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Conducting Tetrahedra

  1. May 18, 2008 #1
    1. The problem statement, all variables and given/known data

    Two conductors, A and B, are each in the shape of a tetrahedron, but of different sizes. They are charged in the following manner:
    1) Tetrahedron A is charged from an electrostatic generator to charge .
    2) Tetrahedron A is briefly touched to tetrahedron B.
    3) Steps 1 and 2 are repeated until the charge on tetrahedron B reaches a maximum value.


    If the charge on tetrahedron B was after the first time it touched tetrahedron A, what is the final charge on tetrahdedron B?

    2. Relevant equations

    V = Cq (C is a constant)

    Va = Vb --> aqA = bqB

    qA/qB = b/a


    3. The attempt at a solution

    in the first portion of the tetrahedra touching a charge of q/4 was added to B and taken from A so I assumed it was left with a charge of 3q/4, I more or less confirmed this when the problem also told me that the ration of charges (the qA/qB formula gave the number 3) so I assumed it was (3/4)/(1/4) = 3.

    As the problem stated this will continue until can no longer accept charges from A, however a hint that the problem came let me find that the max charge on A when B is at it's max is q, I assumed it would've been 0 since when B would have been q A would've run out of charges.

    by the formulas given by the problem qbmax = aqAmax/b, being qbmax = aq/b, but I'm not sure where to go from here, any help is appreciated.
     
  2. jcsd
  3. May 5, 2011 #2
    We determine that (1) qa/qb[the ratio of the charges]=3 because
    (1-1/4)q/ (1/4)q= [ net charge on a after 1/4 q from a flows to b] / [charge on b after first touch] = 3

    Once the potential on each conductor is the same, there is no net flow of charge between them.

    Because step 1 and 2 are repeated unit B reaches its maximum value the charge on a will always be "refilled" back to charge q. Therefore qamax (the charge on a when b is full)=q.

    Finally from equation(1) we know that qbmax=qamax/3 ==> q/3
     
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