1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Conducting Tetrahedra
  1. Thermal conductivity (Replies: 6)

  2. Thermal Conduction (Replies: 7)

  3. Thermal conductivity (Replies: 4)