How Do You Calculate the Radius of a Variable Capacitor's Plates?

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Discussion Overview

The discussion revolves around calculating the radius of the plates of a multi-plate variable capacitor, given its capacitance range and plate separation. Participants explore the relevant equations and calculations, addressing both the geometry of the plates and the electrical characteristics of the capacitor.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes using the formula C = εr εo A (n-1)/d to estimate the area of the plates, leading to a calculation of the radius.
  • Another participant suggests that the correct formula should involve (2n - 1) for the number of plates, indicating a potential misunderstanding in the application of the formula.
  • Several participants calculate the area and radius, with varying results and methods, including the use of semi-circle geometry for the plates.
  • Discrepancies arise regarding the interpretation of the number of plates and how it affects the capacitance calculation.
  • Participants express uncertainty about their calculations and seek confirmation on whether their results are correct.
  • One participant revises their calculations multiple times, reflecting on the need to account for the geometry of the plates accurately.
  • In the second part of the discussion, a participant calculates the current and voltage drop during the discharge of the capacitor, seeking validation of their approach.
  • Another participant points out a potential error in the resistance value used in a calculation, prompting a reevaluation of the results.

Areas of Agreement / Disagreement

Participants express differing views on the correct interpretation of the formula for capacitance and the number of plates involved. While some calculations are confirmed as correct, others remain contested, and no consensus is reached on all aspects of the problem.

Contextual Notes

Some calculations depend on the interpretation of the number of plates and the geometry of the capacitor. Participants also note the importance of using proper notation and definitions in their equations, which may lead to confusion in the results.

Who May Find This Useful

This discussion may be useful for students or individuals interested in capacitor design, electrical engineering principles, and the mathematical relationships governing capacitance and geometry.

jitznerd
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Homework Statement



A multi plate variable capacitor has 4 pair of plates. The plates, when closed, are separated in air by 0.01mm and a capacitance range of 10 to 400pF.

a) Estimate the required radius R of each plate.

b) The capacitor is set to maximum 400pF and is charged to 10V through a 50kΩ resistor. Determine:
i) The initial value of the current flowing.
ii) The time constant for the circuit.


Homework Equations



C = εr εo A (n-1)/d

The Attempt at a Solution



I am still new to all this so please don't go too technical with any replies.
I think the correct formula to work out the radius is C=εr εo A (n-1)/d

Transpose this to make A the subject:

A= Cd/εr εo (n-1)

Which gives me A= (400x10^-12)x(0.00001)/1 x 8.85x10^-12 x 7

Area of plate = 0.00316 (I think this is meter^2)

Then use this answer in A=∏r^2
Transpose for radius = √A/∏

Which gives me an answer of 0.0317 (Again I think this is meters).

bi) The initial value of current flowing.
If t=0 The charging current has its highest value.
Therefore io=V/R
=10/50000
=0.0002 Amps

bii) The time constant t=CR
t=(400x10^-12)x 50000
t=0.00002s

Could you tell me if I am going about these the correct way or am I way off mark thanks.
 
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jitznerd said:

Homework Statement



A multi plate variable capacitor has 4 pair of plates. The plates, when closed, are separated in air by 0.01mm and a capacitance range of 10 to 400pF.

a) Estimate the required radius R of each plate.

<snip>

Homework Equations



C = εr εo A (n-1)/d
Presumably you mean (2n - 1) there, where n is the number of plates. In this case (2n - 1) = 7.

The Attempt at a Solution



I am still new to all this so please don't go too technical with any replies.
I think the correct formula to work out the radius is C=εr εo A (n-1)/d

Transpose this to make A the subject:

A= Cd/εr εo (n-1)

Which gives me A= (400x10^-12)x(0.00001)/1 x 8.85x10^-12 x 7

Area of plate = 0.00316 (I think this is meter^2)
I think you were tripped up by your own lack of parenthesis use. The "7" at the end of the calculation is part of the denominator for the last division.
C = \frac{\epsilon_o A (2n - 1)}{d}
A = \frac{C\;d}{\epsilon_o (2n - 1)}
 
Thanks for that gneil
Yeah the n is for the plates.
Am I right in assuming the answers are in meters^2 for the first part?
Could you also tell me if everything looks ok on the remaining answers.
Many thanks
 
jitznerd said:
Thanks for that gneil
Yeah the n is for the plates.
Am I right in assuming the answers are in meters^2 for the first part?
Could you also tell me if everything looks ok on the remaining answers.
Many thanks

The area will be in m2, yes. But the radius will be in meters.

The remaining answers are correctly done. You might want to use scientific notation for the results, or express currents in mA and time in ms. It just looks neater!
 
I get the radius to equal 6.41x10^-3. For question A, did you get this?

I got area=6.457x10^-5

Then half circle radius of plates = sqrt(2*6.457x10^-5)/(Pi)
 
charger9198 said:
I get the radius to equal 6.41x10^-3. For question A, did you get this?

I got area=6.457x10^-5

Then half circle radius of plates = sqrt(2*6.457x10^-5)/(Pi)

Your area value looks okay. Your formula for the radius does not. Can you show your derivation?
 
Daft me..

Should have been R=SQRT (6.457x10^-3)/(Pi)

Answer 4.53x10^-3 m^2

Apologies for my slack formulas.. Think this is correct now
 
charger9198 said:
Daft me..

