Let's solve each part step by step, with explanations. --- ## **Part (a): Finding the angle θ** ### **Step 1: Set up the forces acting on each sphere** Each 5.-g sphere is in equilibrium under three forces: - Gravitational force downward, \( F_g = mg \) - Tension in the thread, \( T \) - Electrostatic repulsion, \( F_e \), between the two charges The geometry shows the spheres hang at an angle θ from the vertical. The horizontal (electrostatic) and vertical (gravity) components of the tension must balance the respective forces. #### **Balance of forces:** - Vertical: \( T\cos(\theta) = mg \) - Horizontal: \( T\sin(\theta) = F_e \) So, \[ \tan(\theta) = \frac{F_e}{mg} \] --- **Explanation:** We resolve the tension into vertical and horizontal directions and equate them to the gravitational and electrostatic forces, respectively, because the spheres are at rest. --- ### **Step 2: Calculate the electrostatic force and solve for θ** #### **Find the distance between the charges:** Each thread is 1. m, angle from vertical is θ, and the charges are separated by \( 2 \times 1. \sin(\theta) \): \[ r = 2 \times 1. \sin(\theta) \] #### **Electrostatic force:** \[ F_e = k \frac{q^2}{r^2} \] Where \( q = 91 \, \text{nC} = 91 \times 10^{-9} \, \text{C} \), \( k = 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 \). \[ F_e = k \frac{q^2}{(2 \sin(\theta))^2} \] #### **Plug into equilibrium equation:** \[ \tan(\theta) = \frac{F_e}{mg} = \frac{k q^2}{(2 \sin(\theta))^2 mg} \] \[ \tan(\theta) = \frac{k q^2}{4 m g \sin^2(\theta)} \] Now, let's plug in values: - \( m = 5. \text{ g} = .005 \, \text{kg} \) - \( g = 9.8 \, \text{m/s}^2 \) \[ \tan(\theta) = \frac{8.99 \times 10^9 \times (91 \times 10^{-9})^2}{4 \times .005 \times 9.8 \times \sin^2(\theta)} \] \[ \tan(\theta) = \frac{8.99 \times 10^9 \times 8.281 \times 10^{-15}}{.196 \times \sin^2(\theta)} \] \[ \tan(\theta) = \frac{7.445 \times 10^{-5}}{.196 \times \sin^2(\theta)} \] \[ \tan(\theta) \cdot \sin^2(\theta) = \frac{7.445 \times 10^{-5}}{.196} \] \[ \tan(\theta) \cdot \sin^2(\theta) = 3.799 \times 10^{-4} \] This equation is transcendental; you’d solve it numerically. Typical values for small angles: try θ ≈ 1°, 2°, etc. For θ ≈ 2°, - \( \tan(2°) = .0349 \) - \( \sin(2°) = .0349 \) - \( (.0349)^2 = .00122 \) - \( .0349 \times .00122 = 4.26 \times 10^{-5} \) (too small, try larger θ) Try θ ≈ 5°: - \( \tan(5°) = .0875 \) - \( \sin(5°) = .0872 \) - \( (.0872)^2 = .0076 \) - \( .0875 \times .0076 = 6.65 \times 10^{-4} \) (close to \( 3.8 \times 10^{-4} \)) Interpolate between 2° and 5°, θ ≈ 3.5–4°. ### **Final Answer (a):** \[ \boxed{\theta \approx 4^\circ} \] --- ## **Part (b): Force on the charge Q** ### **Step 1: Write the force vectors** Let’s place \( Q \) at the top, and two \( 2\, \text{nC} \) charges below, each at a distance of 1 cm and at 60° from each other. The force due to one charge: \[ F = k \frac{Q \cdot q}{r^2} \] Where \( Q = 5.8\, \text{nC} = 5.8 \times 10^{-9} \, \text{C} \), \( q = 2 \times 10^{-9} \, \text{C} \), \( r = .01 \, \text{m} \) Calculate the force from one charge: \[ F_1 = 8.99 \times 10^9 \cdot \frac{5.8 \times 10^{-9} \cdot 2 \times 10^{-9}}{(.01)^2} \] \[ F_1 = 8.99 \times 10^9 \cdot \frac{11.6 \times 10^{-18}}{10^{-4}} \] \[ = 8.99 \times 10^9 \cdot 1.16 \times 10^{-13} \] \[ = 1.042 \times 10^{-3} \, \text{N} \] --- **Explanation:** We calculated the magnitude of the force from one charge on Q, using Coulomb's law. --- ### **Step 2: Find the net force from both charges** The two forces are symmetric and each makes a 60° angle with the vertical. The vertical (y) components: \[ F_y = 2 F_1 \cos(60^\circ) = 2 F_1 \cdot .5 = F_1 \] The horizontal (x) components cancel out. So, the net force upward: \[ F_{\text{net}} = F_1 \] ### **Final Answer (b):** \[ \boxed{F = 1. \times 10^{-3} \, \text{N}} \] (to 2 significant figures)