2. Two identical spheres are 2 cm apart and carry equal positive charges. The force between them is \( 2,025 \times 10^{8} \mathrm{~N} \). Calculate the size of each charge.

We start with Coulomb’s law for the force between two point charges: \[ F = k\frac{q^2}{r^2} \] where \( F \) is the force between the charges, \( k \) is Coulomb’s constant \(\left( k \approx 8.99\times10^{9}\,\mathrm{N\,m^2/C^2} \right) \), \( q \) is the charge on each sphere, and \( r \) is the distance between the centers of the spheres. Given: \[ F = 2.025 \times 10^{8}\,\mathrm{N}, \quad r = 2\,\mathrm{cm} = 0.02\,\mathrm{m} \] **Step 1. Express \( q \) in terms of the other variables** Rearrange Coulomb’s law to solve for \( q \): \[ q = \sqrt{\frac{F\,r^2}{k}} \] **Step 2. Substitute the given values** First, compute the square of the distance: \[ r^2 = (0.02)^2 = 0.0004\,\mathrm{m}^2 \] Now substitute into the expression for \( q \): \[ q = \sqrt{\frac{(2.025 \times 10^{8}) \times 0.0004}{8.99 \times 10^{9}}} \] **Step 3. Perform the calculations inside the square root** Multiply \( F \) and \( r^2 \): \[ 2.025 \times 10^{8} \times 0.0004 = 2.025 \times 0.0004 \times 10^{8} = 0.00081 \times 10^{8} = 8.1 \times 10^{4} \] Divide by \( k \): \[ \frac{8.1 \times 10^{4}}{8.99 \times 10^{9}} = \frac{8.1}{8.99} \times 10^{4-9} \approx 0.901 \times 10^{-5} = 9.01 \times 10^{-6} \] **Step 4. Take the square root** \[ q = \sqrt{9.01 \times 10^{-6}} = \sqrt{9.01}\times\sqrt{10^{-6}} \approx 3.00 \times 10^{-3}\,\mathrm{C} \] Thus, the charge on each sphere is approximately: \[ q \approx 3.0 \times 10^{-3}\,\mathrm{C} \]