Electric Charge: Finding Q_1 Given Q_2 & r

[Robin04]

Homework Statement

The distance between two small balls of the same radius and charge is r. If we release the balls they approach, touch and then repulse each other. When their distance is 4r the repulsive force is 1/20th of the initial force. What is the charge of the first ball if the second's is Q_2 = 2*10^-5 C?

Homework Equations

F = 9*10^9*Q_1*Q_2 / r^2
Their charge after the collision: Q = (Q_1 + Q_2) / 2

The Attempt at a Solution

In the end I got a quadratic function but the discriminant was negativ.
http://kepfeltoltes.hu/150227/11026697_898760020174352_224095763_n_www.kepfeltoltes.hu_.jpg
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[TSny]

$20 = \frac{16Q_1Q_2}{Q^2}$. Can this equation possibly be true if you consider the signs of the charges?

 

[Robin04]

$20 = \frac{16Q_1Q_2}{Q^2}$. Can this equation possibly be true if you consider the signs of the charges?

You're right. And if I write -20 instead of 20? Because the force in (2) is in the opposite direction as in (1).

 

[TSny]

I think that should work.

 

[Robin04]

I did it. The two solutions are: -4*10^-6 C and -1*10^-4 C. Both of them are negativ. How can I decide which one is correct? Or maybe there's two correct solutions?

 

[TSny]

I get different answers for Q₂. But I think you have it set up correctly (with -20).

To see if both solutions make sense, calculate Q for both answers. Then check if both solutions lead to the correct ratio of forces for the attraction at r and the repulsion at 4r.