Balancing Gravity and Charge (Sears and Zemansky 4th Edition)

[equant]

Homework Statement

A certain metal sphere of volume 1cm^3 has a mass of 7.5g and contains 8.2 x 10^22 free electrons. How many electrons must be removed from each of two such spheres so that the electrostatic force of repulsion between them just balances the force of gravitational attraction? Assume the distance between the spheres is great enough so that the charges on them can be treated as point charges. (Problem 24-5 From University Physics Sears and Zemansky 4th Edition)

Answer in the back of the book is 4E6 electrons

Homework Equations

$F_g = GMm/r^2 = Gm^2/r^2$
$F_e = kQq/r^2 = kq^2/r^2$

The Attempt at a Solution

The charge needed for a balanced situation:

$q = m(G/k)^2 = 6.46 x 10^-13$

$n_{electrons} = q/e = 4x10^6$

... I believe that should be the number of electrons in the sphere. The book gives that number as the answer, but the question asks how many should be removed. Shouldn't the answer be 8.2E22 - 4E6 (per sphere)? Any idea where I'm going wrong?

Thanks.

 

[Spinnor]

Your numbers look good, put them back in your force equations and see if you get roughly the same force. Let R = 1. So two metal spheres that weigh 7.5E-3 kg and each have a net positive charge of 4E6 e will have forces that cancel.