How Does Inserting a Dielectric Affect Voltage in an Uncharged Capacitor?

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Homework Help Overview

The discussion revolves around the effects of inserting a dielectric into an uncharged capacitor and its impact on voltage. Participants explore the relationship between electric fields, capacitance, and voltage in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the decrease in voltage with the insertion of a dielectric, questioning the role of the electric field and the implications of charge distribution. Some mention the relationship between capacitance and voltage, while others seek clarification on the underlying properties of dielectrics.

Discussion Status

The discussion is active, with participants providing insights and questioning each other's reasoning. Some guidance has been offered regarding the relationship between capacitance and voltage, but multiple interpretations of the dielectric's effects are still being explored.

Contextual Notes

There are indications of assumptions regarding the properties of dielectrics and the conditions of the capacitor, such as it being uncharged and not connected to a battery.

Gear300
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For a capacitor not connected to a battery, why does the voltage decrease with the insertion of a dielectric. Would it be because the net electric field between the conductors is decreased?
 
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you just answered your own question. there is also less room for particles to lose potential over.
 
username2 said:
you just answered your own question. there is also less room for particles to lose potential over.

Gear300, your answer is a lot better than this one. Yes, the dielectric decreases the electric field. What do you mean "less room"?
 
alternative method :
Q=CV
V=Q/C

insert a dielectric , keeping the charge accumulated unchanged. the greater the dielectric(coefficient e), the greater the capacitance and the lower the voltage across

but why. it is related to the properties of dielectric ... and polarization makes the V decreases as well as ... that's another story
 
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