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Acceleration under Coloumb's Law |
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| Jun17-12, 07:04 PM | #1 |
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Acceleration under Coloumb's Law
F = q q/r^2
ma = q q / r^2 Under which particle does the mass apply to? |
| Jun17-12, 07:31 PM | #2 |
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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.
So [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 |
| Jun17-12, 08:11 PM | #3 |
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That's not field. Those are equations for the electrostatic force. acceleration is not imparted by mass.
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| Jun17-12, 10:33 PM | #4 |
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Acceleration under Coloumb's LawAcceleration 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. |
| Jun18-12, 09:42 AM | #5 |
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Ok, true, but then what is your question? :)
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| Jun18-12, 10:47 AM | #6 |
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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.
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