1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Some Help! Charged particles

  1. Feb 23, 2005 #1
    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!
  2. jcsd
  3. Feb 23, 2005 #2
    Yes, you should use conservation of energy next.

  4. Feb 23, 2005 #3
    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?
  5. Feb 23, 2005 #4
    Why would this be zero? Do you know an expression for the energy between two charged particles?

  6. Feb 23, 2005 #5
    F=k|q1||q2|/r^2 is this the one you're talking about?
  7. Feb 23, 2005 #6
    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!
  8. Feb 23, 2005 #7
    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 thats negligible in this case.
  9. Feb 23, 2005 #8
    Kinetic Energy?
  10. Feb 23, 2005 #9
    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. [tex] P_{E} = - \frac{kq_{1}q_{2}}{r} [/tex]
  11. Feb 23, 2005 #10
    That's odd, I thought EPE=qV is the electrical potential energy.
  12. Feb 23, 2005 #11
    what is the expression for V??
  13. Feb 23, 2005 #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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?