# Capacitance of two almost-touching hemispheres

This capacitor is composed of two half spherical shelled conductors both with radius $r$. There is a very small space between the two parts seeing to that no charge will exchange between them.

So far, I only know the capacity of a half spherical shell conductor is $r/2k$ and I can't figure out how to use this. How do I really calculate this capacitance?

For reference, the answer to this problem is $2r/k$.

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Hang on... I'm solving the problem right now. I'm thinking, maybe you have to take each hemisphere as a single plate. Then the capacitance of the whole thing as a single capacitor. I'm trying now :D – mikhailcazi Jun 13 '13 at 11:04
Found a simpler way to go about my solution. I'll edit my solution in a while! – mikhailcazi Jun 14 '13 at 12:20

Definition of Capacitance is:

$$C = \frac{Q}{\Delta V}$$

These are the first principles from which we need to work, because only charge and voltage are truly fundamental quantities. When you propose a type of capacitor, you are positing a physical system where two surfaces (or volumes, but surfaces in this case) are equipotential and have some relative voltage difference. The absolute voltage is undefined, which is why we write the voltage with a Delta. All we know is the voltage difference and geometry. We can say the charges are $-\Delta V/2$ and $\Delta V/2$ or that they are $0$ and $\Delta V$. This information thus far is sufficient to find the stored charge. That doesn't mean it's easy.

This can at least be reduced to a 2D problem, where the charge surface density is defined as a function of angle, $\sigma(\theta)$. We only need one function because of symmetry, and we only need it defined from $\theta=0$ to $\pi/2$. I will use the convention of opposite and equal voltages, meaning the charge distribution will also be opposite and equal.

The integral that defines voltage is hairy, as always. The charges on both plates contribute to the electric potential at some point. Label the point's own sphere to be 1, and the opposing sphere to be 2, and take the point's own sphere to be the positive charge.

$$V(\theta) = \frac{\Delta V}{2} = k \left(\int_0^{\pi/2} \frac{\sigma}{d_1^2} dS_1 - \int_0^{\pi/2} \frac{\sigma}{d_2^2} dS_2 \right)$$

Solving this would give us an equation that contains a specific $\theta$ as well as $\sigma(\theta)$. You may recognize that as similar to a differential equation. That could hypothetically be solved for $\sigma(\theta)$, possibly by algebra, although that might not be practical. From there it would be a true statement to say:

$$Q = 2 \int_0^{\pi/2}\sigma dS$$

The two is added because of the two plates. This would go into the previous definition of capacitance and then you have your answer. The hard part is doing it. I can start to slowly fill in the blanks here. While our charge density is a nice one-variable function, the same can not be said about the integral. We have to integrate over two variables, because the distance between points depends on both of them.

$$dS = R \sin(\theta) d\theta d \varphi$$

We have to use a coordinate transformation from cylindrical to rectangular coordinates in order to subtract vectors. I'll take the x-axis to be going through the center of both spheres.

$$r_1(x,y,z) = r(\theta,\varphi) = R \left< \left( 1 - \cos{(\theta)} \right) , \sin{(\theta)} \sin{(\varphi)} , \sin{(\theta)} \cos{(\varphi)} \right>$$

$$r_2(x,y,z) = r(\theta,\varphi) = R \left< -\left( 1 - \cos{(\theta)} \right) , \sin{(\theta)} \sin{(\varphi)} , \sin{(\theta)} \cos{(\varphi)} \right>$$

$$d_1 = \left| r_1(\theta,0) - r_1(\theta',\varphi) \right|$$

$$d_2 = \left| r_1(\theta,0) - r_2(\theta',\varphi) \right|$$

I'll suppress the dependencies a bit now.

$$d_1^2 = \left( x_1 - x_1' \right)^2 + \left( x_1 - x_1' \right)^2 +\left( x_1 - x_1' \right)^2$$

$$d_2^2 = \left( x_1 - x_2 \right)^2 + \left( x_1 - x_2 \right)^2 +\left( x_1 - x_2 \right)^2$$

You should be able to get an explicit form out of this. I was able to write out $d_1^2$ and $d_2^2$ explicitly in terms of the spherical variables. You're still left with the daunting task of integrating over $\theta'$ and $\varphi$. This is done for both the integrals.

Even if you do that, you're nowhere close to finished. You would just have made an equation sufficient to solve for $\sigma(\theta)$, and even that equation you can't just plug into a differential equation solver because it's an integral equation. Perhaps you could make it an integro-differential equation by putting it in terms of $Q$ (although I have not tried this). It's still a valid mathematical specification of the surface charge density.

I think you can get rid of the $\varphi$ integral part. I understand that it only appears in the form of:

$$\int \frac{f(\theta)}{g(\theta) + h(\theta) \cos{(\varphi)}} d\theta d\varphi$$

and

$$\int \frac{f(\theta)}{g(\theta) + h(\theta) \cos{(\varphi)}^2} d\theta d\varphi$$

That can be evaluated such that it becomes explicit in terms of $\varphi$. It's not short, so I won't write it. But doing this gets you to an expression to integrate in terms of $\theta$ alone. With algebra or iterative techniques, that can be used to find the charge density for every $\theta$, which itself can be integrated to get charge, which can be used to obtain the capacitance.

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An astonishingly low assessment of the only correct approach in all this thread. Good. BTW it is, in principle, a mathematical problem from the potential theory. – Incnis Mrsi Oct 24 '14 at 13:24

Here's what I feel could be a way to solve the problem.

