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Electric Potential Energy problem

  1. Apr 10, 2005 #1
    "Two particles each have a mass of 6.1*10-3 kg. One has a charge of +5.0*10-6 C, and the other has a charge of -5.0*10-6 C. They are initially held at rest at a distance of 0.80 m apart. Both are then released and accelerate toward each other. How fast is each particle moving when the separation between them is one-half its initial value?"

    What equations would I need to use in order to solve this problem? I tried using EPE/q=Va-Vb, but the voltage cancels out, so I'm unsure as to where to go from there. Thanks.

    ~Futron
     
  2. jcsd
  3. Apr 10, 2005 #2
    Use kinetic and potential energy relationships. Initially they both have no kinetic energy, but a certain amount of potential energy. When d = d/2 they have a certain about of potential and kinetic energy. Find the potential energy in both cases, any potential energy lost would have been translated into kinetic energy.
     
  4. Apr 10, 2005 #3
    Alright, I tried using EPEi=EPEf+(1/2)mv^2 where EPE=q0*(Va-Vb), but that doesn't seem to work. Any ideas?
     
  5. Apr 10, 2005 #4
    [itex] EPE_f-EPE_i = mv^2/2, and EPE = q(V_b-V_a) [/itex] are the amended versions of your equations. Find the potential with [itex] U = qE\Delta x[/itex]

    This will take you in circles though.

    The force between the charges is given by coulombs law. Find the difference in potential energy at d = r and d = r/2.
     
  6. Apr 11, 2005 #5
    So if I have [itex] EPE_a-EPE_b=q_0(V_A-V_B) [/itex] where [itex] V=kQ/r [/itex], how would I then find the difference between the two when the distance is halved?
     
  7. Apr 11, 2005 #6
    [itex] U = qE\Delta x[/itex]

    The field for point charges is [itex] E = \frac{kq}{r^2} [/itex]

    [tex] \Delta x_i = 0.8m, \Delta x_f = 0.4m [/tex]
     
  8. Apr 11, 2005 #7
    Thanks, but I'm still not getting the correct answer. Which charge should I use for [itex] q [/itex] in [itex] EPE=qE\Delta x [/itex], and once I get that, would it simply be a matter of solving [itex] EPE_f-EPE_i=(1/2)mv^2 [/itex]?
     
  9. Apr 12, 2005 #8
    Can you show some work? In your first post you said the voltages cancel out which isnt correct, this would only happen if [itex] V_b = v_a [/itex] which isnt true.

    Show me where your messing up, all these techniques should give the same answer, if you prefer voltages we can use that.
     
  10. Apr 12, 2005 #9
    Alright, I tried using EPEf-EPEi=(1/2)mv^2 where EPEf=(5x10^-6)((9*10^9)*(-5*10^-6)/0.4)) and EPEi=(5*10^-6)x((9x10^9)(-5x10^-6)/0.8)), but that gives 5.43m/s, which is not the correct answer.
     
  11. Apr 12, 2005 #10

    GCT

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    I think that the potential energy state is [tex]U=k_{e}q_1q_2/r[/tex], you'll need to find [tex] \Delta U=k_eq_1q_2(1/r_{initial}~-1/r_{final})[/tex], this will equal the negative value of the change in kinetic energy [tex]=-mv^{2}_{final}-0[/tex]
     
  12. Apr 12, 2005 #11
    Thanks! I was jumbling too many equations together at once but that made it much clearer.
     
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