# Capacitance of Non-concentric Spheres

We all know how to obtain the capacitance $C=\frac{ab}{b-a}$ (ignoring constants) for two concentric spheres of radii $a,b$.

I was just thinking to myself, what would happen to the capacitance for non-concentric spheres?
Suppose we perturbed the inner sphere off the origin. Does capacitance increase? Ideally I would be looking for an explicit calculation (under sufficiently nice assumptions) but I am also guessing it would be a rather ugly formula.

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We know that the system is unstable. After you perturb the inner sphere, it'll move all the way until it touches the outer sphere. Since the potential energy is $\frac{Q^2}{2C}$, and it's decreasing along the way -> the capacitance increases –  pcr Nov 14 '11 at 6:09
Here is my estimate for the change in potential energy $\approx U_0 (\frac{\delta}{b})^2$ (I got this from the change in $\frac12 E^2 2\pi b^2 \delta$). This gives us $\frac{\delta C}{C}\approx (\frac{\delta}{b})^2$. Not sure –  pcr Nov 14 '11 at 6:38
@pcr: I do not see why the system is unstable... Irregardless, I am defining my region, with boundaries of the conductors! So everything is held fixed! –  Chris Gerig Nov 14 '11 at 20:57
I cheated by using Earnshaw theorem =) (the system is unstable if you don't nail down the conductors). Alternatively, you can also argue by drawing the E-field line: the lines in the "forward pole" region (using Georg's convention) are denser than those in the "backward pole" region, which suggests an attraction. –  pcr Nov 14 '11 at 21:14
Check section III and figures 5 to 9 of this note. –  mmc Nov 14 '11 at 23:25

Important notice: My previous result was a little bit incorrect. I found the factor $1/2$ by comparison with the textbook V.V. Batygin, I.N. Toptygin «Problems in electrodynamics».

Let's denote the radius of the inner sphere $S_1$ as $a$, the radius of the outer sphere $S_2$ as $b$ and the displacement as $c$, so that $c\ll a,b$. We choose the origin of the coordinate system to be in the center of the inner sphere. Then, up to the second order in $c$ the distance from the origin to the outer sphere has the form: $$R\left( \theta\right) =b+c\cos\theta.$$ Therefore, the potential in the space between them can be found as $$\phi=\left( \alpha_{1}+\frac{\beta_{1}}{r}\right) +c\left( \alpha _{2}r+\frac{\beta_{2}}{r^{2}}\right) \cos\theta,$$ where $\alpha_{i}$ and $\beta_{i}$ are constant which should be found from the boundary conditions: $$\left. \phi\right\vert _{S_{1}}=const,\quad\left. \phi\right\vert _{S_{2} }=0,\quad \oint_{S_{1}} \mathrm{d}S\,\mathbf{n}\cdot\mathbf{\nabla}\phi=-4\pi Q,\,$$ where $\mathbf{n}=\mathbf{r}/r$. Hence the potential reads as follows: $$\phi=Q\left( \frac{1}{r}-\frac{1}{b}\right) +\frac{Qc}{b^{3}-c^{3}}\left( r-\frac{a^{3}}{r^{2}}\right) \cos\theta.$$ Therefore the potential on the inner sphere doesn't depend on $c$ up to the second order: $$\left. \phi\right\vert _{S_{1}}=Q\,\frac{b-a}{ab}.$$ The charge distribution on the inner sphere can be found as follows: $$\sigma=-\frac{1}{4\pi}\left( \mathbf{n}\cdot\mathbf{\nabla}\phi\right) =-\frac{1}{4\pi}\left. \frac{\partial}{\partial r}\,\phi\right\vert _{r=a}=\frac{Q}{4\pi a^{2}}\left( 1-\frac{3a^{2}c}{b^{3}-c^{3}}\cos \theta\right) .$$ Hence, the force acting on the inner sphere has the form: $$\mathbf{F}=-\frac{1}{2}\oint_{S_{1}} \mathrm{d}S\,\sigma\mathbf{\nabla}\phi,$$ $$F =-\frac{Q^{2}}{4}\int_{-1}^{1}\mathrm{d}\cos\theta\,\left( 1-\frac{3ca^{2}}{b^{3}-c^{3}}\cos\theta\right) \left. \frac{\partial }{\partial z}\left[ \frac{1}{r}+\frac{c}{b^{3}-c^{3}}\left( 1-\frac{a^{3} }{r^{3}}\right) z\right] \right\vert _{r=a}\\ =-\,\frac{Q^{2}c}{b^{3}-c^{3}},$$ where I use the trivial identity: $$\frac{\partial}{\partial z}\frac{1}{r^{n}}=-\frac{nz}{r^{3}}.$$ The capacity $C$ can be found from the potential energy: $$U=\frac{CV^{2}}{2}\quad\Rightarrow\quad F=-\frac{\Delta U}{\Delta c} =-\frac{\phi^{2}}{2}\frac{\Delta C}{\Delta c},$$ thus $$\frac{\Delta C}{\Delta c}=-\frac{2F}{\phi^{2}}=\frac{2ca^{2}b^{2}}{\left( b^{3}-a^{3}\right) \left( a-b\right) ^{2}}.$$ Finally, we obtain $$C=\frac{ab}{b-a}+\frac{a^{2}b^{2}c^{2}}{\left( b^{3}-a^{3}\right) \left( a-b\right) ^{2}}\quad\quad (1)$$