We can consider each hemisphere to be made up of a number of rings. Each ring will have a 'mirror' image on the other hemisphere, making it a capacitor of two ring-shaped plates. Since the whole hemisphere is at one potential, the capacitors are connected in parallel, and we can find the equivalent capacitance of the system by simple addition.

In parallel combination: $C_{eq} = \sum C$

We know that the capacitance of a capacitor is: $$C = \frac{\epsilon_oA}{d}$$

Therefore the capacitance of a ring with infinitesimal area would be: $$dC = \frac{\epsilon_o.dA}{d}$$

As shown in the figure below, assume a ring at an angle of $\theta$ from the x-axis, with the width subtending an angle of $d\theta$ at the origin.
The width of such a ring would be $rd\theta$, and it's radius - as shown in the figure - would be $rsin\theta$.

Thus we have the area of the ring:
It is equal to the circumference ($2\pi R$) multiplied by the width ($rd\theta$). Here, $R = rsinθ$ (radius of ring).

\begin{eqnarray*} \therefore dA &=& 2 \pi (r\sin\theta) . rd\theta\\ &=& 2\pi r^2 \sin\theta d\theta \end{eqnarray*}

Now to find the distance between the rings on each hemisphere. By geometry, we can conclude:

The distance (d) between the two rings:

$$d = 2r(1-\cos\theta)$$

Substituting area and distance into the equation, we get:

\begin{eqnarray*} dC &=& \frac{\epsilon_o(2\pi r^2 \sin\theta d\theta)}{2r(1-\cos\theta)}\\ &=& \frac{\epsilon_o \pi r \sin\theta d\theta}{1-\cos\theta} \end{eqnarray*}

Now, we just need to add up all the capacitances to find the equivalent capacitance! To do that, we integrate the capacitance:

$$\int dC = \int \frac{\epsilon_o \pi r \sin\theta d\theta}{1-\cos\theta}$$

Since $\epsilon_o \pi r$ is a constant, we take it out of the integral, to get:

$$\int dC = \epsilon_o \pi r \int \frac{\sin\theta d\theta}{1-\cos\theta}$$

Now, the integration of $dC$ is $C$. Applying the limits for $\theta$ as $0$ and $\frac{\pi}{2}$ (since these limits will enable us to cover the entire hemisphere), we get:

$$C = \epsilon_o \pi r \int_0^{\pi /2} \frac{\sin\theta}{1-\cos\theta}.d\theta$$ $$C = \epsilon_o \pi r \bigg[ ln(1-\cos\theta) \bigg]_0^{\pi /2}$$

$\big(ln(1-\cos\theta)$ is the integral of $\frac{\sin\theta}{1-\cos\theta}$, since the derivative of $ln(1-\cos\theta)$ gives us $\frac{\sin\theta}{1-\cos\theta}\big)$.

$$\therefore C = \epsilon_o \pi r \bigg[ ln(1-\cos(\pi / 2)) - ln( 1 - \cos(0)) \bigg]$$

$\because \cos(\pi / 2) = 0$ AND $\cos(0) = 1$

we get:

$$C = \epsilon_o \pi r \bigg[ ln(1 - 0) - ln( 1 - 1) \bigg]$$ $$C = \epsilon_o \pi r \bigg[ ln(1) - ln(0) \bigg]$$

At which point I am stuck, because ln(0) does not exist since it $\rightarrow (-\infty)$

But I think I'm pretty close to the answer! The only thing I need is for $\bigg[ ln(1) - ln(0) \bigg]$ to get the value of 8, because that will give me the answer of $2r/k$!!

Since, $K = \frac{1}{4\pi \epsilon_o}$

And it is quite possible for the value to turn out close to 8, since the natural log of a number JUST bigger than zero is negative, and it may be close to -8. If it is so, we will get the answer 2r/k, but I don't know how we can. Sorry! :) Hope I helped a bit, though!

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if you want to calculate capacitance of a spherical capacitor you can see it here hyperphysics.phy-astr.gsu.edu/hbase/electric/capsph.html – abhinav Jun 13 '13 at 13:45
@abhinav the problem doesn't seem to be of that type. At least from the diagram given, you can make out it isn't two hemispherical capacitors joined together in series. It seems like the two hemispheres are the plates themselves... – mikhailcazi Jun 13 '13 at 14:39
Hmm..yes what you are saying makes sense..my bad :-) – abhinav Jun 13 '13 at 15:00
Division of area by distance and subsequent integration is a good way to obtain a rough estimate, but by no means is it a correct answer. Not pass. – Incnis Mrsi Oct 24 '14 at 13:21

To solve this problem, one needs to realize that 4 surfaces are involved. The outside and inside of hemisphere A, and the same for B.

Given: C = r/2k. This is the capacitance of one hemisphere between/due to the "outside" surface and infinity.
There is also an equal capacitance from the inside surface and infinity. Since these two capacitors are in parallel, hemisphere A has a total capacitance of (r/2k + r/2k = ) r/k.

The same applies to hemisphere B, which is a capacitor in parallel with hemisphere A, so the total capacitance of both hemispheres is ( r/k + r/k = ) 2r/k).

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The original poster clearly specified a capacitor to be composed of two conductors. Theoretically, there will be some (small) capacitance between a hemisphere and infinity, but much less than between two hemispheres, and the former is irrelevant anyway. Guill answered another question (capacitance between two hemispheres combined and infinity), and answered incorrectly because there is no addition of capacitance in these circumstances. – Incnis Mrsi Oct 24 '14 at 13:15

As far as I know this seems as two capacitors connected in series, so

1/c = 1/c1+1/c2

putting values this comes to r/4k.

By the way 1 gives a different formula to calculate capacitance

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