18. a) When a negatively charged electroscope is touched with a finger, the excess electrons on the electroscope are repelled by each other and flow through the human body to the Earth. This process, called earthing or grounding, neutralizes the electroscope, causing its leaves to collapse. 18. b) Given values: $C_1 = 1 \, \mu\text{F}$ $C_2 = 2 \, \mu\text{F}$ $C_4 = 4 \, \mu\text{F}$ $C_5 = 5 \, \mu\text{F}$ $V = 8.0 \, \text{V}$ Circuit Analysis: The 1 µF and 4 µF capacitors are in parallel. This combination is in series with the 5 µF capacitor. This entire series-parallel branch is in parallel with the 2 µF capacitor. The total combination is connected to an 8.0 V battery. (i) The total capacitance ($C_{total}$): Step 1: Calculate the equivalent capacitance of the 1 µF and 4 µF capacitors in parallel. $$C_{14} = C_1 + C_4$$ $$C_{14} = 1 \, \mu\text{F} + 4 \, \mu\text{F} = 5 \, \mu\text{F}$$ Step 2: Calculate the equivalent capacitance of the 5 µF capacitor in series with $C_{14}$. $$\frac{1}{C_{5-14}} = \frac{1}{C_5} + \frac{1}{C_{14}}$$ $$\frac{1}{C_{5-14}} = \frac{1}{5 \, \mu\text{F}} + \frac{1}{5 \, \mu\text{F}} = \frac{2}{5 \, \mu\text{F}}$$ $$C_{5-14} = \frac{5}{2} \, \mu\text{F} = 2.5 \, \mu\text{F}$$ Step 3: Calculate the total capacitance by adding $C_{5-14}$ in parallel with the 2 µF capacitor. $$C_{total} = C_{5-14} + C_2$$ $$C_{total} = 2.5 \, \mu\text{F} + 2 \, \mu\text{F} = 4.5 \, \mu\text{F}$$ The total capacitance is $\boxed{\text{4.5 µF}}$. (ii) The total charge ($Q_{total}$): The total charge is given by $Q_{total} = C_{total} \times V$. $$Q_{total} = 4.5 \, \mu\text{F} \times 8.0 \, \text{V}$$ $$Q_{total} = 4.5 \times 10^{-6} \, \text{F} \times 8.0 \, \text{V}$$ $$Q_{total} = 36 \times 10^{-6} \, \text{C} = 36 \, \mu\text{C}$$ The total charge is $\boxed{\text{36 µC}}$. (iii) Voltage across 4µF ($V_4$): The voltage across the parallel combination ($C_{5-14}$) is equal to the battery voltage because it is in parallel with the 2 µF capacitor, and this entire parallel arrangement is connected directly across the battery. So, the voltage across the series branch containing $C_5$ and $C_{14}$ is $V = 8.0 \, \text{V}$. The charge in this series branch is $Q_{5-14} = C_{5-14} \times V$. $$Q_{5-14} = 2.5 \, \mu\text{F} \times 8.0 \, \text{V} = 20 \, \mu\text{C}$$ This charge is the same for both $C_5$ and the parallel combination $C_{14}$. So, $Q_{14} = 20 \, \mu\text{C}$. Now, find the voltage across $C_{14}$: $$V_{14} = \frac{Q_{14}}{C_{14}}$$ $$V_{14} = \frac{20 \, \mu\text{C}}{5 \, \mu\text{F}} = 4.0 \, \text{V}$$ Since the 1 µF and 4 µF capacitors are in parallel, the voltage across them is the same as $V_{14}$. Therefore, the voltage across the 4 µF capacitor is $V_4 = V_{14}$. $$V_4 = 4.0 \, \text{V}$$ The voltage across the 4µF capacitor is $\boxed{\text{4.0 V}}$. (iv) Energy stored in 4µF ($E_4$): The energy stored in a capacitor is given by $E = \frac{1}{2} C V^2$. Using $C_4 = 4 \, \mu\text{F}$ and $V_4 = 4.0 \, \text{V}$: $$E_4 = \frac{1}{2} C_4 V_4^2$$ $$E_4 = \frac{1}{2} (4 \times 10^{-6} \, \text{F}) (4.0 \, \text{V})^2$$ $$E_4 = \frac{1}{2} (4 \times 10^{-6}) (16)$$ $$E_4 = 2 \times 10^{-6} \times 16$$ $$E_4 = 32 \times 10^{-6} \, \text{J} = 32 \, \mu\text{J}$$ The energy stored in the 4µF capacitor is $\boxed{\text{32 µJ}}$. That's 2 down. 3 left today — send the next one.