• Support PF! Buy your school textbooks, materials and every day products Here!

Capacitor question - Estimate the required radius of each plate

  • Thread starter brenfox
  • Start date
  • #1
71
1

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
Mentor
57,470
7,490

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
71
1
This produces 4 parallel capacitors.
 
  • #4
berkeman
Mentor
57,470
7,490
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:
 
  • Like
Likes 1 person
  • #5
71
1
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
Mentor
20,795
2,773
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
71
1
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
Mentor
57,470
7,490
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:
  • Like
Likes 1 person
  • #9
71
1
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
Mentor
20,795
2,773
It's an area. What are the units?
 
  • Like
Likes 1 person
  • #11
71
1
So A= 6.46*10^-5

A = 6.45*10^-5*10^6= 64.5mm2?
 
Last edited:
  • #12
71
1
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
Mentor
20,795
2,773
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
71
1
ah right. so..

r=√a/∏ so my answer is √64.5/∏ = 4.53mm. thanks for your help.
 
  • #15
3
0
So A= 6.46*10^-5

A = 6.45*10^-5*10^6= 64.5mm2?
Should this not be μm ?
 
  • #16
71
1
My tutor marked my answer as correct.
 
  • #17
57
8
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
Mentor
20,795
2,773
@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.
 

Related Threads on Capacitor question - Estimate the required radius of each plate

Replies
10
Views
12K
  • Last Post
2
Replies
25
Views
19K
Replies
5
Views
349
  • Last Post
Replies
4
Views
626
  • Last Post
Replies
3
Views
2K
Replies
5
Views
814
Replies
5
Views
734
  • Last Post
Replies
0
Views
2K
Replies
3
Views
6K
Replies
8
Views
2K
Top