Capacitor question - Estimate the required radius of each plate

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Discussion Overview

The discussion revolves around estimating the required radius of capacitor plates given their separation and capacitance. Participants explore the calculations involved in determining the area of the plates and subsequently the radius, while addressing potential errors in their computations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Participants attempt to calculate the area of capacitor plates using the formula A = dc/εrεo, but express uncertainty about their results, particularly regarding the units and magnitude of the area.
  • There is a discussion about the number of parallel capacitors formed by the arrangement of the plates, with some participants suggesting it is 4, while others argue it is 7 based on an analogy involving interlaced fingers.
  • One participant suggests that the area calculation should account for the number of pairs of plates, leading to confusion about the correct formula to use.
  • Participants debate the correct approach to finding the radius from the area, with one suggesting the formula for the area of a circle and another proposing a different factor based on the shape of the plates.
  • There is a correction regarding the area calculation, with a participant noting that the area should be calculated without the factor of 0.5, leading to a revised radius calculation.
  • One participant expresses surprise that their answer was marked correct by their tutor, prompting further discussion about the accuracy of the calculations and the interpretation of the problem.
  • A later reply introduces the idea of semicircles in the context of the problem, which raises questions about whether this was an oversight in the original problem statement.

Areas of Agreement / Disagreement

Participants express differing views on the number of parallel capacitors and the correct approach to calculating the radius from the area. The discussion remains unresolved regarding the interpretation of the problem and the accuracy of the calculations.

Contextual Notes

There are limitations in the clarity of the problem statement, particularly regarding the shape of the capacitor plates and the assumptions made in the calculations. Participants also express uncertainty about the units used in their calculations.

brenfox
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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.
 
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brenfox said:

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?
 
This produces 4 parallel capacitors.
 
brenfox said:
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:
 
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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!
 
brenfox said:
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?
 
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.
 
brenfox said:
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...

A = \frac{10^{-5} * 400 * 10^{-12}}{7 * 8.85 * 10^{-12}}
 
Last edited:
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This is where i struggle.My defence is the fact that I am 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
It's an area. What are the units?
 
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  • #11
So A= 6.46*10^-5

A = 6.45*10^-5*10^6= 64.5mm2?
 
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  • #12
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
brenfox said:
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
ah right. so..

r=√a/∏ so my answer is √64.5/∏ = 4.53mm. thanks for your help.
 
  • #15
brenfox said:
So A= 6.46*10^-5

A = 6.45*10^-5*10^6= 64.5mm2?

Should this not be μm ?
 
  • #16
My tutor marked my answer as correct.
 
  • #17
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
@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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