Let's solve each part step by step: --- ## (a) Separation Distance \( d \) Between the Plates The capacitance of a parallel-plate capacitor is given by: \[ C = \epsilon_ \frac{A}{d} \] Where: - \( C = 4.4 \times 10^{-12} \) F (capacitance) - \( A = .025 \) m² (area) - \( \epsilon_ = 8.854 \times 10^{-12} \) F/m (perm. of free space) - \( d = ? \) (separation) **Rearrange for \( d \):** \[ d = \epsilon_ \frac{A}{C} \] **Plug in values:** \[ d = \frac{8.854 \times 10^{-12} \times .025}{4.4 \times 10^{-12}} \] \[ d = \frac{2.2135 \times 10^{-13}}{4.4 \times 10^{-12}} \] \[ d = .0503 \text{ m} = 5.03 \text{ mm} \] --- ## (b) Magnitude of the Electric Field \( |\vec{E}_| \) The electric field between the plates of a parallel-plate capacitor is: \[ E_ = \frac{V}{d} \] Where: - \( V = 12 \) V (battery voltage) - \( d = .0503 \) m (from part a) **Plug in values:** \[ E_ = \frac{12}{.0503} \] \[ E_ \approx 238.6 \text{ V/m} \] --- ## (c) Magnitude of Charge on Each Plate The charge on each plate: \[ Q = CV \] Where: - \( C = 4.4 \times 10^{-12} \) F - \( V = 12 \) V **Plug in values:** \[ Q = 4.4 \times 10^{-12} \times 12 \] \[ Q = 5.28 \times 10^{-11} \text{ C} \] --- ## (d) Energy Stored in the Capacitor The energy stored is: \[ U = \frac{1}{2} C V^2 \] **Plug in values:** \[ U = \frac{1}{2} \times 4.4 \times 10^{-12} \times (12)^2 \] \[ U = 2.2 \times 10^{-12} \times 144 \] \[ U = 3.168 \times 10^{-10} \text{ J} \] --- # **Summary of Answers** **(a)** \( d = \boxed{.0503\ \text{m} = 5.03\ \text{mm}} \) **(b)** \( |\vec{E}_| = \boxed{238.6\ \text{V/m}} \) **(c)** \( Q = \boxed{5.28 \times 10^{-11}\ \text{C}} \) **(d)** \( U = \boxed{3.17 \times 10^{-10}\ \text{J}} \)