omega t) \right] [/tex]

    [tex] \phi = -\frac{qc}{4\pi \epsilon_0} \frac{1}{ c\sqrt{a^2 + z^2} \left[ 1 - \frac{a \Delta x cos(\omega t)}{a^2 + z^2} \right] + a\omega \Delta x sin(\omega t) }[/tex]

    [tex]\phi \approx \frac{-qc}{4\pi \epsilon_0} \frac{1}{c\sqrt{a^2+z^2} + \Delta x \left[ a\omega sin(\omega t) - \frac{ac cos(\omega t)}{\sqrt{a^2 + z^2}} \right] } [/tex]

    [tex] \phi \approx \frac{-qc}{4\pi \epsilon_0} \left[ 1 - \frac{a\omega sin(\omega t) - \frac{ac cos(\omega t)}{\sqrt{a^2+z^2}} }{c\sqrt{a^2+z^2}} \Delta x \right] [/tex]

    [tex]E_x = -\frac{\partial \phi}{\partial x} = \frac{qc}{4\pi \epsilon_0 c^2 (a^2 + z^2) \left[ a\omega sin(\omega t) - \frac{ac cos(\omega t)}{\sqrt{a^2 + z^2 }} \right] } [/tex]

    Can't seem to match the final expression - what worries me is the lack of ##\omega^2## in the answer.
     
  2. jcsd
  3. Jan 14, 2015 #2

    Orodruin

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    What happened to the x-component of the vector potential?
     
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