The total charge on two spheres is 600 µC. The two spheres are placed 0.900m apart and the force of repulsion between the two is 30.0 N. What is the charge on each sphere? (2 equations...)

Question

anonymous · Answer

just let q1=600µC -q2 and sub it into 
$\frac{kq1q2}{r^2} =30$

anonymous · Answer

one equation is enough.

anonymous · Answer

There's supposed to be another equation though because there's two answers.

anonymous · Answer

no, because the equation then results in a quadratic eq so there are two answers.

anonymous · Answer

could you draw how to do this please?

whpalmer4 · Answer

Let's see if I remember my electrostatics!

$$F = \frac{q_0q_1}{4\pi\epsilon_0r^2}$$$$ \epsilon_0 = 8.8542×10^{-12} C^/Nm^2$$$$q_0+q_1 = q_t = 600 * 10^{-6} C$$$$F = 30.0 N, r = 0.900m$$

$$q_0 = q_t - q_1$$$$q_0*q_1 = q_1*q_t - q_1^2$$

$$F=\frac{q_1q_t - q_1^2}{4\pi\epsilon_0r^2}$$$$4{\pi}F\epsilon_0r^2 = q_1q_t-q_1^2$$but that's just a quadratic in $q_1$ with $a=-1, b=q_t, c = -4\pi{F}\epsilon_0r^2$, use the quadratic formula $$q_1 = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ to get your answers.  One charge is much bigger than the other!