Charges on a sphere?

Question

total charge of 7.50 x 10^-6 C is distributed on two different small conducting spheres. when the spheres are 6.00 cm apart, they each feel a repulsive force of 20.0 N. what is the charge on each sphere

Answer 1

Call each charge q1 and q2, measured in coulombs.

The total charge: q1 + q2 = Q = 7.50 x 10^-6

The force between them (assuming you don't have a significant radius) is

F = k q1 q2 /r^2

Fr^2/k = q1 q2

Substituting q2 = Q - q1 from the first equation.

Fr^2/k = q1 (Q - q1)

Fr^2/k = q1 Q - q1^2

q1^2 -Q q1 + Fr^2/k = 0

Solve using the quadratic equation

q1 = [Q ± √(Q^2 - 4Fr^2/k) ]/ 2 and with this

q2 = Q - [Q ± √(Q^2 - 4Fr^2/k) ]/ 2

√(Q^2 - 4Fr^2/k) ]  < Q and is positive, q1 = [Q + √(Q^2 - 4Fr^2/k) ]/ 2 would be greater than Q but this would make q2 negative since q2 = Q - q1.  If one is positive and the other negative, then your original statement of repulsion would be violated since the charges are repelling.

So,

q1= [Q - √(Q^2 - 4Fr^2/k) ]/ 2 which will be positive and less than Q.

q2 = Q - [Q - √(Q^2 - 4Fr^2/k) ]/ 2 =√(Q^2 - 4Fr^2/k) /2 which is also positive.

Plugin your values (Q = 7.5 x 10^-6 C, r = 0.06 m, and F = 20N, k = 9 x 10^9 Nm^2/C^2 ) and you're done!

Hope this helps.