Should have been R=SQRT (6.457x10^-3)/(Pi)

Answer 4.53x10^-3 m^2

Apologies for my slack formulas.. Think this is correct now

No, it still doesn't look right. Write out the (no numbers!) for the area given a radius r. Then show how you rearranged it to find r in terms of area.
 
I think you mean;

Area = Π*r^2
radius = √(area/Π)
 
  • #10
charger9198 said:
I think you mean;

Area = Π*r^2
radius = √(area/Π)

Right. Note that the \pi is inside the square root, which is not where you had it in your posts above.

One more correction is necessary. The capacitor plates of the variable capacitor are in the form of half circles. When they completely overlap for maximum capacitance the total area of the opposing faces is that of a half circle.
 
  • #11
So I now get

radius = √(6.457x10^-5/Π)
=0.004533571 m

However area = full circle so divide 6.457x10^-5 by 2 to get semicircle area

Thus
√(3.228x10^-5/Π)

=0.00320

Or 3.21x10^-3 m

Sorry my physics is as poor as poor can be
 
  • #12
charger9198 said:
So I now get

radius = √(6.457x10^-5/Π)
=0.004533571 m

However area = full circle so divide 6.457x10^-5 by 2 to get semicircle area

Thus
√(3.228x10^-5/Π)

=0.00320

Or 3.21x10^-3 m

Sorry my physics is as poor as poor can be

You're catching on, but it's still not quite right. The area of a plate must turn out to be equal to what you calculated earlier: 6.45 x 10-5 m2. After all, that's the area that the plates are required to have in order to achieve the required capacitance. Since the plate is in the shape of a semi-circle, the full circle would have double that area...
 
  • #13
I think I may be there...so..

Double the area 6.457x10^-3
To give 0.00012914

radius =√(0.00012914/Π)

Gives, 0.00641143 m or 6.41x10^-3 m

This can be confirmed by;

Area = Π*0.00641143^2 which gives the full circle area
And divided by 2 gives 6.456x10^-5

Hope this is it!
 
  • #14
Yes, that looks better!
 
  • #15
Thanks for helping me, appreciate it
 
Last edited:
  • #16
charger9198 said:
Thanks for helping me, appreciate it

Always happy to help! Good luck.
 
  • #17
Sorry to pop up again

A second part asks;

Having fully charged, the capacitor is then discharged through the 50 kΩ resistor. Determine:

- the current flowing when the capacitor has been discharging for 5 μs

I get ;

I = (V/R)e^-0.25 = (10/5000)*(0.7788)=1.5576*10^-4=0.1557*10^-3=0.1557mA

the voltage drop across the resistor when the capacitor has been discharging for 10 μs.

I get;

V=10exp(-0.5) = 10*(0.60653) = 6.0653volts

Am I on the right lines with this? As youll be able to see I lack confidence in physics
 
  • #18
charger9198 said:
Sorry to pop up again

A second part asks;

Having fully charged, the capacitor is then discharged through the 50 kΩ resistor. Determine:

- the current flowing when the capacitor has been discharging for 5 μs

I get ;

I = (V/R)e^-0.25 = (10/5000)*(0.7788)=1.5576*10^-4=0.1557*10^-3=0.1557mA

the voltage drop across the resistor when the capacitor has been discharging for 10 μs.

I get;

V=10exp(-0.5) = 10*(0.60653) = 6.0653volts

Am I on the right lines with this? As youll be able to see I lack confidence in physics

Yes, that's good.
 
  • #19
Thanks :)
 
  • #20
got it now, thanks
 
Last edited:
  • #21
"Presumably you mean (2n - 1) there, where n is the number of plates. In this case (2n - 1) = 7."

I don't think that is the case. 4 plates would equate to 3 individual parallel capacitors which is where the multiplying factor of n-1 was tieing in.
 
  • #22
JimmyTheBlue said:
"Presumably you mean (2n - 1) there, where n is the number of plates. In this case (2n - 1) = 7."

I don't think that is the case. 4 plates would equate to 3 individual parallel capacitors which is where the multiplying factor of n-1 was tieing in.

For this problem, n is the number of pairs of plates. The statement in quotes should have read: "Presumably you mean (2n - 1) there, where n is the number of pairs of plates. In this case (2n - 1) = 7."
 
  • #23
Dammit. I totally didn't read that and regurgitated the formula in the book despite even spotting the half circle capacitor intricacy lol. Cheers.

Recalc time.
 
  • #24
Charger, for your second part, where do you get the value for R=5000? It is a 50 KΩ resistor or did I miss something critical? I get a current value of 0.1125 mA.

Yes I did miss something... critical... my constant! Now it works out. Doh, never too old for schoolboy errors!
 
Last edited:
  • #25
"I = (V/R)e^-0.25 = (10/5000)*(0.7788)=1.5576*10^-4=0.1557*10^-3=0.1557mA"

could someone clarify how to get "e^-0.25" from the above equation? and also "0.7788".

I don't understand the method that's being used here :-(

Thanks
 
  • #26
andytb1232000 said:
"I = (V/R)e^-0.25 = (10/5000)*(0.7788)=1.5576*10^-4=0.1557*10^-3=0.1557mA"

could someone clarify how to get "e^-0.25" from the above equation? and also "0.7788".

I don't understand the method that's being used here :-(

Thanks

It's a simple RC circuit where the capacitor is discharging through the resistor, so the current takes the form ##I_{max}e^{-t/\tau}##.
 

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