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Acceleration under Coloumb's Law

  1. Jun 17, 2012 #1
    F = q q/r^2
    ma = q q / r^2

    Under which particle does the mass apply to?
  2. jcsd
  3. Jun 17, 2012 #2
    Okay I've got it figured out. So like the acceleration field of gravitation, we may choose to define m as m_1 or m_2. And that gives you the equation of what field a mass imparts.

    [tex]a = k \frac{q_1 q_2}{m_2 r^2} [/tex]
    is the field of acceleration imparted by mass of q_1, in other words m_1
    Last edited: Jun 17, 2012
  4. Jun 17, 2012 #3
    That's not field. Those are equations for the electrostatic force. acceleration is not imparted by mass.
  5. Jun 17, 2012 #4
    Yes it isn't an electric field, but it can be easily thought of as an acceleration field. We can think of the equation as the acceleration of the second particle is inversely proportional to the its mass. m_2 is the mass of particle_2 with q_2,

    Acceleration is not imparted by mass, but acceleration lessens when the mass is higher.

    When I say field, I'm talking about an affect that permeates space.
  6. Jun 18, 2012 #5
    Ok, true, but then what is your question? :)
  7. Jun 18, 2012 #6

    My question was partly confused by which particle m should be assigned to under the equation

    [tex] ma = k \frac{q_1 q_2}{r^2}[/tex]

    But I figured it out in the second post. It seems that it can be assigned to which ever one you want, depending on which particle's acceleration you want to measure. I apologize for the confusion. :yuck:
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