This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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b) Observation: The leaves of the electroscope will gradually collapse. Explanation: A sharp pin on a highly charged electroscope creates a very strong electric field at its tip. This strong field ionizes the air molecules around the pin. If the electroscope is positively charged, it attracts electrons from the air, neutralizing its charge. If it is negatively charged, it repels electrons from the air, causing positive ions to be attracted to the pin, also neutralizing its charge. This process, known as corona discharge or point discharge, causes the electroscope to lose its charge to the surrounding air over time, leading to the collapse of its leaves. c) Given: Battery e.m.f., V = 4.5 V Capacitance of C_1, C_1 = 0.6 Capacitance of C_2, C_2 = 1.0 i) Determine the charge on C_1 when both switch A is closed and switch B is open. Step 1: Analyze the circuit configuration. When switch A is closed and switch B is open, capacitor C_1 is connected directly across the battery. Capacitor C_2 is not connected to the circuit. Therefore, the voltage across C_1 is equal to the battery's e.m.f. V_1 = V = 4.5 V Step 2: Use the formula for charge on a capacitor, Q = CV. Q_1 = C_1 V_1 Step 3: Substitute the values and calculate the charge. C_1 = 0.6 = 0.6 × 10^-6 F Q_1 = (0.6 × 10^-6 F) × (4.5 V) Q_1 = 2.7 × 10^-6 C Q_1 = 2.7 The charge on C_1 is 2.7 . That's 2 down. 3 left today — send the next one.