Let's walk through each question step-by-step: --- ## **Given Data** - Plate area, \(A = 13. \, \text{cm}^2 = 13. \times 10^{-4} \, \text{m}^2\) - Plate separation, \(d = 1.50 \, \text{cm} = .015 \, \text{m}\) - Mass of bead, \(m = 1.50 \, \text{mg} = 1.50 \times 10^{-6} \, \text{kg}\) - Charge of bead, \(q = 6.50 \times 10^{-9} \, \text{C}\) - \(g = 9.8 \, \text{m/s}^2\) --- ## **Q1: What is the weight of the bead?** \[ W = mg = (1.5 \times 10^{-6} \, \text{kg}) \times (9.8 \, \text{m/s}^2) = 1.47 \times 10^{-5} \, \text{N} \] **Final answer:** \(1.47 \times 10^{-5} \, \text{N}\) --- ## **Q2: What is the magnitude of the electric force acting on the bead?** Since the bead is suspended and at rest, the electric force must balance the weight: \[ F_E = W = 1.47 \times 10^{-5} \, \text{N} \] **Final answer:** \(1.47 \times 10^{-5} \, \text{N}\) --- ## **Q3: What is the magnitude of the electric field between the plates of the capacitor?** \[ F_E = qE \implies E = \frac{F_E}{q} \] \[ E = \frac{1.47 \times 10^{-5} \, \text{N}}{6.50 \times 10^{-9} \, \text{C}} = 2.26 \times 10^{3} \, \text{N/C} \] **Final answer:** \(2.26 \times 10^{3} \, \text{N/C}\) --- ## **Q4: What is the magnitude of the charge on each plate?** The electric field between parallel plates is: \[ E = \frac{\sigma}{\epsilon_} \] where \(\sigma = \frac{Q}{A}\), so: \[ E = \frac{Q}{A \epsilon_} \implies Q = E A \epsilon_ \] With \(\epsilon_ = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2\): \[ Q = (2.26 \times 10^3) \times (13. \times 10^{-4}) \times (8.85 \times 10^{-12}) \] \[ Q = 2.26 \times 10^3 \times 1.3 \times 10^{-3} \times 8.85 \times 10^{-12} \] \[ Q = 2.938 \times 10^{-9} \, \text{C} \] **Final answer:** \(2.94 \times 10^{-9} \, \text{C}\) (rounded to 3 significant figures) --- ## **Q5: In what direction do electric field lines between the plates point? Up or Down?** Since the force must oppose gravity (upwards), and the bead is **positively charged**, the electric field must be **upward**. **Final answer:** **Up** --- ## **Q6: Which plate of the capacitor is positively charged? Top or bottom?** Electric field goes from positive to negative plate. Since the field is upward, the **bottom plate is positive**. **Final answer:** **Bottom** --- ## **Summary Table** | Question | Answer | |----------|---------------------------------------| | Q1 | \(1.47 \times 10^{-5}\) N | | Q2 | \(1.47 \times 10^{-5}\) N | | Q3 | \(2.26 \times 10^{3}\) N/C | | Q4 | \(2.94 \times 10^{-9}\) C | | Q5 | Up | | Q6 | Bottom | Let me know if you need further clarifications!