# Capacitance of a 2D capacitor

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!

• Could you explain the difference in what you mean by 2d capacitor as opposed to 3d capacitor? – d_b Oct 18 '19 at 4:07

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.

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