1. Homework Statement
"Demonstrate that the capacitance of a conductor is always smaller than or equal to that of a conductor which completely surrounds it."
2. Homework Equations
- Gauss' law
## \int_S E \cdot d\vec{s} = \frac{Q}{\epsilon_0}##
- Surface of conductor is an equipotential: ##\varphi_1##
- Electric field is normal to surface conductor
- Electric field at the surface of a conductor is:
##E_{surface} = \frac{\sigma}{4\pi \epsilon_0}##
- ##\sigma = \frac{Q}{A_{surface}}##
- Capacitance:##C = \frac{Q}{\varphi_0}##
3. The Attempt at a Solution
We are trying to solve for the self-capacitance of the capacitors and show that the larger conductor has a larger self-capacitance than the smaller one. In order to compare the two conductors we can assume that either the conductors have equal charges distributed over them, or that they have equal potentials. Since C = Q/V, if we assume the former then we need to show that the larger conductor has a lower potential, and if we assume the latter, then we need to show that the larger conductor has a smaller charge.
Let's consider each conductor separately. In general, the potential at any point, ##\vec{r}##, is
\varphi(\vec{r}) = \int^{\infty}_\vec{r} \frac{\rho(\vec{r})}{|\vec{r} - \vec{r}'|}dV'
If we assume that the conductors have the same charge, then we can solve the above for both of them as:
\varphi_{in}(\vec{r}) = \int^{\infty}_\vec{r} \frac{\sigma_{in} \delta^3(g_{in}(\vec{r}'))}{|\vec{r} - \vec{r}'|}dV'
\varphi_{out}(\vec{r}) = \int^{\infty}_\vec{r} \frac{\sigma_{out} \delta^3(g_{out}(\vec{r}'))}{|\vec{r} - \vec{r}'|}dV'
where ##g_{in}(\vec{r}')## and ##g_{out}(\vec{r}')## represent the surfaces on which the charge distributions are distributed.
Then
\varphi_{in}(\vec{r}) - \varphi_{out}(\vec{r})=\int^{\infty}_\vec{r} \frac{\sigma_{in} \delta^3(g_{in}(\vec{r}')) - \sigma_{out} \delta^3(g_{out}(\vec{r}'))}{|\vec{r} - \vec{r}'|} dV'
This seems to be getting out of hand though, but at least surely ##\sigma_{in} > \sigma_{out}##, so if I could show that either the remaining part of the integrand is equal to 1, or that:
\int^{\infty}_\vec{r} \frac{ \delta^3(g_{in}(\vec{r}'))}{|\vec{r} - \vec{r}'|}dV' > \int^{\infty}_\vec{r} \frac{ \delta^3(g_{out}(\vec{r}'))}{|\vec{r} - \vec{r}'|}dV'
then I'm done, but showing even this seems intractable.
Ok so what if they have the same potential? I haven't thought about this too much, but I think one can approach this using Gauss' law as:
\frac{Q{in}}{\epsilon_0} = \int_{S} \vec{E} \cdot d\vec{s}= -\int_{S} \nabla \varphi \cdot \frac{\nabla f}{|\nabla f|} ds\\<br />
-\int_{S} \nabla \varphi \cdot \frac{\nabla f}{|\nabla f|}ds = -\int_{S} \frac{1}{|\nabla f|}\nabla \cdot (\varphi \nabla f)ds + \int_{S}\frac{1}{|\nabla f|}\varphi \nabla^2fds \\= -\int_{S} \frac{1}{|\nabla f|}\nabla \cdot (\varphi \nabla f)ds - \varphi_0\int_{S}\frac{1}{|\nabla f|} \nabla^2fds
Thus
\frac{Q{in}}{\epsilon_0} = -\int_{S_{in}} \frac{1}{|\nabla f|}\nabla \cdot (\varphi \nabla f)ds + \varphi_0\int_{S_{in}}\frac{1}{|\nabla f|} \nabla^2fds
and for the outer conductor it follows similarly that
\frac{Q{out}}{\epsilon_0} = -\int_{S_{out}} \frac{1}{|\nabla g|}\nabla \cdot (\varphi \nabla g) ds + \varphi_0\int_{S_{out}}\frac{1}{|\nabla g|} \nabla^2g ds
where ##f## and ##g## represent the inner and outer surfaces respectively; note the have that ##\varphi_{in} = \varphi_{out} = \varphi_0## has been used. I don't know how to proceed with this solution either.
