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Rate of decreasing static field - is it measurable?

  1. Mar 18, 2010 #1
    If a static charge is placed on an object, and that object begins to move through space, a magnetic field will form around the object (relative to a stationary observer).

    At the same time, the static field around the object will begin to reduce. The faster the object moves, the stronger the magnetic field will become and the weaker the static force.

    How do we measure that rate of the changing static force? Is this rate completely theoretical (I believe it's something like P=E.H.sineo), or has anyone managed to actually measure the static force at high speeds to 'actually' measure this change? Where could I find those results?

    Thanks!
    :)
     
    Last edited: Mar 18, 2010
  2. jcsd
  3. Mar 18, 2010 #2

    collinsmark

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    Actually, I don't that's quite right. The (static) electric field is not a function of velocity.

    [tex] \vec E = \frac{1}{4 \pi \epsilon _0}\frac{q}{r^2} \hat r [/tex]

    [tex] \vec F_E = q\vec E [/tex]

    The magnetic field is, on the other hand.

    [tex] \vec B = \frac{\mu _0}{4 \pi}\frac{q (\vec v \times \hat r)}{r^2} [/tex]

    [tex] F_M = q(\vec v \times \vec B) [/tex]

    Thus the total force is,

    [tex] \vec F = \vec F_E + \vec F_M [/tex]

    [tex] = q \vec E + q(\vec v \times \vec B) [/tex]

    If you calculate the relative strengths between the electric and magnetic force (as a function of v) you get an interesting result. Note that Maxwell's equations tell us that the speed of light in a vacuum is [tex] c = \frac{1}{\sqrt{\epsilon _0 \mu _0}} [/tex].
     
  4. Mar 20, 2010 #3
    Were not E and H multiplied by each other supposed to equal a constant?
     
  5. Mar 20, 2010 #4

    collinsmark

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    Something like

    [tex] \vec E \cdot \vec H = \vec E' \cdot \vec H' [/tex]

    Yeah, but I think that's more to do with the geometry changing when you go from one from one inertial frame to another inertial frame (at a different velocity) due to Lorenz transformations.

    Yeah, I think I see what you mean now. But I'll let others comment from here.
     
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