Spherical Shell Capacitance

  • #1
What is the capacitance C of a capacitor that consists of two concentric spherical shells, the inner of radius [tex]r_{1}[/tex] and the outer of radius [tex]r_{2}[/tex]? What would the limit be if [tex]r_{2}-r_{1}<<r_{1}[/tex]. What would happen if [tex]r_{2}[/tex] goes to infinity (assume the potential there is zero)?

Some equations:

[tex]
SA=4\pi r^{2}[/tex]
[tex]
C=Q/V[/tex]
[tex]
V=V_{R_{2}}-V_{R_{1}}[/tex]


So, I was looking at a similar example in the book that had to do with two cylindrical shells with inner Radius r1 and outer radius r2 and length L. It started out by showing the net flux through another cylinder in which they give the radius greater than r1, less than r2 and length [tex]l[/tex] is basically [tex]E_{R}[/tex] times the curved surface area of the cylinder. I am having trouble, applying this to a sphere. Once they get the net flux, [tex]E_{R}2\pi R l[/tex] they equate it to [tex]\frac{Q_{inside}}{\epsilon_{0}}[/tex] then [tex]\frac{Q_{inside}}{\epsilon_{0}}=\frac{l}{L}[/tex] and then [tex]Q_{inside}=\frac{l}{L}Q[/tex]. Then they plug in [tex]Q_{inside}[/tex] to the previous equation where [tex]E_{R}2\pi R l=\frac{Q_{inside}}{\epsilon_{0}}[/tex], solve for [tex]E_{R}[/tex] then plug [tex]E_{R}[/tex] into [tex]dV=-E_{R}dR[/tex]. Integrate and then you get V. Plug V into C=Q/V and there you go..

I am having issues figuring out how to get [tex]E_R[/tex] for a sphere from that... Whats throwing me off is the use of the third cylinder in the example problem. I tried to figure out the [tex]E_{R}[/tex] for a sphere with a radius greater than r1 and less than r2 but what do I do for the [tex]l[/tex] that they give with the cylinder problem? How do I find [tex]Q_{inside}[/tex] thus allowing me to find what [tex]E_{R}[/tex] is?



Once I figure that out, then I should be able to plug it into [tex]dV=-E_{R}dR[/tex] and solve for V, thus answering the first part of the question, right? Then the second and third part, I need some help conceptualizing.


*** After looking online for a bit, I have found this equation:

For a sphere [tex]E=\frac{Q}{4\pi r^{2}\epsilon_{0}}[/tex]

Using that [tex]dV=\frac{Q}{4\pi r^{2}\epsilon_{0}}dR[/tex]

Then [tex]V=\int_{r_{1}}^{r_{2}}\frac{Q}{4\pi r^{2}\epsilon_{0}}dR[/tex]

[tex]V=\frac{Q}{4\pi \epsilon_{0}}(-\frac{1}{r_{2}}+\frac{1}{r_{1}})[/tex]

Now that I have V I can Solve for C
[tex]

C= Q/V[/tex]

[tex]
C= \frac{Q}{\frac{Q}{4\pi \epsilon_{0}}(-\frac{1}{r_{2}}+\frac{1}{r_{1}})}[/tex]


Which leaves me with
[tex]
C=\frac{4\pi \epsilon_{0}}{(-\frac{1}{r_{2}}+\frac{1}{r_{1}})}[/tex]

If this is right, then if the distance between the two spheres is much smaller than the radius of the first sphere can I approximate and say that r2 and r1 are the same and then that fraction becomes zero, making the equation undefined/ approaches infinity?

As r2 goes to infinity then then that fraction goes to zero so it becomes just 4 pi e_0 over 1/r1 ?

Thanks
 
Last edited:

Answers and Replies

  • #2
kuruman
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If this is right, then if the distance between the two spheres is much smaller than the radius of the first sphere can I approximate and say that r2 and r1 are the same and then that fraction becomes zero, making the equation undefined/ approaches infinity?
Nope. Subtract the fractions first, then use the approximation r = r1 = r2. Substitute in the numerator d = difference r2-r1 (d is a small number but not zero).

As r2 goes to infinity then then that fraction goes to zero so it becomes just 4 pi e_0 over 1/r1 ?
Yup.
 
  • #3
Can some one just verify that I have the right equation for E for concentric spherical shells? How was that E derived?

Thanks!
 
  • #4
kuruman
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Can some one just verify that I have the right equation for E for concentric spherical shells? How was that E derived?

Thanks!
It is correct. Derived from Gauss's Law where the Gaussian surface is a concentric sphere between the two shells. You can verify it yourself.
 
  • #5
It is correct. Derived from Gauss's Law where the Gaussian surface is a concentric sphere between the two shells. You can verify it yourself.
Thanks!
 
  • #6
For the second part of the question: What would the limit be if [tex]r_2 - r_1 << r_1[/tex].

I have [tex]

C=\frac{4\pi \epsilon_{0}}{(-\frac{1}{r_{2}}+\frac{1}{r_{1}})}
[/tex]

then I manipulate the right side of the equation to get [tex]

C=\frac{4\pi \epsilon_{0}}{(\frac{r_{2}-r_{1}}{r_{1}})(\frac{1}{r_{2}})}
[/tex]

For [tex]r_2 - r_1 << r_1[/tex] The limit of [tex] \frac{r_{2}-r_{1}}{r_{1}}[/tex] goes to zero, right?

As it is going to zero the entire equation goes to infinity, right?













"Nope. Subtract the fractions first, then use the approximation r = r1 = r2. Substitute in the numerator d = difference r2-r1 (d is a small number but not zero)."

I get what you are saying but is that the limit? If I make the substitution r1=r2 then I inevitably get a zero in the denominator.
 
  • #7
kuruman
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I get what you are saying but is that the limit? If I make the substitution r1=r2 then I inevitably get a zero in the denominator.
Look, r1 gets to be close to r2, but it will never be equal to r2. If it did, you wouldn't have a capacitor, just a single shell and the problem would be different. Just take the difference first then let r1=r2. If you don't believe us, do a numerical calculation with r1=1, r2=1.001 using your calculator to find

[tex]\frac{1}{1}-\frac{1}{1.001}[/tex]

then see how close this is to

[tex]\frac{0.001}{1}[/tex]

Repeat by adding an extra zero after the decimal and calculate

[tex]\frac{1}{1}-\frac{1}{1.0001}[/tex]

and see how close this is to

[tex]\frac{0.0001}{1}[/tex]

Keep on adding zeroes and recalculating until you get it.
 
  • #8
^Thanks, had a bit of a brain fart there. I get it now.
 
  • #9
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0
This topic is really useful,I just finished the same question on my assignment.
 

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