What Happens to Voltage When a Dielectric is Inserted into a Charged Capacitor?

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Homework Help Overview

The discussion revolves around a problem involving a parallel plate capacitor that has been disconnected from a battery and has a dielectric inserted between its plates. The subject area includes concepts related to capacitance, charge, and voltage in the context of dielectrics.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the new voltage after inserting a dielectric but questions their previous calculation. Some participants suggest considering the dielectric constant in their reasoning, while others seek clarification on the complete problem statement.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the dielectric constant on voltage. There is a request for more information to clarify the problem, indicating that the discussion is still in a formative stage.

Contextual Notes

Participants note that parts of the original question are missing, which may be contributing to confusion in the calculations and interpretations being discussed.

whoknows12345
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A very large parallel plate capacitor has two plates, each with an area of 9m^2, and a separation of 3nm between them.

If the capacitor was then disconnected from the battery (12V), and then a dielectric of k=4 inserted between the plates, fnd the new values for the following (after system has reached equilibrium)

C= 0.1062 F
Q= .3186 C
V= ____ V

I did V=Q/C = 3V, but I got it wrong. I don't see what I did wrong really.
 
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Take the dielectric constant of k=4 into consideration.
 
so it would be 3v/4? and that would give me the answer?
 
It would help if you could post the exact text of the question. You've left some parts out, and that is probably where the error happens. Can you please post the whole question verbatim?
 

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