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I would like to know how to calculate the capacitance of a 2D capacitor and the derivation of the formula.

I have been looking all over the place but could not find an answer to this.

Thank you!

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    $\begingroup$ Could you explain the difference in what you mean by 2d capacitor as opposed to 3d capacitor? $\endgroup$ – d_b Oct 18 '19 at 4:07
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The Equation is:

$$C= \frac{K\epsilon_0}{A}$$

where $\epsilon_0= 8.854\times10^{-12}$

$K$ is the dielectric constant of the material*

$A$ is the overlapping surface area of the plates (plate area) ($\rm{mm}^2$)

$d$ is the distance between the plates ($\rm mm$)

$C$ is capacitance ($\rm F$)

*Note: All materials have a relative permeability, $k > 1$, thus the capacitance can be increased by inserting a dielectric. Sometimes $k$ is referred to as the dielectric constant of the material.

If you want to derive the capacitance which depends on the area of the plates $A$ and their separation $d$, we start with the electric field between two plates:

$$E = \frac{Q}{\epsilon_0A} \to E\,d = V = \frac{Qd}{\epsilon_0A}$$

and since $$V = \frac{Q}{C}\to C =\frac {\epsilon_0A}{d}$$

If dielectric material is inserted between the plates, we have the practical case. Materials have a permeability E that is sometimes given by the relative permeability $K$, $E=KE_0$. The capacitance is then given by:

$$C = \frac{EA}{d} = \frac{K\epsilon_0A}{d}$$

which is our initial formula.

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  • $\begingroup$ So basically in the 2d case the same formula applies as for 3d plate capacitors? $\endgroup$ – Lt_Peanutbutter May 7 '18 at 15:30
  • $\begingroup$ For 3D you have a different type of surface disposition, so you'll have to adjust considering the exact structure of the capacitor. $\endgroup$ – Overmind May 8 '18 at 5:55
  • $\begingroup$ Wouldn't the area of the overlapping plates be zero in 2D? $\endgroup$ – jim Mar 20 at 19:58

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