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Electric Potential: Two identical + 8.5mu C point charges

  1. Aug 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Two identical + 8.5mu C point charges are initially 2.5 cm from each other. If they are released at the same instant from rest, how fast will each be moving when they are very far away from each other? Assume they have identical masses of 2.0mg .


    2. Relevant equations

    ΔKE + ΔPE = 0
    ΔPE = qV


    3. The attempt at a solution

    I went to a tutoring session today and the tutor told me ΔPE = -kq^2/r. I have no idea from where this derivation came. I'm looking at my book right now and it says ΔPE = qV. Also, she said since there are two balls ΔKE = mv^2 as opposed to 1/2mv^2, which I understand. If I'm going to use ΔPE = qV, I'm unsure how to find the electric potential difference.
     
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  3. Aug 28, 2012 #2

    SammyS

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    Re: Electric Potential

    ]
    What is the electric potential due to only one of the charges, at a distance of 2.5 cm from that charge?
    That's your V.

    Then the initial potential energy of the system of two charges is qV, for that V.​
     
  4. Aug 28, 2012 #3
    Re: Electric Potential

    These two equations are all you need to solve the problem, keeping in mind that all of the potential energy will be transferred to kinetic energy.

    Potential energy of 2 point charges q1 and q2:
    [itex]\Large U_{E}=\frac{kq_{1}q_{2}}{r}[/itex]

    Kinetic energy:
    [itex]\Large K_{E}=\frac{1}{2}mv^{2}[/itex]

    Both of these equations are basic equations that you'll find in any physics textbook, by the way.
     
  5. Aug 28, 2012 #4
    Re: Electric Potential

    Now what I've done is:

    U initial = (8.5*10^-6)(8.99*10^9 N*m/C^2) / (0.025 m)^2

    U initial = -1.0392*10^3 J

    -1.0392*10^3 J +(2)1/2mv^2 = 0

    (0.002g) v^2 = 1.0392*10^3 J

    v^2 = 5.196*10^5 m^2/s^2

    v= 720 m/s

    Hmm.
     
  6. Aug 28, 2012 #5
    Re: Electric Potential

    Your initial potential should be positive, not negative. The ΔU is negative because it goes from the initial value to 0. Otherwise, it looks alright. Your answer shouldn't change.
     
  7. Aug 28, 2012 #6

    SammyS

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    Re: Electric Potential

    [itex]\displaystyle U_{\text{initial}}=\frac{k\,q^2}{r}[/itex]

    You have [itex]\displaystyle U_{\text{initial}}=\frac{k\,q}{r^2}\,,[/itex] which is incorrect.
     
  8. Aug 29, 2012 #7
    Re: Electric Potential

    I really hope I'm not doing something silly. Does everyone else get the initial U value to be 25.98 J?

    (8.5*10^-6 C)^2 * (8.99*10^9 Nm/C^2) / (0.025 m) = 25.98 J

    Then from there get the final answer to be ~ 114 m/s?
     
  9. Aug 29, 2012 #8
    Re: Electric Potential

    Oh right, I missed that. Sorry.

    Your potential energy (25.98 J) is correct. The final answer seems to be a bit different. Try solving the kinetic energy equation for v again, and double check that you're using the right values for KE and m.
     
  10. Aug 30, 2012 #9
    Found my problem; I was using "g" instead of "kg" for mass.
     
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