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Electric Potential Conductor Problem

  1. Jan 2, 2007 #1
    1. The problem statement, all variables and given/known data

    Two insulating spheres having radii 0.300 cm and 0.500 cm, masses 0.100 kg and 0.700 kg, and charges -2.000 micro-C and 3.00 micro-C are released from rest when their centers are separated by 1.00 m. (a) How fast will each be moving when they collide? (Hint: Consider conservation of energy and linear momentum.) (b) If the spheres were conductors would the speeds be larger or smaller than those calculated in part (a)? Explain.

    2. Relevant equations
    [tex]U = K_e \frac{q_1 q_2}{r_{12}}[/tex]
    [tex]E_1 = U = E_2 = \frac{1}{2} m v^2[/tex]

    3. The attempt at a solution

    Let the radii be r_1 and r_2, masses m_1, m_2, charges q_1, q_2, and distance separated d.

    so we have:

    [tex]U = K_e \frac{q_1 q_2}{r_{12}} = K_e \frac{q_1 q_2}{d + r_1 + r_2}[/tex]

    Since the charges are at rest at first, they have only electrical potential energy (given above) and they're the same, and when they collide, I assume they have only kinetic energy.

    [tex]U = KE_1 = \frac{1}{2}m_1 v_{1}^2[/tex]
    [tex]K_e \frac{q_1 q_2}{d + r_1 + r_2} = \frac{1}{2}m_1 v_{1}^2[/tex]

    solving for V we have [tex]v = \sqrt { \frac{2 K_e q_1 q_2}{(d + r_1 + r_2) m}[/tex]

    but this fails to provide the proper solution. what's wrong with it? How is linear momentum involved?
  2. jcsd
  3. Jan 3, 2007 #2
    so, [tex]q_1 = -2.00 \times 10^{-6} C, r_1 = 0.003 m, m_1 = 0.100 kg, q_2 = 3 \times 10^{-6} C, r_2 = 0.005 m, m_2 = 0.700 kg[/tex]
    Hence for the first particle:

    [tex]v_1 = \sqrt{ \frac{2 \cdot 8.99 \times 10^9 \cdot 2 \times 10^{-6} \cdot 3 \times 10^{-6} }{(1 + 0.003 + 0.005) \cdot 0.100}} = 1.033 m/s[/tex]
    [tex]v_2 = .390 m/s[/tex]

    but these are incorrect; the book has: 10.8 m/s and 1.55 m/s
  4. Jan 3, 2007 #3
    Reconsider the net energy before they collide and after they collide , initially they will have only potential energy (both electrical as well as Gravitational P.E)
    , when they collide they will still have some Potential energy (since the two centers are separated by some distance) along with K.E , and on both sides of equality , you have to consider potential energy of both spheres together , vice versa for terms on right hand side.

    Then use C.O.Momentum , which have simple 'm(i)v(i)' format terms on both sides of the equality.

    And while considering the initial C.O.Energy , remember that Gravitational P.E will have a -ve sign , as compared to the Electrical POTential energy which always depends on the sign of the charges.
  5. Jan 3, 2007 #4


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    There is no need to consider the gravitational PE - it is several orders of magnitude smaller than the electrostatic PE, and can be neglected.
  6. Jan 3, 2007 #5
    So we have

    [tex]E_{1i} = E_{2i} = k_e \frac{ q_1 q_2 }{(d + r_1 + r_2)}[/tex]

    [tex]E_{1f} = \frac{1}{2} m_1 v_1^2 + k_e \frac{ q_1 q_2 }{(r_1 + r_2)}[/tex]

    [tex]E_{2f} = \frac{1}{2} m_2 v_2^2 + k_e \frac{ q_1 q_2 }{(r_1 + r_2)}[/tex]

    therefore [tex]E_{1i} = E_{1f}[/tex] and [tex]E_{2i} = E_{2f}[/tex] solve for v then that's it?

