Plate Capacitor: Find Voltage with 4\muC & 0.4m2 Area

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Homework Help Overview

The problem involves a plate capacitor with a specified area and separation distance, where one plate is charged. The goal is to determine the voltage across the capacitor using the given charge and the properties of the capacitor.

Discussion Character

  • Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss calculating capacitance using the formula involving area and distance, and then using that capacitance to find voltage. There are questions about the accuracy of calculations and the method of presenting them.

Discussion Status

Some participants have shared their calculations and expressed uncertainty about their accuracy. There is an ongoing exploration of the correct approach to solving the problem, with no explicit consensus reached yet.

Contextual Notes

One participant mentions difficulties with the calculation format on the website, indicating potential constraints in communication or presentation of their work.

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Homework Statement



The plates of a capacitor have an area of 0.4m2 and are separated by a distance of 0.5 micrometers. The material between the two plates is air with a dielectric constant of 1.0 One plate of the capacitor is given the charge of 4[tex]\muC[/tex]. The voltage across the capacitor is, in volts:

Homework Equations




C=[tex]\epsilon[/tex](A/d) C=q/[tex]\Delta[/tex]V

The Attempt at a Solution


So far I've been trying to calculate the capacitance with the first equation and then use the second equation for finding the change in voltage. I keep coming up with the wrong answer, though. Help please!
 
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Show your calculations.
 
(.4m2/.5[tex]\mu[/tex]m)x8.85E-12=C

C=q/[tex]\Delta[/tex]V
[tex]\Delta[/tex]V=q/C=4[tex]\mu[/tex]C/Capacitance
 
I just looked at my calculations and cnai just say sorry if they are a little garbled. I'm having trouble getting used to the way you do calculations on the website. Let me know if you would like me to do them again. Thanks!
 

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