Let's break down the problem step by step: **Given:** - Mass of each sphere, \( m = 2.2 \text{ g} = 2.2 \times 10^{-3} \text{ kg} \) - Separation, \( r = 3.7 \text{ cm} = 0.037 \text{ m} \) - Acceleration, \( a = 481 \text{ m/s}^2 \) - The charges are equal (let's call it \( q \)), and **sign is the same** (so they repel). - Coulomb's constant, \( k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \) --- ### **Step 1: Write the equations** **Coulomb's Law (force between the two charges):** \[ F = k \frac{q^2}{r^2} \] **Newton’s 2nd Law (relating force and acceleration):** \[ F = m a \] Set the two forces equal, since the force causing the acceleration is the electrostatic repulsion: \[ k \frac{q^2}{r^2} = m a \] --- ### **Step 2: Solve for \( q \)** \[ k \frac{q^2}{r^2} = m a \implies q^2 = \frac{m a r^2}{k} \implies q = \sqrt{\frac{m a r^2}{k}} \] --- ### **Step 3: Substitute values** - \( m = 2.2 \times 10^{-3} \) kg - \( a = 481 \) m/s\(^2\) - \( r = 0.037 \) m - \( k = 8.99 \times 10^9 \) N·m\(^2\)/C\(^2\) Plug in: \[ q = \sqrt{ \frac{(2.2 \times 10^{-3}) (481) (0.037)^2} {8.99 \times 10^9} } \] --- ### **Step 4: Calculate numerator and denominator** #### (a) \( (0.037)^2 = 0.001369 \) #### (b) Multiply numerator: \[ 2.2 \times 10^{-3} \times 481 \times 0.001369 \\ = (2.2 \times 481) \times 10^{-3} \times 0.001369 \\ = 1058.2 \times 10^{-3} \times 0.001369 \\ = 1.0582 \times 0.001369 \\ = 0.001448 (approx) \] #### (c) Now, denominator is \( 8.99 \times 10^9 \) #### (d) Fraction: \[ \frac{0.001448}{8.99 \times 10^9} = \frac{0.001448}{8.99 \times 10^9} \] --- ### **Step 5: Compute the value** First, divide: \[ 0.001448 \div 8.99 = 0.0001611 \] So, \[ \frac{0.001448}{8.99 \times 10^9} = 0.0001611 \times 10^{-9} = 1.611 \times 10^{-7} \times 10^{-9} = 1.611 \times 10^{-10} \] Oops, let's double-check: \[ 0.001448 \div 8.99 = 0.0001611 \] But since denominator is \(8.99 \times 10^9\): \[ \frac{0.001448}{8.99 \times 10^9} = \frac{0.001448}{8.99} \times 10^{-9} = 0.0001611 \times 10^{-9} = 1.611 \times 10^{-13} \] --- ### **Step 6: Square root to find \(q\)** \[ q = \sqrt{1.611 \times 10^{-13}} \] Calculate: - \(\sqrt{1.611} \approx 1.27\) - \(\sqrt{10^{-13}} = 10^{-6.5} = 10^{-6} \times 10^{-0.5} = 10^{-6} \times 0.3162 = 3.162 \times 10^{-7}\) So, \[ q = 1.27 \times 3.162 \times 10^{-7} \] \[ = 4.02 \times 10^{-7} \text{ C} \] --- ### **Step 7: Convert to nanoCoulombs (nC)** \[ 1 \text{ nC} = 1 \times 10^{-9} \text{ C} \] \[ q = \frac{4.02 \times 10^{-7}}{1 \times 10^{-9}} = 402 \text{ nC} \] --- ## **Final Answer** \[ \boxed{q = 4.0 \times 10^{2} \ \text{nC}} \] --- ### **Summary of steps:** 1. Equate electrostatic force and \( ma \): \( k \frac{q^2}{r^2} = ma \) 2. Solve for \(q\): \( q = \sqrt{\frac{ma r^2}{k}} \) 3. Substitute values and solve. 4. Convert to nC. --- **The magnitude of the charge on one of the spheres is** \[ \boxed{4.0 \times 10^2\ \text{nC}} \]