Dimensions of Parallel Plate Capacitors

  • #1

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!
 

Answers and Replies

  • #2
lightgrav
Homework Helper
1,248
30
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.)
 
  • #3
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
 
  • #4
lightgrav
Homework Helper
1,248
30
Yes. You cannot actually determine the gap distance, because you're not told the voltage change rate.
 

Related Threads on Dimensions of Parallel Plate Capacitors

  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
0
Views
972
  • Last Post
Replies
3
Views
864
  • Last Post
Replies
1
Views
1K
Top