This computer science problem involves algorithmic thinking and programming concepts. The solution below explains the approach, logic, and implementation step by step.
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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 \, C_2 = 2 \, C_4 = 4 \, C_5 = 5 \, V = 8.0 \, 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 \, + 4 \, = 5 \, Step 2: Calculate the equivalent capacitance of the 5 µF capacitor in series with C_14. (1)/(C_5-14) = (1)/(C_5) + (1)/(C_14) (1)/(C_5-14) = (1)/(5 \, ) + (1)/(5 \, ) = (2)/(5 \, ) C_5-14 = (5)/(2) \, = 2.5 \, 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 \, + 2 \, = 4.5 \, The total capacitance is 4.5 µF. (ii) The total charge (Q_total): The total charge is given by Q_total = C_total × V. Q_total = 4.5 \, × 8.0 \, V Q_total = 4.5 × 10^-6 \, F × 8.0 \, V Q_total = 36 × 10^-6 \, C = 36 \, The total charge is 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 \, V. The charge in this series branch is Q_5-14 = C_5-14 × V. Q_5-14 = 2.5 \, × 8.0 \, V = 20 \, This charge is the same for both C_5 and the parallel combination C_14. So, Q_14 = 20 \, . Now, find the voltage across C_14: V_14 = Q_14C_14 V_14 = 20 \, 5 \, = 4.0 \, 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 \, V The voltage across the 4µF capacitor is 4.0 V. (iv) Energy stored in 4µF (E_4): The energy stored in a capacitor is given by E = (1)/(2) C V^2. Using C_4 = 4 \, and V_4 = 4.0 \, V: E_4 = (1)/(2) C_4 V_4^2 E_4 = (1)/(2) (4 × 10^-6 \, F) (4.0 \, V)^2 E_4 = (1)/(2) (4 × 10^-6) (16) E_4 = 2 × 10^-6 × 16 E_4 = 32 × 10^-6 \, J = 32 \, The energy stored in the 4µF capacitor is 32 µJ. That's 2 down. 3 left today — send the next one.