What is the final charge on sphere 2?

[soccerj17]

[SOLVED] electric charge problem

Homework Statement

Two identical small metal spheres with q1>0 and magnitude of q1>magnitude q2 attract each other with a force of magnitude 74.9mN. They are 2.47 m apart. The spheres are then brought together until they are touching. At this point, the spheres are in electrical contact so that the charges can move from one sphere to the other until both spheres have the same final charge, q. After the charges have come to equilibrium, the spheres are moved so that they are again 2.47 m apart. Now the spheres repel each other with a force of magnitude 14.98 mN. What is the final charge on sphere 2? This is the first part of the problem.

It also gives the radius of the spheres, which is 28e-6 m, but I don't think you need that for this first part because the 2.47 m is from center to center of the spheres.

Homework Equations

Coulomb's Law: F = kq1q2/r^2
k = 9e9
1 mN = 10e-3 N

The Attempt at a Solution

I used Coulomb's law F=kq1q2/r^2 and the charges of the two spheres are the same, so F=kq^2/r^2. So then q = (Fr^2/k)^1/2
I converted the mN into N and got q = (14.98e-3*2.47^2/9e9)^1/2 = 3e-6 C
I submitted this answer online on the site we have to do our homework for, and it said it was incorrect. What did I do wrong?

 

[Doc Al]

What's the sign of the charge?

 

[soccerj17]

I tried submitting both positive and negative answers, so i don't think that's the problem.

 

[Doc Al]

How many significant figures did you leave in your submitted answer. (I hope you didn't round off to just one!)

 

[soccerj17]

Thanks! That was the problem. My calculator had given me the answer 0.000006 so I had to convert everything to the -6 power to get the more specific answer.