First box choices are 4, 3, 6, 1, 2 (V) the second box choices are 1, 8, 3, 1.5, 6 (nJ) and the third box pick from I-V **Educational Content: Capacitors and Dielectrics**A fully charged parallel-plate capacitor with air between the two plates has a charge of $6.0 \, \text{nC}$ and is disconnected from the battery after charging. Its capacitance is $3.0 \, \text{nF}$. Now an insulating material with a dielectric constant $\kappa = 2$ is inserted into the capacitor and occupies the whole space between the two plates.**Questions:**1. What is the new voltage across the two plates after the insertion? [Select]2. What is the new electric potential energy stored in the capacitor? [Select]**In the process of inserting the dielectric material, which one of the following statements regarding the stored potential energy change is correct? [Select]**- **I.** There is no change of the electric potential energy in the capacitor due to energy conservation. - **II.** The electric potential energy doubles, since additional potential energy (the same as the original potential energy) is gained with the negative work being done on the dielectric material by an external pushing force other than the electric force during the insertion process.- **III.** The electric potential energy doubles, since additional potential energy (the same as the original potential energy) is gained with the positive work being done on the dielectric material by an external pushing force other than the electric force during the insertion process. - **IV.** The electric potential energy halves, since half of the original potential energy is used to do negative work by the electric force on the dielectric material during the insertion process. - **V.** The electric potential energy halves, since half of the original potential energy is used to do positive work on the dielectric material during the insertion process.This exercise explores the effect of inserting a dielectric in a capacitor on both the voltage and the stored potential energy, emphasizing concepts like dielectric constants and energy conservation laws in capacitors.

Required : New potential across the plates.

A fully charged parallel-plate capacitor with air between the two plates has charge 6.0 nC and is disconnected from the battery after charging. Its capacitance is 3.0 nF. Now an insulating material with dielectric constant K 2 is inserted into the capacitor and occupies the whole space between the two plates. What is the new voltage across the two plates after the insertion? [Select]

A fully charged parallel-plate capacitor with air between the two plates has charge 6.0 nC and is disconnected from the battery after charging. Its capacitance is 3.0 nF. Now an insulating material with dielectric constant K 2 is inserted into the capacitor and occupies the whole space between the two plates. What is the new voltage across the two plates after the insertion? [Select]

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

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

First box choices are 4, 3, 6, 1, 2 (V) the second box choices are 1, 8, 3, 1.5, 6 (nJ) and the third box pick from I-V

**Educational Content: Capacitors and Dielectrics**

A fully charged parallel-plate capacitor with air between the two plates has a charge of $6.0 \, \text{nC}$ and is disconnected from the battery after charging. Its capacitance is $3.0 \, \text{nF}$. Now an insulating material with a dielectric constant $\kappa = 2$ is inserted into the capacitor and occupies the whole space between the two plates.

**Questions:**

1. What is the new voltage across the two plates after the insertion? [Select]

2. What is the new electric potential energy stored in the capacitor? [Select]

**In the process of inserting the dielectric material, which one of the following statements regarding the stored potential energy change is correct? [Select]**

- **I.** There is no change of the electric potential energy in the capacitor due to energy conservation.

- **II.** The electric potential energy doubles, since additional potential energy (the same as the original potential energy) is gained with the negative work being done on the dielectric material by an external pushing force other than the electric force during the insertion process.

- **III.** The electric potential energy doubles, since additional potential energy (the same as the original potential energy) is gained with the positive work being done on the dielectric material by an external pushing force other than the electric force during the insertion process.

- **IV.** The electric potential energy halves, since half of the original potential energy is used to do negative work by the electric force on the dielectric material during the insertion process.

- **V.** The electric potential energy halves, since half of the original potential energy is used to do positive work on the dielectric material during the insertion process.

This exercise explores the effect of inserting a dielectric in a capacitor on both the voltage and the stored potential energy, emphasizing concepts like dielectric constants and energy conservation laws in capacitors.

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