Confused on this dielectric capacitance problem

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Homework Help Overview

The discussion revolves around a problem involving the capacitance of a capacitor with a dielectric material (paper with a dielectric constant K=3.75). The original poster presents the electric field, plate separation, and charge on the plates, seeking guidance on how to determine both the capacitance and the area of the plates.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between electric field, charge, area, and capacitance. The original poster considers using the formula C=Q/V but expresses uncertainty about its applicability to dielectrics. Others suggest using the equation E = Q/(ε₀A) to find the area, but the original poster reports difficulties with this approach.

Discussion Status

The discussion is ongoing, with participants exploring different equations and questioning the assumptions about the applicability of capacitance formulas in the context of dielectrics. There is no clear consensus yet, as some participants are still seeking alternative suggestions.

Contextual Notes

The problem involves specific constraints such as the known electric field, plate separation, and charge, but there is a lack of clarity on how to connect these to find the required capacitance and area. The original poster expresses confusion about the definitions and relationships involved.

supermenscher
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The electric field between the plates of a paper-separated (K=3.75) capacitor is 9.41E4 V/m. The plates are 1.75mm apart and the charge on each plate is 0.775E-6C. Determine the capacitance of the capacitor and the area of each plate?

I got the voltage by V=Ed. and that came out to 164.675

Where do i go from there because in order to get capacitance you need to know area, and finding the area of each plate is also a question in the problem. I thought about using C=Q/V, but that doesn't apply for dielectrics. Can someone helpme out?
 
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I think you should be able to use the fact that [itex]E = \frac{Q}{\epsilon_0A}[/itex]. This gives you an equation for the area which you can use to determine the capacitance.
 
I tried that and that did not work out either, do you have anymore suggestions?
 
supermenscher said:
Where do i go from there because in order to get capacitance you need to know area, and finding the area of each plate is also a question in the problem. I thought about using C=Q/V, but that doesn't apply for dielectrics.
Who says that C = Q/V doesn't apply for dielectrics? It's the definition of capacitance.
 

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