[SOLVED] Charged Spheres 1. The problem statement, all variables and given/known data Two small aluminum spheres, each of mass 0.0250 kilograms, are separated by 80.0 centimeters. How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude 1.00x10^4 N (roughly one ton)? Assume that the spheres may be treated as point charges. 2. Relevant equations F = kq1q2/r^2 q = ne 3. The attempt at a solution In a previous question, it was worked out that each sphere has 7.25x10^24 electrons. It's confirmed that this answer is correct. F = kq1q2/r^2. We're taking an amount from the first, and giving it to the second. I declared variable 'x' to be this amount. Since each sphere has the same number of electrons, we can use a common variable 'n', instead of n1 and n2. F = kq1q2/r^2 F = k(ne)(ne)/r^2 F = k((n-x)e)((n+x)e)/r^2, introducing the variable x, taking x electrons from one, and giving them to the other, then solving for F = 10^4. Fr^2/k = e^2(n^2 - x^2) Fr^2/(ke^2) = n^2 - x^2 (Fr^2/(ke^2)) - n^2 = -x^2 -(Fr^2/(ke^2)) - n^2 = x^2 Sqrt(-(Fr^2/(ke^2)) - n^2) = x Sqrt(-((10000)(0.8)^2/((8.988*10^9)(1.60217646*10^-19)^2)) - (7.25*10^24)^2) = x x = Not a real value. I took the absolute value of the huge number above before square rooting, and it came out to 7.25*10^24... which would mean that x = n, therefore to give the said force, 100% of the electrons would need to be transfered. I tried this answer, and was told I was wrong.