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.


  • 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.

  • $\begingroup$ Related to Capacitance of two non parallel plates $\endgroup$ – sammy gerbil Apr 16 '18 at 19:17
  • $\begingroup$ Mathematics SE is a better place to get help with integration. The substitution $x=r\cos y$ should work. $\endgroup$ – sammy gerbil Apr 16 '18 at 19:22
  • $\begingroup$ @sammygerbil I see you added the homework tag, this is not actually homework (thankfully I'm 10 years past that stage). I don't mind the tag, if you think it'll attract the right people to answer though. I did see the linked question, but I think it just uses the same approach I was already attempting so it didn't help much. I have also tried trigonometric substitutions without success. I'll add that to the question. $\endgroup$ – Jack B Apr 16 '18 at 20:51
  • $\begingroup$ The tag is not only for "homework", as the name suggests. Sorry, my mistake, the substitution is not as useful as I thought. Given particular values of $d$ and $\theta$ you can get a numerical value from Wolfram Alpha. $\endgroup$ – sammy gerbil Apr 16 '18 at 22:38
  • $\begingroup$ This seems related to loss tangent and this 'paywall answer' at Chegg which hints: "Imagine the capacitor as many infinitesimal capacitors in parallel". $\endgroup$ – Rob Apr 17 '18 at 0:21

Let's start with the small-angle approximation $\sin\theta\rightarrow\theta$. We get


You say that $d\ll r$: this is important in the sense that it's useful for analyzing any typical parallel-plate capacitor, where we want to ignore the fringe field. On the other hand, what is special to this problem is how $d/\theta$ compares with $r$.

Assuming $d/\theta\gg r$, we can take the first terms of the Taylor expansion of the denominator: $$C=\epsilon_0\int_{-r}^rdx\sqrt{r^2-x^2}\Big(1-\frac{\theta}{d}x+\big(\frac{\theta}{d}x\big)^2-...\Big).$$ The odd powers integrate to zero. Taking only the first three non-zero terms we get $$C\approx \frac{\epsilon_0\pi r^2}{d} \Big(1+\frac{1}{4}\big(\frac{r}{d/\theta}\big)^2+\frac{1}{16}\big(\frac{r}{d/\theta}\big)^4\Big).$$

By the way, I don't think the shape of the plate played much of a role here -- I believe a more general formula would be

$$C\approx \frac{\epsilon_0 A}{d} \Big(1+\big(\frac{l/4}{d/\theta}\big)^2+\big(\frac{l/4}{d/\theta}\big)^4\Big),$$

where $A$ is the area of the plate and $l$ is the "length" of the plate.

In case $d/\theta$ is comparable to $r$ (it cannot be less than $r$), this whole approach might break down, since the surface charge density will start to become significantly non-uniform (note that the original integral already makes an approximation).

  • $\begingroup$ Thanks. I'm interested in case 1, but to a higher accuracy. I might try a series expansion about that point. If I pick a Taylor expansion the linear and cubic terms will disappear, and the quadratic term should be tractable in the integral... $\endgroup$ – Jack B Apr 17 '18 at 9:51
  • $\begingroup$ @JackB I think that's a good idea, so edited accordingly. Apart from the fact that you're not interested in it, I think this method of calculating capacitance probably doesn't work in the second case I had earlier, so I dropped it. $\endgroup$ – ostrichCamel Apr 18 '18 at 16:32

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.

  • $\begingroup$ That sounds interesting, but I'm not sure how to set up the problem for the multipole expansion. Can you elaborate? $\endgroup$ – Jack B Apr 18 '18 at 14:46
  • $\begingroup$ @JackB, I tried working it out but ran into some problems. I would be happy to discuss on chat some time. $\endgroup$ – ostrichCamel Apr 18 '18 at 15:22
  • $\begingroup$ @JackB I think the main benefit of this technique is that you would also be able to handle the situation where $d/\theta$ is comparable to $r$. $\endgroup$ – ostrichCamel Apr 18 '18 at 16:42

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