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]

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Chapter26: Capacitance And Dielectrics
Section: Chapter Questions
Problem 26.12OQ: (i) Rank the following five capacitors from greatest to smallest capacitance, noting any cases of...
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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

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 |
What is the new electric potential energy stored in the capacitor? I Select J
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.
Transcribed Image Text: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 | What is the new electric potential energy stored in the capacitor? I Select J 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.
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