Here are the solutions to the capacitor problems: b) (i) Calculate the effective capacitance of the arrangement. Step 1: Identify the parallel combination. The $3 \text{ µF}$ and $1 \text{ µF}$ capacitors are connected in parallel. The equivalent capacitance for parallel capacitors ($C_p$) is the sum of their individual capacitances. $$ C_p = C_1 + C_2 $$ $$ C_p = 3 \text{ µF} + 1 \text{ µF} $$ $$ C_p = 4 \text{ µF} $$ Step 2: Identify the series combination. The $4 \text{ µF}$ capacitor is in series with the equivalent parallel capacitance ($C_p = 4 \text{ µF}$). The effective capacitance for series capacitors ($C_{eff}$) is given by: $$ \frac{1}{C_{eff}} = \frac{1}{C_a} + \frac{1}{C_b} $$ $$ \frac{1}{C_{eff}} = \frac{1}{4 \text{ µF}} + \frac{1}{4 \text{ µF}} $$ $$ \frac{1}{C_{eff}} = \frac{2}{4 \text{ µF}} $$ $$ \frac{1}{C_{eff}} = \frac{1}{2 \text{ µF}} $$ $$ C_{eff} = \boxed{\text{2 µF}} $$ b) (ii) Calculate the charge on the 3 µF capacitor. Step 1: Calculate the total charge stored in the circuit. The total charge ($Q_{total}$) is stored by the effective capacitance ($C_{eff}$) at the total voltage ($V_{total}$). $$ Q_{total} = C_{eff} \times V_{total} $$ $$ Q_{total} = 2 \text{ µF} \times 12 \text{ V} $$ $$ Q_{total} = 24 \text{ µC} $$ Step 2: Determine the charge and voltage for the series components. In a series circuit, the charge on each component is the same as the total charge. So, the charge on the $4 \text{ µF}$ capacitor is $24 \text{ µC}$, and the charge on the parallel combination ($C_p = 4 \text{ µF}$) is also $24 \text{ µC}$. Step 3: Calculate the voltage across the parallel combination. The voltage across the parallel combination ($V_p$) is: $$ V_p = \frac{Q_p}{C_p} $$ $$ V_p = \frac{24 \text{ µC}}{4 \text{ µF}} $$ $$ V_p = 6 \text{ V} $$ Step 4: Calculate the charge on the 3 µF capacitor. Since the $3 \text{ µF}$ capacitor is part of the parallel combination, the voltage across it is $V_p = 6 \text{ V}$. $$ Q_{3\mu F} = C_{3\mu F} \times V_p $$ $$ Q_{3\mu F} = 3 \text{ µF} \times 6 \text{ V} $$ $$ Q_{3\mu F} = \boxed{\text{18 µC}} $$ c) Find the separation distance between two plates if the capacitance between them is $8.0 \times 10^{-12} \text{ F}$ and the area enclosed is $2.0 \text{ cm}^2$ (take $\epsilon_o = 8.85 \times 10^{-12}$). Step 1: State the formula for capacitance of a parallel plate capacitor. $$ C = \frac{\epsilon_o A}{d} $$ Where $C$ is capacitance, $\epsilon_o$ is the permittivity of free space, $A$ is the area of the plates, and $d$ is the separation distance. Step 2: Convert the given area to square meters. $$ A = 2.0 \text{ cm}^2 = 2.0 \times (10^{-2} \text{ m})^2 = 2.0 \times 10^{-4} \text{ m}^2 $$ Step 3: Rearrange the formula to solve for $d$. $$ d = \frac{\epsilon_o A}{C} $$ Step 4: Substitute the given values and calculate $d$. $$ d = \frac{(8.85 \times 10^{-12} \text{ F/m}) \times (2.0 \times 10^{-4} \text{ m}^2)}{8.0 \times 10^{-12} \text{ F}} $$ $$ d = \frac{17.7 \times 10^{-16}}{8.0 \times 10^{-12}} \text{ m} $$ $$ d = 2.2125 \times 10^{-4} \text{ m} $$ Rounding to three significant figures: $$ d = \boxed{\text{2.21} \times 10^{-4} \text{ m}} $$ Last one today — make it count tomorrow, or type /upgrade for unlimited.