Capacitance of a spherical capacitor

  • #1
Homework Statement:
A spherical capacitor is formed by two thin conductive layers, spherical and concentric, of radius [itex] R_1 [/itex] and [itex] R_2>R_1 [/itex], between which we have placed a dielectric material of relative permittivity [itex] \varepsilon_r [/itex]. Knowing that the inner layer has an [itex] Q [/itex] charge, idetermines the capacity of the capacitor and the total energy stored.
Relevant Equations:
Gauss Law
When I try to do Gauss, the permeability is not always that of the free space, but it varies: up to a certain radius it is that of the void and then it is the relative one. How can I relate them? I'm trying to calculate the capacity of a spherical capacitor.

The scheme looks like this: inside I have the free space and between the plates of the capacitor I have the dielectric material.
67D98992-D759-4157-BF39-9E70AA815294.jpeg


The broken lines represent the Gaussian surface.
 

Answers and Replies

  • #2
hutchphd
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What is the result if it were space ( ε0) between the spherical shells?
 
  • #3
kuruman
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The vacuum doesn't matter because it contains no charge. The capacitor consists of two conducting plates with the space between them filled completely with the dielectric. Use a Gaussian surface completely inside the dielectric. Or you can find the capacitance with no dielectric between the shells and then multiply it by the dielectric constant.
 
  • #4
I think I have the solution. Is that right?
[tex]
\left.
\phi =\oint \vec{E}\cdot d\vec{S}=\oint E\cdot dS\cdot \underbrace{\cos 0}_1=E\oint dS=E\cdot S \atop
\phi =\dfrac{Q_{enc}}{\varepsilon_0 \varepsilon_r}=\dfrac{Q}{\varepsilon_0 \varepsilon_r}=\dfrac{\sigma \cdot S}{\varepsilon_0 \varepsilon_r}
\right\} E\cdot S=\dfrac{\sigma S}{\varepsilon_0 \varepsilon_r}\rightarrow E=\dfrac{\sigma}{\varepsilon_0 \varepsilon_r}=\dfrac{Q}{4\pi R^2 \varepsilon_0}
[/tex]
[tex] C=\dfrac{Q}{V_2-V_1} [/tex]
[tex] V_2-V_1=-\int_{R_1}^{R_2} \vec{E}\cdot d\vec{l}=-\int_{R_1}^{R_2}E\cdot \overbrace{dl\cdot \cos \theta}^{dR}=-\int_{R_1}^{R_2}\dfrac{Q}{4\pi R^2 \varepsilon_0 \varepsilon_r}dR=\dfrac{Q}{4\pi \varepsilon_0 \varepsilon_r}-\int_{R_1}^{R_2} \dfrac{1}{R^2}dR [/tex]
because [itex]Q[/itex] is constant as it has been transferred to us by an external field/generator. Therefore, it is invariant. [itex]V[/itex] varies due to distance. Then
[tex]V_2-V_1=\dfrac{Q}{4\pi \varepsilon_0 \varepsilon_r}\left( -\dfrac{1}{R_2}+\dfrac{1}{R_1}\right) \rightarrow C=\dfrac{4\pi \varepsilon_0}{\left( -\frac{1}{R_2}+\frac{1}{R_1}\right)}=\boxed{4\pi \varepsilon_0 \varepsilon_r\dfrac{R_2R_1}{R_2-R_1}} [/tex]
 
  • #5
kuruman
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That looks about right.
 
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