What Have I Learned So Far? 1. Solve for the capacitance of a capacitor if it stores a charge of \( 1.3 \times 10^{4} \mathrm{C} \) when delivered with a voltage of \( 3 \times 10^{3} \mathrm{~V} \). 2. How much charge can a \( 4.00-\mu \mathrm{F} \) capacitor store on its plates when it is connected to a 12.0 - V battery? 3. A capacitor with air between its plates has a capacitance of \( 5 \mu \mathrm{~F} \). What is its capacitance when dry wax with a dielectric constant of 2.7 is placed between the plates? Compute for the charge on each plate of a \( 0.07 \mu \mathrm{~F} \) capacitor when the potential difference between the plates is 210 V . A capacitor is charged with 8.79 nC and has a \( 150-\mathrm{V} \) potential difference between its terminals. What is the capacitance offered by the capacitor? 4.

Sure! Let's tackle each of these capacitor-related problems step by step. --- ### **1. Calculating Capacitance** **Problem:** Solve for the capacitance of a capacitor that stores a charge of \( 1.3 \times 10^{4} \, \mathrm{C} \) when a voltage of \( 3 \times 10^{3} \, \mathrm{V} \) is applied. **Solution:** The relationship between charge (\( Q \)), capacitance (\( C \)), and voltage (\( V \)) is given by: \[ C = \frac{Q}{V} \] Plugging in the given values: \[ C = \frac{1.3 \times 10^{4} \, \mathrm{C}}{3 \times 10^{3} \, \mathrm{V}} = \frac{1.3 \times 10^{4}}{3 \times 10^{3}} \, \mathrm{F} = 4.333... \, \mathrm{F} \] **Answer:** The capacitance is approximately **4.33 farads (F)**. --- ### **2. Determining Stored Charge** **Problem:** How much charge can a \( 4.00 \, \mu\mathrm{F} \) capacitor store on its plates when connected to a 12.0 V battery? **Solution:** Using the formula: \[ Q = C \times V \] Where: - \( C = 4.00 \times 10^{-6} \, \mathrm{F} \) - \( V = 12.0 \, \mathrm{V} \) \[ Q = 4.00 \times 10^{-6} \, \mathrm{F} \times 12.0 \, \mathrm{V} = 4.8 \times 10^{-5} \, \mathrm{C} \] **Answer:** The capacitor stores **4.8 × 10⁻⁵ coulombs (48 µC)** of charge. --- ### **3. Capacitance with a Dielectric** **Problem:** A capacitor with air between its plates has a capacitance of \( 5 \, \mu\mathrm{F} \). What is its capacitance when dry wax with a dielectric constant of 2.7 is placed between the plates? **Solution:** The capacitance with a dielectric (\( C' \)) is related to the original capacitance (\( C \)) by the dielectric constant (\( \kappa \)): \[ C' = \kappa \times C \] Given: - \( \kappa = 2.7 \) - \( C = 5 \times 10^{-6} \, \mathrm{F} \) \[ C' = 2.7 \times 5 \times 10^{-6} \, \mathrm{F} = 13.5 \times 10^{-6} \, \mathrm{F} = 13.5 \, \mu\mathrm{F} \] **Answer:** The new capacitance is **13.5 µF**. --- ### **4. Computing Charge on a Capacitor** **Problem:** Compute the charge on each plate of a \( 0.07 \, \mu\mathrm{F} \) capacitor when the potential difference between the plates is 210 V. **Solution:** Using the formula: \[ Q = C \times V \] Where: - \( C = 0.07 \times 10^{-6} \, \mathrm{F} \) - \( V = 210 \, \mathrm{V} \) \[ Q = 0.07 \times 10^{-6} \, \mathrm{F} \times 210 \, \mathrm{V} = 14.7 \times 10^{-6} \, \mathrm{C} = 1.47 \times 10^{-5} \, \mathrm{C} \] **Answer:** Each plate holds **1.47 × 10⁻⁵ coulombs (14.7 µC)** of charge. --- ### **5. Determining Capacitance from Charge and Voltage** **Problem:** A capacitor is charged with 8.79 nC and has a 150 V potential difference between its terminals. What is the capacitance offered by the capacitor? **Solution:** Using the formula: \[ C = \frac{Q}{V} \] Where: - \( Q = 8.79 \times 10^{-9} \, \mathrm{C} \) - \( V = 150 \, \mathrm{V} \) \[ C = \frac{8.79 \times 10^{-9} \, \mathrm{C}}{150 \, \mathrm{V}} = 5.86 \times 10^{-11} \, \mathrm{F} = 58.6 \, \mathrm{pF} \] **Answer:** The capacitance is approximately **58.6 picofarads (pF)**. --- If you have any further questions or need clarification on any of these problems, feel free to ask!