Acceleration under Coloumb's Law

  • Thread starter Thread starter Nano-Passion
  • Start date Start date
  • Tags Tags
    Acceleration Law
AI Thread Summary
The discussion focuses on the application of mass in the context of Coulomb's Law and its relation to acceleration. It clarifies that acceleration can be defined in terms of either mass m_1 or m_2, depending on which particle's acceleration is being measured. While acceleration is not directly imparted by mass, it is inversely proportional to the mass of the second particle, m_2. The conversation emphasizes the distinction between electric fields and acceleration fields, noting that the latter can be conceptualized in relation to the effects of charge. Ultimately, the confusion regarding which mass to assign in the equations was resolved, highlighting the flexibility in choosing the mass based on the particle of interest.
Nano-Passion
Messages
1,291
Reaction score
0
F = q q/r^2
ma = q q / r^2

Under which particle does the mass apply to?
 
Physics news on Phys.org
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
a = k \frac{q_1 q_2}{m_2 r^2}
is the field of acceleration imparted by mass of q_1, in other words m_1
 
Last edited:
That's not field. Those are equations for the electrostatic force. acceleration is not imparted by mass.
 
xAxis said:
That's not field. Those are equations for the electrostatic force. acceleration is not imparted by mass.

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.
 
Ok, true, but then what is your question? :)
 
xAxis said:
Ok, true, but then what is your question? :)

Hey,

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

ma = k \frac{q_1 q_2}{r^2}

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.
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top