A total charge of \(7.50 \times 10^{-6} \mathrm{C}\) is distributed on two different small metal spheres. When the spheres are \(6.00 \mathrm{cm}\) apart, they each feel a repulsive force of \(20.0 \mathrm{N} .\) How much charge is on each sphere?

Let's define the variables: - \(Q_1\) is the charge on the first sphere. - \(Q_2\) is the charge on the second sphere. - \(Q_t = 7.5 \times 10^{-6} \mathrm{C}\) is the total charge. - \(F = 20.0 \mathrm{N}\) is the force between the two spheres. - \(r = 6.00 \times 10^{-2} \mathrm{m}\) is the distance between the two spheres. - \(k\) is the Coulomb constant, \(k = 8.987 \times 10^{9} \mathrm{N m}^2 \mathrm{C}^{-2}\). We will use Coulomb's law: $$F = k \frac{Q_1 Q_2}{r^2}$$ Since the total charge is given and both spheres have positive charges, we also have: $$Q_t = Q_1 + Q_2$$

First, let's solve for \(Q_1\) from the total charge equation: $$Q_1 = Q_t - Q_2$$ Now substitute \(Q_1\) into the Coulomb's law equation: $$F = k \frac{(Q_t - Q_2) Q_2}{r^2}$$

Now, multiply both sides by \(r^2\) and divide both sides by \(k\): $$Q_2^2 - Q_t Q_2 = \frac{F r^2}{k}$$ Next, plug in the given values for \(F\), \(r\), \(k\), and \(Q_t\): $$Q_2^2 - (7.5 \times 10^{-6} \mathrm{C})Q_2 = \frac{(20.0 \mathrm{N})(6.0 \times 10^{-2} \mathrm{m})^2}{(8.987 \times 10^{9} \mathrm{N m}^2 \mathrm{C}^{-2})}$$ Simplify the equation on the right side: $$Q_2^2 - 7.5 \times 10^{-6}Q_2 = 8.01 \times 10^{-16} $$

We can use the quadratic formula to solve for \(Q_2\): $$Q_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ In our case, \(a = 1\), \(b = -7.5 \times 10^{-6}\), and \(c = -8.01 \times 10^{-16}\). Plug these values into the quadratic formula: $$Q_2 = \frac{7.5 \times 10^{-6} \pm \sqrt{(-7.5 \times 10^{-6})^2 - 4(1)(-8.01 \times 10^{-16})}}{2}$$ Solve for \(Q_2\): $$Q_2 \approx 1.33 \times 10^{-6} \mathrm{C} \quad \text{or} \quad Q_2 \approx 6.17 \times 10^{-6} \mathrm{C}$$

Now we will find the corresponding values of \(Q_1\) for each possible value of \(Q_2\) using the equation \(Q_1 = Q_t - Q_2\): 1) When \(Q_2 \approx 1.33 \times 10^{-6} \mathrm{C}\): $$Q_1 \approx 7.5 \times 10^{-6} \mathrm{C} - 1.33 \times 10^{-6} \mathrm{C} \approx 6.17 \times 10^{-6} \mathrm{C}$$ 2) When \(Q_2 \approx 6.17 \times 10^{-6} \mathrm{C}\): $$Q_1 \approx 7.5 \times 10^{-6} \mathrm{C} - 6.17 \times 10^{-6} \mathrm{C} \approx 1.33 \times 10^{-6} \mathrm{C}$$ Both cases give valid solutions, so the charges on the two spheres can be either \(Q_1 \approx 1.33 \times 10^{-6} \mathrm{C}\) and \(Q_2 \approx 6.17 \times 10^{-6} \mathrm{C}\), or \(Q_1 \approx 6.17 \times 10^{-6} \mathrm{C}\) and \(Q_2 \approx 1.33 \times 10^{-6} \mathrm{C}\).