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    1. #1
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      Electrostatic Force Problem

      Two spheres are brought into contact:

      Sphere one:
      r= 17.5 cm = 0.175 m
      m= 150 g = 0.150 kg
      q = -11.0 microcoloumbs = -1.1 * 10^-5 Coloumbs

      Sphere two:
      r= 30.5 cm = 0.305 m
      m= 46.0 g = 0.046 kg
      q= -28.0 microcoloumbs = -2.8 * 10^-5 Coloumbs

      (r = radius, m = mass, q = charge )

      The two spheres are then released, what is the maximum acceleration and velocity achieved by both spheres?

      I've calculated the maximum accelerations because I calculated the electrostatic force between them, and I know the acceleration will be greatest when they are first released since the force is proportional to 1/distance^2.

      I am unsure how to find the maximum velocity achieved, since the acceleration won't be constant. Is integration needed for this part of the problem?

    2. #2
      Xei
      UnitedKingdom Xei is offline
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      Weird, I'm not sure how to solve this.

      You've basically got two differential equations here,

      d2x1/dt2 = -k1/(x2 - x1)^2,
      d2x2/dt2 = k2/(x2 - x1)^2

      where x1 and x2 are the positions of particles 1 and 2.

      I'm not sure what you can do with these equations though. Each one is intrinsically linked to the other...

    3. #3
      Xei
      UnitedKingdom Xei is offline
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      Actually there is kind of a way which bypasses my lack of knowledge about solving the above (which I'm learning about at the moment actually), as you can prove with vector calculus that the electrostatic potential energy of two particles is q1*q2/(4*pi*epsilon0*r).

      The particles will have maximum velocity when all of this has been converted to KE (at a separation of infinity), so equate the sum of their final kinetic energies to the above quantity, and then you can use the conservation of momentum (which sums to 0 here), sub that in to eliminate the other velocity, and hence get the final velocity of one of the particles.

      My final result is v1 = sqrt(q1*q2*m2 / (2*pi*epsilon0*r*m1*(m1+m2))).

      And obviously the symmetrical result for v2.

    4. #4
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      But as the seperation distance approaches infinity that equation will go to zero, no?

    5. #5
      Xei
      UnitedKingdom Xei is offline
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      No, r is the initial separation which is all that matters because you're calculating the energy of the whole system which is constant.

    6. #6
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      I plugged in the numbers but the velocities that come out of it are too small I think. I got 13.9 m/s for the lighter sphere and 4.25 m/s for the heavier sphere.

      However, the accelerations were 261 m/s^2 and 80.1 m/s^2.

    7. #7
      Xei
      UnitedKingdom Xei is offline
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      Why do you think those values are incorrect?

      Given those large accelerations, the particles only need to accelerate for a fraction of a second before the electrostatic force has decreased by a large factor and their accelerations become neglibible.

      I could of course have slipped up in deriving my formula (I get the same numerical answers as you). Try it yourself and check that it comes out okay.

    8. #8
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      Hey Xei what you said makes sense I didn't really think of that :/

      It turns out those values are correct, thanks a lot for the help!

    9. #9
      Xei
      UnitedKingdom Xei is offline
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      Any time. I needed something to warm up with today anyway.

    10. #10
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      I approached this question another way after seeing an example in one of my textbooks, using conservation of mechanical energy:

      K1 + U1 = K2 + U2

      Where K represents the kinetic energy of the system and U is the electric potential energy of the system.

      Since K1 will be 0 (velocity is zero), and U2 will be zero (at seperation of infinity), I get the equation:

      U1 = K2

      q1q2/4pi*r*epsilonnaught = 1/2mv^2

      When I solve for velocity I get similar values to those I got after plugging in the values in the equation that you derived, but they are slightly off. Can you see any problems with this derivation? It seems like it should work, unless there is a problem assuming that the electric potential energy will be zero for U2. In the books example they were given initial conditions so that K1 and U2 were not zero.

    11. #11
      Xei
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      That's actually the same method I used.

      However you're doing it wrong with the whole K2 thing; K1 + U1 = K2 + U2 is an identity which is true for the entire system at any time. We can find K1 + U2 at t = 0, as you have done. This represents the total energy of the system.

      However I think your next step has been to equate the kinetic energy of one of the masses with the total energy of the system at t = infinity, which is a mistake. U2 is indeed 0 here, but K2 represents the total kinetic energy of the system, not just one particle, so on the right you should have 1/2*m1*v1^2 + 1/2*m2*v2^2.

      This looks unsolvable because we have two variables and only one equation, but as I detailed above, you can use the law of conservation of momentum,

      m1v1 + m2v2 = c

      and looking at t = 0 we can deduce that c = 0, so then you can express v2 explicitly in terms of v1 and plug this into the first equation to derive an explicit formula for v1.

      Is this making more sense?

    12. #12
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      Ooooooh, ok, it all just came together, I see where I was going wrong now.

      Thanks again

    13. #13
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      Wow - we can talk physics here too. Awesome.

      Interesting problem, but it's kind of ambiguous. A mix of EM and classical Newtonian mechanics, a little weird. Let me see if I can understand the problem a little more clearly here. It seems we have two point charges coming into contact and then repelling due to a repulsive net electrostatic force, or equivalent charges. You want to determine the maximum velocity of the point charges as they repel. So, first we need to determine the magnitude of the average electrostatic force that acts between the two particles,

      F_e = [1/(4*pi*epsilon_0)][(e^2)/(r^2)]

      and then translate that into a vector form of Coulomb's law. Since the force is repulsive, assuming the particles have the same sign, the position vectors must be parallel. We can designate that as a vector R and a vector F, and then take into account Newtons third law since gravity is obviously at play.

      F = [1/4*pi*epsilon][(q_1*q_2)/(R^2)]R,

      F_g = G[(m_e*m_p)/(r^2)]

      Sadly, this is rather pointless because even though the vector form of coulomb's law carries within it directional information about the two vectors, and whether they are attractive or repulsive, it's really only critical when taking into account a system of more than two charges. In this case, the above equation would hold for every pair of charges, and the total force on any one charge would be found by taking the vector sum of the forces due to each of the other charges. Incidentally, the equation utilized for that purpose is identical to the mathematical representation of the principle of superposition applied to electric forces.

      What it seems like you're really asking though is, what would be the maximum velocity and maximum acceleration of two small inactive magnet spheres placed side by side, then switched on, and taking account for the amperes per second, be after a certain period of time? Great question, but I too am not sure, since Coulomb's law,

      F ∝ (|q_1|*|q_2|)/(r^2)

      Generally holds only for point charges.

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