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A 90 pF capacitor is charged to a potential difference of 74 V, and the charging battery is disconnected. The capacitor is then connected in parallel with a second (initially uncharged) capacitor. If the measured potential difference across the first capacitor drops to 45 V, what is the capacitance of this second capacitor? Number i Units

A 90 pF capacitor is charged to a potential difference of 74 V, and the charging battery is disconnected. The capacitor is then connected in parallel with a second (initially uncharged) capacitor. If the measured potential difference across the first capacitor drops to 45 V, what is the capacitance of this second capacitor? Number i Units

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
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**Capacitance Problem Overview** **Problem Statement:** A 90 pF (picofarad) capacitor is charged to a potential difference of 74 V, and the charging battery is disconnected. The capacitor is then connected in parallel with a second (initially uncharged) capacitor. If the measured potential difference across the first capacitor drops to 45 V, what is the capacitance of this second capacitor? **Diagram and Explanation:** 1. **Initial Conditions:** - Capacitor 1 (C1) = 90 pF - Initial Voltage (V1_initial) = 74 V 2. **Final Conditions:** - Voltage across Capacitor 1 after connecting in parallel (V1_final) = 45 V 3. **Calculation:** - Initial charge on Capacitor 1 (Q1_initial) = C1 * V1_initial - After connection, the total charge is conserved and shared between Capacitor 1 and Capacitor 2 (C2). - The final voltage (V_final) across both capacitors is the same (45 V). **Required:** - Calculate the capacitance of the second capacitor (C2). **Input Fields:** - **Number**: [User Input Field] - **Units**: [Drop-down Selection] **Resources:** - **eTextbook and Media**: [Link to educational resources for further reading and study] **Further Study:** - Relate concepts of charge conservation and parallel capacitors to solve similar capacitance problems.
Transcribed Image Text:**Capacitance Problem Overview** **Problem Statement:** A 90 pF (picofarad) capacitor is charged to a potential difference of 74 V, and the charging battery is disconnected. The capacitor is then connected in parallel with a second (initially uncharged) capacitor. If the measured potential difference across the first capacitor drops to 45 V, what is the capacitance of this second capacitor? **Diagram and Explanation:** 1. **Initial Conditions:** - Capacitor 1 (C1) = 90 pF - Initial Voltage (V1_initial) = 74 V 2. **Final Conditions:** - Voltage across Capacitor 1 after connecting in parallel (V1_final) = 45 V 3. **Calculation:** - Initial charge on Capacitor 1 (Q1_initial) = C1 * V1_initial - After connection, the total charge is conserved and shared between Capacitor 1 and Capacitor 2 (C2). - The final voltage (V_final) across both capacitors is the same (45 V). **Required:** - Calculate the capacitance of the second capacitor (C2). **Input Fields:** - **Number**: [User Input Field] - **Units**: [Drop-down Selection] **Resources:** - **eTextbook and Media**: [Link to educational resources for further reading and study] **Further Study:** - Relate concepts of charge conservation and parallel capacitors to solve similar capacitance problems.
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