Dimensions of Parallel Plate Capacitors

AI Thread Summary
The discussion revolves around calculating the dimensions of parallel plate capacitors given a changing electric field and displacement current. The relevant equations include the relationship between displacement current, electric field change, and capacitance. The user initially struggled with the problem but eventually derived the area of the plates as approximately 6.03 x 10^-4 m², leading to a radius of about 1.4 cm. However, it was noted that the gap distance between the plates cannot be determined without additional information about the voltage change rate. The conversation emphasizes the importance of understanding the relationships between electric field, displacement current, and capacitance in capacitor calculations.
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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