    What do you mean by
    Last edited: Jan 3, 2007
  7. Jan 4, 2007 #6


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    You can not speak of PE for each object. The PE is only defined for the entire system, and not for individual parts of the system. PE = KqQ/x is the PE of the system consisting of the both charges, Q and q.

    So, you have E(i) = KE(i) + PE(i) = KE1(i) + KE2(i) + PE(i) = PE(i) since initial KEs are both zero.

    E(f) = E(i) = KE(f) + PE(f) = KE1(f) + KE2(f) + PE(f)

    Fianlly, note that energy conservation requires that the total energy of the whole system be conserved: (E(i) = E(f)

    That will give you one equation.

    As for the other, you have momentum conservation to write down.
  8. Jan 4, 2007 #7
    For the conservation of linerar momentum, can we do:
    [tex]P_i = m_1 v_{1i} + m_2 v_{2i} = 0[/tex]

    [tex]P_f = m_1 v_{1f} + m_2 v_{2f}[/tex]

    then [tex]P_i = P_f[/tex]
  9. Jan 4, 2007 #8
    Wow that's it!! thanks ever so much! Here's what I did:

    [tex]E_i = k_e \frac{q_1 q_2}{d + r_1 + r_2}[/tex]

    [tex]E_f = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 + k_e \frac{q_1 q_2}{r_1 + r_2}[/tex]

    [tex]P_i = 0[/tex]

    [tex]P_f = m_1 v_1 + m_2 v_2[/tex]

    Using conservation of energy laws:

    [tex]E_1 = E_f[/tex] and [tex]P_i = P_f[/tex]

    For (b), "If the spheres were conductors would the speeds be larger or smaller than those calculated in part (a)?"

    So since they're conductors, we wouldn't take into account their radius? That is to say,

    [tex]E_i = k_e \frac{q_1 q_2}{d}[/tex]

    [tex]E_f = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 + ???[/tex]

    Would there be any potential if the two charges are conductors and touching?
  10. Jan 4, 2007 #9


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    This is close, but there's a mistake in it. The distance between the spheres is just d, not d+r1+r2. We can treat the spheres as simple point charges located at the centers of the spheres (keeping in mind that finally, the point charges can not get any closer than r1+r2).

    Good - this is right.

    Also correct, except that only the equation on the left is called the energy conservation law - the one on the right is called the momentum conservation law.

    No, this isn't right. Doing an actual calculation with conducting balls is pretty complicated (which is why the question doesn't ask for one). What's important to recognize in a conductor is that the charges are themselves free to move. So, when conducting spheres of opposite charge are placed in the vicinity of each other, the charges on one sphere, being attracted to the charges on the other sphere will redistribute themselves within the sphere. If you want to be careful, you'd keep in mind also, that the charges within a sphere want to repel each other (but for now, consider this only a minor detail).

    Ask yourself how the charge would redistribute and what this redistribution does to the effective distance between the two centers of charge. And you need not worry about what happens after the spheres touch each other...only what happens until the instant they touch.
  11. Jan 4, 2007 #10
    Oh yes, the question states: "when their centers are separated by 1.00 m" - I was adding the radius of the spheres to the distance between the surface of the spheres to get the entire distance between the centers of the spheres - but this was wrong (the question gives that value directly).

    When the spheres are conductors, the charges within the spheres will try to get as close to the other as possible (negatives on one sphere, positive on the other), so:

    [tex]E_i = k_e \frac{q_1 q_2}{d - r_1 - r_2}[/tex]

    So when they touch, there is no potential energy between the spheres?

    Thanks for your help.
    Last edited: Jan 4, 2007
  12. Jan 4, 2007 #11


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    That's basically the right idea, but the actual number in the denominator will be a little bigger than d-r1-r2 (though, as you've correctly figured, it will be less than d). If it were to be exactly d-r1-r2, that would mean that all the charge on each sphere would have to be concentrated at a single point. Naturally, this can't happen...but the general redistribution is towards this direction.

    Think again. The zero is in the denominator!
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