Capacitor question - Estimate the required radius of each plate

  • Thread starter brenfox
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  • #1
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Homework Statement


Estimate the required radius of each plate. The plates are separated in air by 0.01mm and has a capacitance of 400pF. There are 4 pairs of plates.


Homework Equations


c= εrε0*A(n-1)/ d


The Attempt at a Solution

Find area. Transposing for A

A = dc/εrεo
A= 1*10^-5*400*10^-12/1*8.85*10^-12*7

I am getting 6.457^-27mm.
I am going wrong somewhere, pretty sure the equation is correct though.
 

Answers and Replies

  • #2
berkeman
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Homework Statement


Estimate the required radius of each plate. The plates are separated in air by 0.01mm and has a capacitance of 400pF. There are 4 pairs of plates.


Homework Equations


c= εrε0*A(n-1)/ d


The Attempt at a Solution

Find area. Transposing for A

A = dc/εrεo
A= 1*10^-5*400*10^-12/1*8.85*10^-12*7

I am getting 6.457^-27mm.
I am going wrong somewhere, pretty sure the equation is correct though.

If yoiu have 4 pairs of plates, how many parallel capacitors does that form?
 
  • #3
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This produces 4 parallel capacitors.
 
  • #4
berkeman
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This produces 4 parallel capacitors.

Nope.

Hold your two hands out in front of you, with the fingers of each hand interlacing (no thumbs). That puts two sets of 4 fingers interlacing. Count how many gaps there are between opposing hand/fingers. The answer is not n. :smile:
 
  • #5
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So this produces 7 parallel capacitors. Makes sense using that analogy. Thats why the equation is n-1. eg 8-1 equals 7. My answer still doesn`t add up though!
 
  • #6
gneill
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So this produces 7 parallel capacitors. Makes sense using that analogy. Thats why the equation is n-1. eg 8-1 equals 7. My answer still doesn`t add up though!

can you show your new calculations?
 
  • #7
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A= dc/εrεo*7. This is the equation i get to find area.
So.. A= 1*10^-5*400*10^-12 / 1*8.85*10^-12*7

which equates to 6.456^-27!...somethings wrong with that answer.
 
  • #8
berkeman
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A= dc/εrεo*7. This is the equation i get to find area.
So.. A= 1*10^-5*400*10^-12 / 1*8.85*10^-12*7

which equates to 6.456^-27!...somethings wrong with that answer.

It's just an error with how you are entering it into your calculator. You have the equation correct.

If you write the equation out like this, you can see that the 10^-12 terms will cancel...

[tex]A = \frac{10^{-5} * 400 * 10^{-12}}{7 * 8.85 * 10^{-12}}[/tex]
 
Last edited:
  • #9
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This is where i struggle.My defence is the fact that im an electrician trying to achieve an HND!. So... the 10^-12 cancels each other out leaving 10-5*400/7*8.85. Which equates to. 0.0006? Which feels too small a number.
 
  • #10
gneill
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It's an area. What are the units?
 
  • #11
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So A= 6.46*10^-5

A = 6.45*10^-5*10^6= 64.5mm2?
 
Last edited:
  • #12
71
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To find radius. A = 0.5*3.14*r^2
Transpose to r = √ a/∏*0.5

r= √64.5/*0.5

r = 3.2mm?
 
  • #13
gneill
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To find radius. A = 0.5*3.14*r^2
Transpose to r = √ a/∏*0.5

r= √64.5/*0.5

r = 3.2mm?

Lose the 0.5; The area of a circle is ##A = \pi r^2##.
 
  • #14
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ah right. so..

r=√a/∏ so my answer is √64.5/∏ = 4.53mm. thanks for your help.
 
  • #16
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My tutor marked my answer as correct.
 
  • #17
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Brenfox,

I'm surprised that was marked correct, looks to me like it should be:

r=√(2A/∏)
=√(2*64.5/∏)
= 6.4097mm

The 2A is for the fact that they are semi-circles, I know the question but that's been missed out.

Just wondering if someone can explain if I've construed this incorrectly!

Regards,
Jason
 
  • #18
gneill
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@Jason-Li: Please note:

1) This thread is old, from 2014. The Original Poster (OP) has likely moved on to other things by now.

2) Nowhere in the original problem statement was there mention of semicircles.
 

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