UPDATE: The comments given above give the reference to the article of «Capacitance Bounds for Geometries Corresponding to an Advanced Simulator Design» by M.I. Sancer and A.D. Varvatsis. The article in turn contains the reference to the book:

W. R. Smythe, Static and Dynamic Electricity, McGraw-Hill, New York, 1950

where the following exact result for the capacitance is presented: $$C=ab\sinh\alpha\sum_{n=1}^{\infty}\frac{1}{b\sinh n\alpha-a\sinh\left( n-1\right) \alpha},\quad\quad\left( 2\right)$$ so that $$\quad\cosh\alpha=\frac{a^{2}+b^{2}-c^{2}}{2ab}.$$ Sancer and Varvatsis claim that they found the approximation of the exact result in the $c\rightarrow0$ limit: $$C=\frac{ab}{b-a}\left[ \frac{1}{2}\left( \sqrt{\frac{1-y^{2}/\left( 1+x\right) ^{2}}{1-y^{2}/\left( 1-x\right)^{2}}}+1\right) + \frac{x}{2}\left( \sqrt{\frac{1-y^{2}/\left( 1+x\right) ^{2}}{1-y^{2}/\left( 1-x\right) ^{2}}}-1\right) \right] ,\quad\quad\left( 3\right)$$ where $$x=\frac{a}{b},\quad y=\frac{c}{b}.$$ It is easy to see that the expansion of the result (3) doesn't coincide with mine result (1). The numerical comparison of all three results presented in the figure below:

One can see that the result (3) of Sancer and Varvatsis is incorrect.

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In one of the above comments, a paper was referenced, and it shows that: Letting $d$ denote the distance between the spheres' centers, $C\sim 4\pi\varepsilon\frac{ab}{b-a}\cdot F$, where $F=\frac{1}{2}[\sqrt{\frac{1-(\frac{d}{a+b})^2}{1-(\frac{d}{b-a})^2}}+1]+\frac{1‌​}{2}\frac{a}{b}[\sqrt{\frac{1-(\frac{d}{a+b})^2}{1-(\frac{d}{b-a})^2}}-1]$. Does your approximation fit with this? Sorry I'm being lazy. –  Chris Gerig Jul 31 '12 at 1:11
@ChrisGerig I added the comparison with the article you mentioned. –  Grisha Kirilin Aug 1 '12 at 14:27
AWESOME, thanks for the comparison! –  Chris Gerig Aug 1 '12 at 17:02
+1, nice answer! Although you leave the limit of large c undiscussed. –  Ron Maimon Aug 11 '12 at 10:06

The whole system is unstable (you can convince yourself by Earnshaw theorem or by drawing the field lines). After you perturb the inner sphere, it'll move all the way until it touches the outer sphere. Since the potential energy is $\frac{Q^2}{2C}$, and it's decreasing along the way -> the capacitance increases.

Here is my estimate for the change in potential energy $\approx -U_0 \frac{b-a}{a}(\frac{\delta}{b})^2$ .

I got this from the change in bulk electrostatic energy $\frac12 E^2 2\pi b^2\delta$

Here I shift the outer conductor because it is easier to follow my calculation this way. The shift is exaggerated for clarity's sake.

Initially, the electrostatic energy comes from region 1 and 2. After the shift, it comes from region 2 and 3. We can roughly see that the energy in region 3 is less than the one in region 1: region 1 is nearer to the inner conductor than region 3.

Clearly the change in energy is

$$U_3 - U_1 = \frac12 (E_3^2 -E_1^2) 2\pi b^2 \delta$$ with $$E_3^2 - E_1^2 \approx 2E_1 \frac{dE}{dr} \delta \approx -4 E_1^2 \frac{\delta}b$$

we get $$U_3 - U_1 = - \frac{e^2}{4\pi b} (\frac{\delta}b)^2= U_0 \frac{b-a}a (\frac{\delta}b)^2$$

This gives us $-\frac{\Delta U}{U}=\frac{\Delta C}C\approx\frac{b-a}{a}(\frac{\delta}{b})^2$.

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See my comment above, I do not see how Earnshaw is sufficiently applied, because there are not only electrostatic forces here... And the field lines argument is not rigorous so it could be incorrect. But in addition, I would like to see your energy change calculation as well. –  Chris Gerig Nov 14 '11 at 21:40
So yea, I am convinced that this argument is not sound (see comments under the question). –  Chris Gerig Nov 15 '11 at 1:14
@pcr, I erased my comment when I realized that Chris knows better than any answer, especially when he wrote ""especially when these are rings moving in weird ways (or just rings moving towards the other sphere?) and not planes moving closer to each other..."". He lacks imagination of the one-dimensional nature of the problem. –  Georg Nov 15 '11 at 11:36