# Capacitive pressure gauge under a change of pressure

I have seen in an article, that a capacitive pressure gauge made of two parallel plates with a dielectric material between with constant $K$ under an increase of pressure $\Delta P$, a Young´s modulus $Y$, and area of the plates $A$ implies the following changes in length and capacity: $$\Delta l=l_0\cdot \dfrac{\Delta P}{Y}$$ $$\Delta C=C_0 \cdot\dfrac{\Delta P}{Y-\Delta P}$$

But I can't find the way to get there.

I also wonder whether $C_0$ used is referring to the first capacitance without the pressure increase or without the dielectric.

The first equation you have just comes from strength of materials. It's equivalent to $\epsilon=\frac{\sigma}{Y}$, where $\epsilon$ is strain which is $\frac{\Delta l}{l_0}$. This is valid as long as the material is in the range of small deformations.
The $C_0$ and $l_0$ should be the values at equilibrium with no additional external pressure. The reason is that $\epsilon=\frac{\sigma}{Y}$ is based of equilibrium length. However, the $\Delta l$ values we work with are usually very small, so the error you would introduce by taking $C_0$ and $l_0$ at some initial pressure $P_0$ instead of equilibrium should be small.
With an increase in pressure the new total length will be $l_0-\Delta l$ since we expect length to decrease. The capacitance of this parallel plate capacitor at equilibrium is simply $\frac{\epsilon A}{l_0}$. So if the original capacitance is $C_0$, the new capacitance $C_1$ after pressure is added will be $$\frac{\epsilon A}{l_0-\Delta l}= \frac{\epsilon A}{l_0(1-\frac{\Delta P}{Y})}.$$ The change in capacitance is given by $C_1-C_0$. So $$\Delta C= \frac{\epsilon A}{l_0} - \frac{\epsilon A}{l_0(1-\frac{\Delta P}{Y})}=$$ $$\frac{\epsilon A}{l_0}\left(1- \frac{1}{1-\frac{\Delta P}{Y}}\right)= C_0\left( \frac{-\Delta P}{Y-\Delta P}\right).$$ The negative shows up because I wrote $C_1-C_0$. With a shrinking distance you get a smaller capacitance if all else is equal.
• I have seen another results, and what they say is that $\Delta C=-2 C_0 P/Y$, are this two equivalent? Commented Mar 9, 2015 at 2:02
• I am not sure where the -2 would come from in that equation. But for quite a few situations $Y-P=Y$ is a good approximation. The reason is that $\epsilon=\frac{P}{Y}$ is usually only valid for small deformations. For many materials if $P$ is even two or three percent of Y, you have failure. Commented Mar 9, 2015 at 2:40