Capacitance of a Spherical Capacitor?

In summary, the radii of a spherical capacitor are 51.8 mm and 55.0 mm, and the capacitance is calculated to be 98.86 pF. The plate area of a parallel-plate capacitor with the same plate separation and capacitance is 358 cm2. The program may require specific units or number of significant figures for the answer to be accepted.
  • #1
Destroxia
204
7

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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  • #2
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].
 
  • #3
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 ?
 
  • #4
Your capacitance value looks okay. Is your program expecting some particular units or number of significant figures?
 
  • #5
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?
 
  • #6
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!
 
  • #7
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: )
 

1. What is capacitance?

Capacitance is the ability of a system to store an electric charge. It is measured in Farads (F) and represents the ratio of the stored charge to the potential difference between two conductors.

2. How is the capacitance of a spherical capacitor calculated?

The capacitance of a spherical capacitor can be calculated using the formula C = 4πε₀r, where C is the capacitance, ε₀ is the permittivity of free space, and r is the radius of the spherical conductors.

3. What is the relationship between the capacitance and the distance between the spherical conductors?

The capacitance of a spherical capacitor is inversely proportional to the distance between the conductors. This means that as the distance between the conductors increases, the capacitance decreases.

4. How does the dielectric material between the conductors affect the capacitance?

The dielectric material between the conductors has a direct effect on the capacitance. It increases the capacitance by reducing the electric field between the conductors, allowing for a higher charge to be stored.

5. What is the significance of the capacitance of a spherical capacitor?

The capacitance of a spherical capacitor is an important factor in the design and functionality of electronic circuits. It determines the amount of charge that can be stored and the strength of the electric field between the conductors. It is also used in various calculations and equations in electrical engineering.

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