Dimensions of Parallel Plate Capacitors

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Homework Help Overview

The problem involves determining the dimensions of the plates of a parallel plate capacitor based on a changing electric field and a given displacement current. The context is within the subject area of electromagnetism and capacitor theory.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the displacement current, the changing electric field, and the area of the capacitor plates. There is an exploration of how the area relates to the radius of the plates. Some participants express uncertainty about the approach and seek guidance on the problem.

Discussion Status

The discussion is ongoing, with some participants providing insights into the equations involved and questioning the assumptions made regarding the gap distance and voltage change rate. There is no explicit consensus on the solution, but guidance has been offered regarding the relevant equations.

Contextual Notes

Participants note that the gap distance cannot be determined without additional information about the voltage change rate, which is not provided in the problem statement.

David Truong
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Homework Statement


The electric field between two circular plates of a capacitor is changing at a rate of 1.5 x 10^6 V/m per second. If the displacement current at this instant is Id = 0.80 x 10^-8A, find the dimensions of the plate.

Homework Equations



Id = ΔQ/Δt = ε0(ΔΦE/Δt)
ΦE = EA
Q = CV
C = ε0AE
ε0 = 8.85 x 10^-12

The Attempt at a Solution



I am unsure how to go about this question. Any guidance is appreciated!
 
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You're told Id and ΔE/Δt ... I'm pretty sure you can look up ε in the textbook front cover. So, how does A depend on radius?
(btw, your Capacitance equation should be εA/d, where d is the gap distance.)
 
So I made this question out to be harder it actually was. I was thinking about the changing electric field in the wrong way.

So here's my work:

Id = 0.80 x 10^-8 A
dE/dt = 1.5 x 10^6 V/m
ΦE = EA

Id = ε0(ΔΦE/Δt)
Id = ε0(A)(ΔE/Δt)
A = Id/[ε0 x (ΔE/Δt)]
A = 0.80 x 10^-8/(8.85 x 10^-12)(1.5 x 10^6)
A = πr2 = 6.03 x 10^-4 m2
r = 1.4 x 10^-2 m or 1.4 cm
 
Yes. You cannot actually determine the gap distance, because you're not told the voltage change rate.
 

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