Capacitance of a Spherical Capacitor?

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SUMMARY

The capacitance of a spherical capacitor with inner radius 51.8 mm and outer radius 55.0 mm is calculated using the formula Csphere = 4πε0 * [(a*b)/(b-a)], resulting in a value of 98.86 pF. For the parallel-plate capacitor with the same plate separation of 3.2 mm, the area required to achieve the same capacitance is 358 cm². The discussion highlights a common issue with inputting significant figures into calculation programs, which can lead to incorrect results despite accurate calculations.

PREREQUISITES
  • Understanding of capacitance formulas for spherical and parallel-plate capacitors
  • Familiarity with the constants such as ε0 (permittivity of free space)
  • Ability to convert units between millimeters and meters
  • Knowledge of significant figures in scientific calculations
NEXT STEPS
  • Review the derivation of the capacitance formula for spherical capacitors
  • Learn about the implications of significant figures in scientific computing
  • Explore the use of calculation software for electrical engineering problems
  • Investigate the differences in capacitance between various capacitor types
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Students studying electrical engineering, physics enthusiasts, and professionals working with capacitors and electrical circuits will benefit from this discussion.

Destroxia
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Homework Statement



The plates of a spherical capacitor have radii 51.8 mm and 55.0 mm. (a) Calculate the capacitance. (b) What must be the plate area of a parallel-plate capacitor with the same plate separation and capacitance?

(givens)

a = 51.8 mm
b = 55.0 mm
d = 3.2 mm
ε0 = 8.85E-12

(unknown)

Part a:
Csphere = ?

Part b:
A (plate area of parallel-plate capacitor) = ?

Homework Equations



(Spherical Capacitor) Csphere = 4πε0 * [(a*b)/(b-a)]

(Parallel Capacitor) Cparallel = (ε0*A)/d

Csphere = Cparallel

The Attempt at a Solution



Csphere = 4π(8.85E-12[F/m]) * ((51.8[mm]*55.0[mm]*E-6)/((55.0[mm] - 51.8[mm])*E-3))

= 9.896E-11 F

= 98.86 pF

I could easily solve the second part just by plugging in this capacitance for the parallel-plate capacitor, but the program I am plugging the answer into tells me my answer is wrong. I have double checked the answer, even triple checked the answer, it seems to be coming down to some calculation error?
 
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I don't think you've done the calculation wrong it looks correct the way you have written it, except the units inside your equation are in meters since you have already converted to that unit. Are you out by a factor of tens/tens. If that's not the issue then I don't know what the problem is.
Just to clarify you have written for example 51.8[mm] when it should be 51.8[m].
 
Hello Ryan, welcome to PF :smile: !

Clear posting ! But difficult to help: I followed your working and:

Funny. I get the 99.0 pF too. And A = 358 cm2. Not much we can do then, eh ?
 
Your capacitance value looks okay. Is your program expecting some particular units or number of significant figures?
 
RyanTAsher said:
I could easily solve the second part just by plugging in this capacitance for the parallel-plate capacitor, but the program I am plugging the answer into tells me my answer is wrong. I have double checked the answer, even triple checked the answer, it seems to be coming down to some calculation error?
It would help to know the area you got, to check that.
It might require some specific units as input?
 
BvU said:
Hello Ryan, welcome to PF :smile: !

Clear posting ! But difficult to help: I followed your working and:

Funny. I get the 99.0 pF too. And A = 358 cm2. Not much we can do then, eh ?

I tried your answer of 99.0 pF and it worked! I looked into it, and apparently even though the program usually takes +/- 1 digit in the 2nd significant digit, this one was to the 3rd significant digit... It appears it was my problem for not looking into the programs syntax for that particular problem.

AlexCdeP said:
I don't think you've done the calculation wrong it looks correct the way you have written it, except the units inside your equation are in meters since you have already converted to that unit. Are you out by a factor of tens/tens. If that's not the issue then I don't know what the problem is.
Just to clarify you have written for example 51.8[mm] when it should be 51.8[m].

gneill said:
Your capacitance value looks okay. Is your program expecting some particular units or number of significant figures?

mfb said:
It would help to know the area you got, to check that.
It might require some specific units as input?

Thank you all for your help! Stupid program mistake on my end!
 
I did notice a small change in your post #1 going from = 9.896E-11 F to = 98.86 pF

(and subtly typed 99.0 to leave you room to discover the typo :smile: )
 

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