Let's solve each question step by step, **with explanations after each step**. --- ## a) Find the capacitance of the air-filled capacitor ### **Step 1: Write the formula for the capacitance of a parallel-plate capacitor** \[ C = \epsilon_ \frac{A}{d} \] Where: - \(\epsilon_ = 8.85 \times 10^{-12}~\text{F/m}\) (permittivity of free space) - \(A = .020~\text{m}^2\) - \(d = 3.~\text{mm} = 3. \times 10^{-3}~\text{m}\) --- ### **Step 2: Substitute values and calculate** \[ C = (8.85 \times 10^{-12}) \frac{.020}{3. \times 10^{-3}} \] \[ C = (8.85 \times 10^{-12}) \times \frac{.020}{.003} \] \[ C = (8.85 \times 10^{-12}) \times 6.67 \] \[ C \approx 5.9 \times 10^{-11}~\text{F} \] \[ C \approx 59~\text{pF} \] **Final Answer (a):** \[ \boxed{C = 59~\text{pF}} \] --- Let me know if you'd like to proceed with part (b)!

Let's solve all parts step by step, providing explanations after each step. --- ## a) Find the capacitance of the air-filled capacitor ### **Step 1: Write the formula for the capacitance of a parallel-plate capacitor** \[ C = \epsilon_0 \frac{A}{d} \] Where: - \(\epsilon_0 = 8.85 \times 10^{-12}~\text{F/m}\) (permittivity of free space) - \(A = 0.020~\text{m}^2\) - \(d = 3.0~\text{mm} = 3.0 \times 10^{-3}~\text{m}\) ### **Step 2: Substitute values and calculate** \[ C = (8.85 \times 10^{-12}) \frac{0.020}{3.0 \times 10^{-3}} \] \[ C = (8.85 \times 10^{-12}) \times \frac{0.020}{0.003} \] \[ C = (8.85 \times 10^{-12}) \times 6.67 \] \[ C \approx 5.9 \times 10^{-11}~\text{F} \] \[ C \approx 59~\text{pF} \] **Final Answer (a):** \[ \boxed{C = 59~\text{pF}} \] --- ## b) Determine the time constant of the circuit ### **Step 1: Write the formula for the time constant** The time constant \(\tau\) for an RC circuit is given by: \[ \tau = R \cdot C \] Where: - \(R = 150~\Omega\) - \(C = 5.9 \times 10^{-11}~\text{F}\) ### **Step 2: Substitute values and calculate** \[ \tau = 150 \cdot (5.9 \times 10^{-11}) \] \[ \tau \approx 8.85 \times 10^{-9}~\text{s} \] **Final Answer (b):** \[ \boxed{\tau \approx 8.85~\text{ns}} \] --- ## c) Determine how long it will take the capacitor to fully charge ### **Step 1: Understand the charging process** A capacitor charges to approximately 63.2% of the full voltage after one time constant \(\tau\). It takes about \(5\tau\) to effectively reach full charge. ### **Step 2: Calculate the time for effective charge** \[ t \approx 5 \cdot \tau = 5 \cdot (8.85 \times 10^{-9}) \approx 4.43 \times 10^{-8}~\text{s} \] **Final Answer (c):** \[ \boxed{t \approx 44.3~\text{ns}} \] --- ## d) What is the charge on the capacitor once fully charged? ### **Step 1: Write the formula for charge** The charge \(Q\) on a capacitor is given by: \[ Q = C \cdot V \] Where: - \(V = 12.0~\text{V}\) (voltage from the power supply) - \(C = 5.9 \times 10^{-11}~\text{F}\) ### **Step 2: Substitute values and calculate** \[ Q = (5.9 \times 10^{-11}) \cdot 12.0 \] \[ Q \approx 7.08 \times 10^{-10}~\text{C} \] **Final Answer (d):** \[ \boxed{Q \approx 708~\text{pC}} \] --- ## e) A dielectric is now inserted into the capacitor (k=3), what happens to the voltage of the capacitor? ### **Step 1: Understand the effect of a dielectric** When a dielectric material is inserted, the capacitance increases by a factor of \(k\): \[ C' = k \cdot C \] The voltage across the capacitor is given by: \[ V' = \frac{Q}{C'} \] ### **Step 2: Calculate the new capacitance and voltage** New capacitance: \[ C' = 3 \cdot (5.9 \times 10^{-11}) = 1.77 \times 10^{-10}~\text{F} \] Using the previously calculated charge: \[ V' = \frac{Q}{C'} = \frac{7.08 \times 10^{-10}}{1.77 \times 10^{-10}} \approx 4.0~\text{V} \] **Final Answer (e):** The voltage decreases to approximately 4.0 V. \[ \boxed{V' \approx 4.0~\text{V}} \]