No. All you need to know is that the ball is positive, and what like charges do to each other.
Two charges, A and B are on opposite sides of a stretched wire measuring 100cm. The charge on A is +1nC and the charge on B is +2nC. On the wire between A and B is a little plus charged plastic ball that can slide without friction. Where does the plastic ball reach a steady position?
I suspect that I need to know the charge on the plastic ball to calculate this. Am I right?
Close. Think about Coulomb's law. That unknown charge of the ball? We'll leave that as variable q.
Now what condition in the problem needs to be satisfied, and what does it say can change to try to satisfy it?
The ball is repelled by both ends, and will come to equilibrium according to your equation ((1nC*q)/rA^2)*k+((2nC*q)/rB^2)*k=total electrical force. Now, at the equilibrium point, the total electric force will be zero; substitute this into your equation, while noting that that the length of the wire is l =rA + rB. Good luck, let us know how it goes.
In this case, I think it is more accurate to say that the electric field is zero at some equilibrium point. The total electric field at any one point is the sum of the electric fields from all charges, and this is E = E_A + A_B. However, since the charges repel there will be a point in between where the field will be zero, i.e. E_A + E_B = 0. That is,
((1nC)/rA^2)*k+(- (2nC)/rB^2)*k = 0
Note that the second term is negative (the same has to be true for Force above) because the field lines at the equillibrium are in opposite directions. Now, solve for rA or rB and make use that rA + rB = 1m.