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## 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.