The two capacitors are part of a capacitive pressure sensor, I am curious with reference to electrostatic theories how would capacitance be found in respect to the dielectrics.

The two capacitors have 2 different dielectrics within them too. There is a diaphragm in between (moved by pressure of one capacitor and pressuring the other.)

enter image description here

  • \$\begingroup\$ is the diaphragm what is connected to "Output"? \$\endgroup\$ – Marcus Müller Jun 27 at 19:52

If the diaphragm is metal (i.e. well-conducting), you can really look at the top capacitor independently from the bottom one: the electric field between the top conductor and the diaphragm only exists above the diaphragm.

You know the formula for capacitance of a plate capacitor:

$$C=\epsilon \frac Ad\text,$$

where \$A\$ is the area of the plates, \$d\$ is their distance, and \$\epsilon\$ the permittivity of the material between them.

Since a capacitor's plates are well conducting, you can just cut them at any point and model them as multiple smaller capacitors wired in parallel.

Now, cut the lower capacitor into two parts: one part with the blue dielectric in there, with an \$\epsilon_\text{blue}\$, and the part with the white dielectric, \$\epsilon_\text{white}\$.

This works because the electric field will be, in a pretty good approaximation, be exclusively between the plates, and exclusively perpendicular to the plates. So, there's no "boundary effects" between the blue and white dielectric.

From there, it's pretty clear that

$$ C_\text{total} = C_\text{blue}+C_\text{white} = \frac 1d\left(\epsilon_\text{blue}A_\text{blue} + \epsilon_\text{white}A_\text{white}\right)\text.$$

Assuming the blue dielectric has a permittivity that is very high compared to the white, i.e. \$\epsilon_\text{blue}\gg\epsilon_\text{white}\$, $$C_\text{total}\approx \epsilon_\text{blue} \frac{A_\text{blue}}{d}\text;$$ meaning that the capacitance of the lower capacitor is a function of the area that the compressed dielectric covers, and of the distance. However, assuming there's a fixed volume of blue dielectric that gets squeezed, it follows geometrically that $$A_\text{blue}=\frac{V_\text{blue}}{d}\text.$$

Inserting that into the above formula yields an inverse quadratic dependence of capacitance to distance, which yields an excellent sensitivity as a measurement device.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.