Force between two charged particles

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SUMMARY

The discussion focuses on calculating the mass of a second charged particle based on the acceleration of two similarly charged particles. Given the first particle's mass of 6.3*10^-7 kg and its acceleration of 7.0 m/s², along with the second particle's acceleration of 9.0 m/s², the derived formula for the second particle's mass is m2 = 4.31*10^-7 kg. The participants emphasize the importance of using Newton's second law and the relationship between the forces acting on the particles, clarifying that the forces are equal and opposite due to their charges.

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  • Knowledge of electrostatic force, F=k*q^2/r^2
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hk4491
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Homework Statement



Two particles with a similar charge are held at a distance of 3.2*10^-3 m and then released. The acceleration of the first particle is measured at 7.0 m/s^2, and for the second at 9.0 m?s^2. The mass of the first particle is 6.3*10^-7. What is the mass of the second particle?


Homework Equations



F=ma F=G*m1*m2/r^2


The Attempt at a Solution



F1=m1a1 F2=m2a2 (where F1 and F2 are the forces exerted by each particle respectively)

F1 - F2 = G*m1*m2/r^2

(since the particles are similarly charged, they would have forces pointing in opposite directions)

after substitution and making m2 the subject of the formula:

m2 = (m1*a1*r^2)/(Gm1 + a2r^2)

which gives: m2 = 4.31*10^-7

I am not so sure if the method I used is completely correct, can someone please tell me?
 
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hk4491 said:

Homework Equations



F=ma F=G*m1*m2/r^2
The particles are charged. Do you really think that they will primarily feel a gravitational attraction? And to you really need to calculate the value of the force, or isn't there one of Newton's law you can use?
 
Hi, thanks for replying. Are the forces produced by the two particles equal? Because then I can use Newton's second law as such:

m1a1=m2a2

Would this be correct?
 
Yes, this is what I was hinting at. If particle 1 feels a force from particle 2, then particle two must feel an equal and opposite force from particle 1.
 
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In the second part they're asking me to find the charge of each particle. I know I should use this formula:

F = k.q^2/r^2

and solve for q. As my force should I substitute 2*ma, since there are two particles, or is one enough?
 
hk4491 said:
In the second part they're asking me to find the charge of each particle. I know I should use this formula:

F = k.q^2/r^2

and solve for q. As my force should I substitute 2*ma, since there are two particles, or is one enough?
In essence, there are not two forces, but one that affects two particles, albeit in different directions. So no factor of two.
 
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