Question Relating Electric Field and Capacitance

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Homework Help Overview

The discussion revolves around a problem involving a parallel-plate capacitor, specifically focusing on the relationship between electric field strength, capacitance, and charge. Participants are examining the calculations related to the charge required to achieve a specified electric field between the capacitor plates.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formula for charge in relation to electric field and capacitance, questioning why one participant consistently arrives at half of the expected answer. There are suggestions that the issue may stem from considering the electric field produced by only one plate or potential misunderstandings regarding dielectric materials.

Discussion Status

The conversation is ongoing, with participants offering insights and hypotheses about the calculations. Some guidance has been provided regarding the potential influence of dielectric constants, but no consensus has been reached on the exact cause of the discrepancy in the calculations.

Contextual Notes

There is a mention of the possibility that the capacitor may contain a dielectric, which could affect the calculations, though the original poster confirms that no dielectric constant was provided in the problem statement.

rabcdred
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Homework Statement


A parallel-plate capacitor has a plate area of 0.2 m2 and a plate separation of 0.1 mm. To obtain an electric field of 2.0 × 10^6 V/m between the plates, the magnitude of the charge on each plate should be


Homework Equations



Q=epsilon(E)(A)

The Attempt at a Solution


I keep getting half of the correct answer. Does anyone know why I am off a factor of 2? Cheers.
 
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Q=CV
C=epsilon*Area/separation
 
rabcdred said:
I keep getting half of the correct answer. Does anyone know why I am off a factor of 2?

No, without seeing your solution.

ehild
 
My guess would be that you are only considering the electric field caused by one of the plates. ?
 
Without seeing your solution, my guess is that the correct answer is twice what your answer is.
 
Sorry about that.
Q=CV, C=(epsilon)A/d, V=Ed/ By substituting these last 2 equations into the first, the result is Q=(epsilon)A(E)-->8.85e-12(0.2)(2e6)= 3.54e-6. However, this is half of the correct answer. What am I doing wrong?
 
That is correct with the given data. But it is possible that the volume between the plates was filled by a dielectric, with dielectric constant of 2. Are you sure that the dielectric constant was not given?

ehild
 
Yes. The question I gave you was a copy paste from the document.
 
I have no more idea. It happens quite often that wrong solutions are given. ehild
 

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