Two particles, each of charge q and mass m, are held in fixedpositions at a dist

Question

Two particles, each of charge q and mass m, are held in fixedpositions at a distance d from each other. A third identicalparticle is then brought in from infinity and placed so that aperfect equlateral triangle is formed by the particles, as depictedin the diagram. Assume that the third particle was at rest when itwas infinitely far away. a ) Determine the work which must be done by an external agentto move the third particle into its position while the other twoparticles remain in place. b) Suppose that the three particles are suddenly released(from rest) and allowed to move apart from each other, out towardinfinity. Determine their final speeds when they are infinitely faraway from each other. Hint: Consider the symmetry of the situation to determine howthe final speeds of the particles are related to each other. Please provide step by step solutions please...

Explanation / Answer

work that must be done is the same as the increase inpotential energy. Initially, charge 3 has zero potential energy.When it is in the triangle, it has PE with each of the othertwo. . PE of a pair of charges is    k Q q /r        so . work = PE of q3 and q1 + PE ofq3 and q2 =   k q1 q3 / d   + k q2q3 / d   = .           = 8.99 x 109 * 2.0 x 10-6 * 2.0 x10-6 / 0.02   +   8.99 x109 * 2.0 x 10-6 * 2.0 x 10-6 /0.02    = .           = 3.596Joules . (b) The total potential energy of all the particlestogether is simply the sum of the PE of each pair. There are threepairs: .         1&2  2&3   1&3 . Each pair has PE of     1.798Joules   (note this is half of the calculation above).So .    total PE = 3 * 1.798 = 5.394 Joules . Due to symmetry, each particle will use 1/3 of this totalpotential energy for its kinetic energy. So .    when very far away, KE of one particle = 1.798 Joules . Sofinally...           KE = (1/2) m v2 . 1.798 = (1/2) * 0.005 *v2              v = 26.82 m/s   is the final speed of eachparticle

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