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Electrostatic Problem - finding initial separation of two particles

  1. Aug 26, 2009 #1
    1. The problem statement, all variables and given/known data
    One particle has a mass of [itex]3.00 \times 10^{-3}[/itex] kg and a charge of [itex]+8.00 \mu[/itex] C. A second particle has a mass of [itex]6.00 \times 10^{-3}[/itex] kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.100 m, the speed of the [itex]3.00 \times 10^{-3}[/itex] kg particle is 125 m/s. Find the initial separation between the particles.


    2. Relevant equations
    [itex]V = \frac{kq}{r}, E = \frac{F}{q} or E=\frac{kq}{r^2}, U = qV, F=\frac{kq_1q_2}{r^2}, \frac{1}{2}mv^2_A + U_A = \frac{1}{2}mv^2_B + U_B[/itex]


    3. The attempt at a solution
    I can easily calculate the force of repulsion between the two particles and the electric field vectors from each particle. I don't know how to include the speed to find the initial distance.

    Thank you to anyone who can help me.
     
    Last edited: Aug 26, 2009
  2. jcsd
  3. Aug 26, 2009 #2

    ideasrule

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    Try using the last equation, remembering that the velocity of both particles is initially 0.

    Also, because no external forces act on the two particles, there's a quantity other than energy that's conserved. What is it?
     
  4. Aug 26, 2009 #3
    Momentum.

    [itex]\begin{aligned}\rho_{before} &= \rho_{after}\\ mv_1 + mv_2 &= mv_1 + mv_2\\ 0 &= -125 \times 3.00 \times 10^{-3} + 6.00 \times 10^{-3}v_2\\v_2 &= \frac{125 \times 3.00 \times 10^{-3}}{6.00 \times 10^{-3}}\\ &=62.5\,m/s \end{aligned}[/itex]

    How can I use the conservation of energy to work out the distance? I don't know the potentials.
     
  5. Aug 26, 2009 #4
    You've written down the necessary equations to work out the potentials in your first post!
     
  6. Aug 26, 2009 #5
    Is this correct?
    [itex]
    \begin{aligned}
    \frac{1}{2}mv^2_A + U_A &= \frac{1}{2}m^2_B + U_B\\
    0 + U_A & = \left(\frac{1}{2}mv^2_1 + \frac{1}{2}mv^2_2 \right) + U_B
    \end{aligned}
    [/itex]
    where [itex]\begin{aligned} U_B &= qV\\ &=q \times \frac{kq}{r}\\ &= \frac{8.99 \times 10^9}{0.100}\end{aligned}[/itex]
    Since I know v1 = 125 m/s and v2 = 62.5 m/s and if U_B is correct, then all I have to do is find U_A and then use U = qV and find r?
     
  7. Aug 26, 2009 #6

    ideasrule

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    You seem to have forgotten to multiply by q2. Other than that, yes, all you have to do is find U_A and use U=qV again.
     
  8. Aug 26, 2009 #7
    Ok. To continue where I left off:
    [itex]
    \begin{aligned}
    U_A &= (0.5 \times 3.00 \times 10^{-3} \times (-125)^2 + 0.5 \times 6.00 \times 10^{-3} \times 62.5^2) + \frac{8.99 \times 10^9 \times (8.00 \times 10^{-6})^2}{0.100}\\
    &= 2.53 \times 10^7\,J\\
    \\
    U_A &= qV\\
    V &= \frac{U_A}{q}\\
    &= \frac{2.53 \times 10^7}{8.00 \times 10^{-6}}\\
    &= 3.16 \times 10^{12}\\
    \\
    V &= \frac{kq}{r}\\
    r&=\frac{kq}{V}\\
    &= \frac{8.99 \times 10^9 \times 8.00 \times 10^{-6}}{3.12 \times 10^{12}}\\
    &= 2.28 \times 10^{-8}\,m\\
    &= 22.8\,nm
    \end{aligned}
    [/itex]
    Is that a reasonable answer? Is it correct? It seems far too small to me.
     
  9. Aug 26, 2009 #8

    ideasrule

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    You made a math error on this line; the answer should be much smaller.
     
  10. Aug 26, 2009 #9
    Thanks. I must have missed a bracket or something.
    [itex]U_A = 40.9\,J[/itex]
    The answer should now be 0.014 m or 1.4 cm. This seems a bit more reasonable.
     
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