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Radius of variable capacitor

  • Thread starter jitznerd
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  • #1
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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 dont 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.
 

Answers and Replies

  • #2
gneill
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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.

<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 dont 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.
[tex] C = \frac{\epsilon_o A (2n - 1)}{d}[/tex]
[tex] A = \frac{C\;d}{\epsilon_o (2n - 1)} [/tex]
 
  • #3
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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
 
  • #4
gneill
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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!
 
  • #5
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)
 
  • #6
gneill
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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?
 
  • #7
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
 
  • #8
gneill
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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.
 
  • #9
I think you mean;

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

Area = Π*r^2
radius = √(area/Π)
Right. Note that the [itex]\pi[/itex] 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
gneill
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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
gneill
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Yes, that looks better!
 
  • #15
Thanks for helping me, appreciate it
 
Last edited:
  • #16
gneill
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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
gneill
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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.
 
  • #20
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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
gneill
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"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
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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 dont understand the method thats being used here :-(

Thanks
 

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