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Unfixed charges released from rest

  1. Mar 3, 2015 #1

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    1. The problem statement, all variables and given/known data
    a.)
    A proton and an "positron" (identical to an electron, except positively charged) are brought ##5µm## apart and released from rest.

    What is the initial potential energy stored by this system?

    b.) In all of the previous problems on this homework, the system's collective potential energy has been assigned to a single charge which is moveable. In this problem, we'll let BOTH charges move. The collective kinetic energy of the charges ##K_{sys}## will be drawn from ##U_{sys}##. However, since both charges are moving, they will need to "share" this energy.

    Once in motion, what percentage of ##K_{sys}## does each charge have at any given moment?
    proton: K = ?% of ##K_{sys}##
    positron: K = ?% of ##K_{sys}##
    (In this scenario, how valid is our usual approximation to assign the entire Usys to one particle? Which particle should it be assigned to?)

    What is the speed of each charge after they have repelled a long distance apart?
    proton: ##v_f = ? m/s##
    positron: ##v_f = ? m/s##

    2. Relevant equations

    $$U=\frac{kQq}{r}$$
    $$p=m_1v_1+m_2v_2$$
    $$U_{sys}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$


    3. The attempt at a solution

    I solved a.) and got ##4.613137\times10^{-23}## which is the correct answer.

    When solving for the allotted distribution of kinetic energy, I thought about using the conservation of linear momentum ##p=m_1v_1+m_2v_2## where ##p=0## initially. From this I find that the ratio of masses is proportional to the ratio of velocities such that
    $$-\frac{m_+}{m_p}=\frac{v_p}{v_+}$$
    where "##+##" indicates the positron.

    I wasn't sure where I could go from there, so I decided to substitute ##-\frac{m_+}{m_p}v_+=v_p## into the kinetic energy equation.

    $$U_{sys}=\frac{1}{2}m_p(\frac{m_+}{m_p}v_+)^2+\frac{1}{2}m_+v_+^2$$

    I'm not entirely sure if this is the right approach, but any help would be greatly appreciated.
     
  2. jcsd
  3. Mar 3, 2015 #2

    Merlin3189

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    b) Your momentum idea is fine. Notice that you are asked, not what the KE is at any time, but what the % distribution between them is. Percentage is just a sort of ratio: instead of saying A:B is something, you are saying A as % = ##\frac{A}{A+B}## x 100
    If you find the ratio A:B, you just rewrite it as %.

    BTW in a) you say, "I solved a.) and got 4.613137×10−23 which is the correct answer. " I don't want to be mean, but I would not consider an answer like this correct, because the unit is not stated. You know what it is of course and so does your tutor, but I think we should always put it, just in case were talking to Americans about sending spacecraft to Mars.
     
  4. Mar 3, 2015 #3

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    Thanks, Merlin! I'm glad I was on the right track.

    I was able to get the correct answers to the rest of the problem thanks to your help.

    Best!
     
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