TITLE: Capacitance of tilted circular plate to ground
QUESTION [0 upvotes]: I am trying to calculate the capacitance between a circular plate of radius $r$ and infinite ground plane, where the circular plate is tilted at an angle $\theta$ to the ground plane. The aim is to better understand capacitive displacement sensors, and an analytical result would be better than a numerical one for that.
Assumptions:

The circular plate is surrounded by a guard electrode held at the same potential as the plate. So there are no fringing fields and if the plates were parallel then the capacitance would be $\varepsilon\pi r^2/d$
We can consider $\theta$ to be small in the $\tan\theta\approx\theta$ sense, if that is useful
We can consider the separation $d$ to be small compared to the radius $r$
The plate does not collide with the ground plane.

My approach was to place the origin at the centre of the plate, and choose $x$  in the direction the plate is tilted.  I can then divide the plate into strips perpendicular to $x$, which are at constant distance from the ground plane. These have width $\mathrm{d} x$ and length $2\sqrt(r^2-x^2)$. So I write the total capacitance as:
$$ C= \varepsilon \int^r_{-r} \frac{2\sqrt{r^2-x^2}}{d_0 + x\tan\theta} \mathrm{d}x$$
But try as I might, I cannot evaluate the integral. I have a feeling that the key is to use the fact that the denominator is always positive to simplify things, but I don't see how.
So how do I solve that? Or is there a different way to set up the problem which avoids this integral? Or should I just make do with a numerical solution?
I did try a trigonometric substitution: $x=r\cos u $ (and writing $\tan \theta = t$ for simplicity)
$$C=2\varepsilon \int \frac{-r^2 \sin^2u}{d + r t \cos u}\mathrm{d}u$$
Which seems like an improvement. Other trigonometric and hyperbolic substitutions also yield similar forms, but I can't see a next step for any of them.

REPLY [0 votes]: Here's another approach that I like more: you could use the method of images. Keep the first circular plate tilted at $\theta$, above the infinite conducting plane; replace the infinite conducting plane with another circular plate below it and tilted at angle $-\theta$. The charge distribution on the new plate will be identical to that on the first one but with opposite charge.
The nice thing is that so far we have made no approximation at all. The tricky part is to now figure out the charge distribution on the plates that minimizes potential energy. For this I suggest using a multipole expansion. Since the charge distributions are two-dimensional, I believe going up to the quadrupole term will give an accurate answer.
The capacitance of this system is twice that of the original one.
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