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An electroscope has a sharp pin attached to its top. Part (b) asks for observations and explanations of what happens when the electroscope is charged. Part (c) involves calculating the charge on a cap

Physics

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Answer

c(i)

2.7 μC\text{2.7 } \mu\text{C}

Step-by-step solution

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Part 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.

Part c(i)

2.7 μC\text{2.7 } \mu\text{C}

Determine the charge on C1C_1 when both switch A is closed and switch B is open.

Given:
Battery e.m.f., V=4.5 VV = 4.5 \text{ V}
Capacitance of C1C_1, C1=0.6μFC_1 = 0.6 \mu\text{F}
Capacitance of C2C_2, C2=1.0μFC_2 = 1.0 \mu\text{F}

  1. Analyze the circuit configuration.

    When switch A is closed and switch B is open, capacitor C1C_1 is connected directly across the battery. Capacitor C2C_2 is not connected to the circuit. Therefore, the voltage across C1C_1 is equal to the battery's e.m.f.

    V1=V=4.5 VV_1 = V = 4.5 \text{ V}

  2. Use the formula for charge on a capacitor, Q=CVQ = CV.

    Q1=C1V1Q_1 = C_1 V_1

  3. Substitute the values and calculate the charge.

    C1=0.6μF=0.6×10−6 FC_1 = 0.6 \mu\text{F} = 0.6 \times 10^{-6} \text{ F} Q1=(0.6×10−6 F)×(4.5 V)Q_1 = (0.6 \times 10^{-6} \text{ F}) \times (4.5 \text{ V}) Q1=2.7×10−6 CQ_1 = 2.7 \times 10^{-6} \text{ C} Q1=2.7μCQ_1 = 2.7 \mu\text{C}

    The charge on C1C_1 is 2.7 μC\boxed{\text{2.7 } \mu\text{C}}.

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