# What is the equation for capacitance with aysmmetric plates?

The general equation for capacitance:

$C = \frac{\varepsilon_0 \ * A }{d}$

• C capacitance
• $\varepsilon_0$ permitivity of free space
• A area of equally sized plates
• d seperation distance between plates

I cant seem to find an equation whereby the equation features two areas i.e A1 and A2 (representative areas of each plate)

Does such an equation exist (I'm sure it does, somewhere)?

Failing that how reasonable a result would I expect to find if assuming A as the smallest area of the two plates?

Background: Working on implementing my own proximity sensor whereby the first plate is the sensor head itself, but the second plate is the object the sensor detects i.e. the object is bigger than the sensor head.

• That isn't a general formula for capacitance. It's a formula for one specific geometry, and it's only an approximation to any real physical situation. Mar 10, 2018 at 1:42

I'd consider using the microstrip model and then think of ways to develop it: -

Where the formulas quoted on this website calculator are: -

This is a starting point and I would urge you to look at other website calculators and compare formulas. There should be a site that gives a derivation of proof that might allow you to reduce the ground plane from being infinitely wide/long to something more akin to your problem.

The model above assumes an infinite ground plane so it will give a slightly higher value for capacitance.

• Unfortunately, these formulas only work well for w ~= h. If h is much smaller than w (likely for the described situation), the result depends heavily on the exact value of h and don't reflect the reality. Feb 26, 2018 at 15:18
• @Janka I agree and, if h is small, the solution might as well be calculated as parallel plates with same area. Feb 26, 2018 at 15:21
• Yeah, exactly what I was suggesting in my answer comments. Feb 26, 2018 at 15:21
• I'm hesitant to mark any as a correct answer for this as none are absolutley definitive, for the purposes of my project though, this answer and the assumption of smallest area plate is used for the equation i gave, works for me. Mar 1, 2018 at 10:27

If you have an array of parallel plates with different areas, distances and permittitivies, you can solve this by assuming they are paralleled capacitors with different parameters. Simply add their capacitances. This is accurate enough almost always.

If you had a more complicated setup, e.g. non-parallel plates or a non-constant permittivity, you had to solve the differential equations for the E/D-field then. That's very complicated to do analytically, so any engineer would use a finite-element software and simulate it.

• I may be missing something, but i dont feel this is answering my question. I have two plates of different areas, it is the capacitance of these two plates I need an equation for, not a summation of an array of parralel plates (of which i wouldn't know the capacticance of each individual parallel pair anyway) Feb 26, 2018 at 12:58
• I've updated question to make clearer... perhaps Feb 26, 2018 at 12:59
• If the two plates of one capacitor are different in size (but still parallel), the E-field is non-uniform and the capacitance depends heavily on the offset/rotation/distance relation of the plates. One could try to make a formula for it, but any reasonable engineer simulates this setup. Feb 26, 2018 at 13:06
• If you are bold, simply disregard all the area not overlapping. This should give you a reasonable result already. Feb 26, 2018 at 13:07
• What research have you done already? Similar question addressed in physics stack exchange last month. Feb 26, 2018 at 13:13

I think your question is discussed here: What is the capacitance of a parallel plate capacitor with different areas? One of the answers gives an according equation.