Dielectric inserted into a parallel plate capacitor

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SUMMARY

The discussion focuses on the effect of inserting a dielectric into a parallel plate capacitor that is disconnected from a voltage source. The new potential between the plates is given by the formula Vf = V/d(d - t + t/Er), where V is the initial voltage, d is the thickness of the capacitor, t is the thickness of the dielectric, and Er is the relative permittivity of the dielectric. Participants express confusion regarding the assumption of a constant electric field and the implications of changing capacitance on voltage. The key takeaway is that while charge remains constant, the introduction of the dielectric alters the capacitance and thus the voltage across the capacitor.

PREREQUISITES
  • Understanding of parallel plate capacitors
  • Knowledge of dielectric materials and their properties
  • Familiarity with electric fields and potential difference
  • Basic grasp of capacitance formulas, specifically C = Q / V
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  • Learn about the relationship between electric field strength and voltage in capacitors
  • Investigate the concept of charge conservation in isolated systems
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Homework Statement


The capacitor (of thickness d) is disconnected from a potential source of V and a dielectric of thickness t is inserted and it has relative permitivity Er. Find the new potential between the plates

Homework Equations


[/B]
This is the answer : Vf = V/d(d - t + t/Er)

The Attempt at a Solution



I don't get why this is it looks like they assumed that the electric field would remain constant and just added the potential drops. We know electric field isn't constant so what give?
 
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The charge is isolated and so is the only thing that remains constant.
The voltage changes as a result of the changing capacitance. C = Q / V.
 

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