Capacitor question - Estimate the required radius of each plate

In summary: Furthermore, the problem statement explicitly states that there are "4 pairs of plates". That would seem to indicate that there are 4 completely separate capacitors. That would also imply that there are 4 completely separate areas.3) It appears that the OP and the other poster, "Brenfox", came to an agreement on the answer, so what's the issue?4) The problem statement just asks for the radius, not the diameter. So why are you using "d" in your formula?5) Finally, the OP was originally asking for a summary of the conversation, not a continuation of it. Let's not start this up again, OK?
  • #1
brenfox
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.
 
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  • #2
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?
 
  • #3
This produces 4 parallel capacitors.
 
  • #4
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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  • #5
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
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?
 
  • #7
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
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...

[tex]A = \frac{10^{-5} * 400 * 10^{-12}}{7 * 8.85 * 10^{-12}}[/tex]
 
Last edited:
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  • #9
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?
 
Last edited:
  • #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.
 

Related to Capacitor question - Estimate the required radius of each plate

What is a capacitor?

A capacitor is an electronic component that stores electrical energy in the form of an electric field. It is composed of two conductive plates separated by an insulating material called a dielectric.

What is the purpose of estimating the required radius of each plate?

Estimating the required radius of each plate is important in designing a capacitor that can store a desired amount of electrical energy. It helps determine the size and dimensions of the capacitor that will meet the specifications of a particular circuit or application.

What factors affect the required radius of each plate in a capacitor?

The required radius of each plate in a capacitor is affected by the capacitance value, the dielectric material used, and the distance between the plates. A higher capacitance value requires larger plates, while a higher dielectric constant can allow for smaller plates. The distance between the plates also plays a role in determining the required radius.

How can the required radius of each plate be calculated?

The required radius of each plate can be calculated using the formula for capacitance: C = εA/d, where C is the capacitance, ε is the dielectric constant, A is the area of the plates, and d is the distance between the plates. Rearranging the formula to solve for A (area), we get A = Cd/ε. The radius of each plate can then be calculated using the formula for the area of a circle: A = πr^2, where r is the radius.

What are some common dielectric materials used in capacitors?

Some common dielectric materials used in capacitors include ceramic, paper, plastic, and electrolytic materials. Each material has different dielectric constants, which affect the required radius of each plate in a capacitor. For example, ceramic capacitors have a high dielectric constant, allowing for smaller plates and a higher capacitance value.

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