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EDIT: title changed to be more descriptive.

I would like to calculate the capacitance of a material (in plate form with thickness d and a known permittivity $\varepsilon_r$ ) between two sphere electrodes of the same radius.

I could calculate the capacitance using the plate model of a capacitor: $$ \frac {\varepsilon_0 \varepsilon_r A}{d}$$ but how do I determine the effective area A between the sphere electrodes?

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  • $\begingroup$ What you mean by "capacitance of a material"? A material by itself has no capacitance but a relative dielectric constant assigned to it. $\endgroup$
    – Markoul11
    Commented Feb 20, 2022 at 17:27
  • $\begingroup$ @Markoul11 if that were true then an isolated sphere wouldn't be considered a capacitor, and yet it is due to self-capacitance $\endgroup$
    – Triatticus
    Commented Feb 21, 2022 at 1:18
  • $\begingroup$ @Triatticus Of course you can represent almost any material with an electric components circuit equivalent like a transmission line related to its electrical properties where you have also capacitors involved to various paths but we don't use the term "capacitance of material". Especially for electrical insulator materials when referring to the electrical properties of the material to characterize it we use its dielectric constant related to capacitance. $\endgroup$
    – Markoul11
    Commented Feb 21, 2022 at 6:57
  • $\begingroup$ Answer was updated. $\endgroup$
    – Markoul11
    Commented Feb 21, 2022 at 7:11
  • $\begingroup$ @Markoul11 yes you're right about a material being characterized with its dielectric constant and not capacitance. Context: at a given voltage V, I need to calculate the impedance of my material and the power (S = U*I) The capacitance was just relevant to me in order to determine impedance. $\endgroup$
    – LTJ_1212
    Commented Feb 21, 2022 at 7:15

1 Answer 1

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For two identical sphere electrodes of radius $a$:

two spheres capacitor

$$C=\frac{2 \pi \epsilon_{0} \epsilon_{r} a}{1-\left(\frac{a}{d-a}\right)}$$

where $ε_{r}$ is the dielectric constant or relative permittivity of your dielectric material, $\epsilon_{0}$ the permittivity of vacuum space and $d$ the axial distance between the centers of the two spheres. The derivation is complicated and demonstrated in this video here:

https://www.youtube.com/watch?v=k8Sp7r3rQBc

Assuming parallel plates your initial electric field $E_{0}$ on air would be:

$$E_{0}=\frac{V}{d}$$

where $V$ the applied voltage on the plates.

Therefore the final $E$ field value in your dielectric center will be:

$$E=\frac{E_{0}}{k}$$

where kappa is the dielectric constant or relative permittivity of your dielectric material $ε_{r}$.

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  • $\begingroup$ Answer was updated. $\endgroup$
    – Markoul11
    Commented Feb 20, 2022 at 19:01
  • $\begingroup$ Interesting. This would answer the question about the capacitance of the dielectric between two sphere electrodes. However, assuming that epsilon_r is the permittivity of the material between the electrodes, is it possible that the dielectric constant epsilon_0 is missing from the first formula? $\endgroup$
    – LTJ_1212
    Commented Feb 21, 2022 at 7:45
  • $\begingroup$ @LTJ_1212 You are absolutely correct, $\epsilon_{0}$ is needed in order to get Farad units. I made the necessary corrections in the answer. $\endgroup$
    – Markoul11
    Commented Feb 21, 2022 at 11:03

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