How Does Charge Redistribution Work Between Two Tetrahedron Conductors?

[clope023]

Homework Statement

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?

Homework Equations

V = Cq (C is a constant)

Va = Vb --> aqA = bqB

qA/qB = b/a

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.

 

[quark_boi]

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