Calculating Initial Separation of Charged Particles Using Conservation Laws

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The discussion revolves around calculating the initial separation of two charged particles using conservation laws. The first step involved applying conservation of momentum to determine the velocity of the second particle after release. Participants then discussed the need to use conservation of energy, specifically focusing on kinetic and electric potential energy. The final calculations led to the conclusion that the initial separation between the particles is approximately 0.01407 meters. The conversation highlights the importance of understanding the relationship between electric potential energy and kinetic energy in solving the problem.
NotaPhysicsMan
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Ok here's the question:

One particle has a mas of 3.00 x10^-3 kg and a charge of +8.00uC. A second particle has a mass of 6.00x10^-3 kg and the same charge. The two particles are initally held in place and then released. The particles fly apart, and when the separation between them is 0.100m, the speed of the 3.00x10^-3kg particle is 125m/s. Find the inital separation between the particles.

Ok, I think I can start by using conservation of momentum to solve for V2.

So m1V1 + m2V2= m1V1(final)+m2V2(Final)

Since my inital velocities are 0, I'm left with the two final terms.

I want V2, so V2=(-m1/m2)V1.

Ok so V2=-62.5m/s.

Now I'm stuck.
Should I be using the conservation of energy next?

Thanks for the input!
 
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Yes, you should use conservation of energy next.

--J
 
Ok so I don't have rotational, gravitational, or spring but I do have kinetic and Electric Potential energy.

So: Inital(1/2m1v1^2+EPE1)+(1/2m2v2^2+EPE2)=Final(1/2m1v1^2+EPE1)+(1/2m2v2^2+EPE2)

Since I have 0m/s from rest, my inital terms are no more for kinetic energy.

The only thing I can find for this is

Delta EPE initial - Delta EPE final= 1/2mv1^2+1/2mv2^2.

So delta EPEi-EPEf=1/2(3.00 x10^-3 kg)(125m/s)^2 +1/2( 6.00x10^-3 kg )(-62.5m/s)^2.

Ok so I have the Delta EPEi-EPEf value as 0, which is unsual I can't use it anywhere.

Maybe I should have found the Voltage instead?
 
NotaPhysicsMan said:
Ok so I have the Delta EPEi-EPEf value as 0, which is unsual I can't use it anywhere.

Why would this be zero? Do you know an expression for the energy between two charged particles?

--J
 
F=k|q1||q2|/r^2 is this the one you're talking about?
 
Ok I'm lost now.

I've used cons of momentum to find V2. Where am I to go now?

KEf +EPEf=KEi+EPEi?

Ok I can use this to solve for? Since KEi can be eliminated, we have EPEi=KEf+EPEf.

So qVi=1/2mv^2 + qVf. Ok now extremely lost!
 
Ok do you know how to calculate electric potential energy??

Wht is the initial potential energy?
and the final?
Intial = final + something

figure out what the something is. The something is an aenergy and its not gravitational becuase that's negligible in this case.
 
Kinetic Energy?
 
NotaPhysicsMan said:
Kinetic Energy?
yes and it is the kinetic energy of both the particles

intial = final + kinetic energies of both and solve for the initial distance

do you know that the electric potential energy is just like gravitational potential energy? i.e. P_{E} = - \frac{kq_{1}q_{2}}{r}
 
  • #10
That's odd, I thought EPE=qV is the electrical potential energy.
 
  • #11
NotaPhysicsMan said:
That's odd, I thought EPE=qV is the electrical potential energy.
what is the expression for V??
 
  • #12
aahh yes, qV and V=kq/r.

I see and yes my answer to the question came out to be 0.01407 m just by following what you had said before:

PEi=PEf+(KE1+KE2) I just solved for the r in the PEi.

Thanks.
 

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