# Conductor completely surrounds another conductor

1. Dec 6, 2015

### Loonuh

1. The problem statement, all variables and given/known data

"Demonstrate that the capacitance of a conductor is always smaller than or equal to that of a conductor which completely surrounds it."

2. Relevant equations/3. The attempt at a solution

Solving this problem for concentric spherical conductors is easy enough, but I am unable to see how one should solve for irregularly shaped conductors. Can anyone offer advice?

2. Dec 6, 2015

### Staff: Mentor

The problem seems underdefined so far. Is there any more to the full problem statement? Is there a figure that goes with the problem?

What does it mean "capacitance of a conductor"? Capacitance to what? You need two things to define a capacitance, no?

3. Dec 7, 2015

### vela

Staff Emeritus
The problem is referring to self-capacitance of a conductor. You can think of it as the capacitance of the conductor and a conductor at infinity.

4. Dec 7, 2015

### Staff: Mentor

Thanks vela. I thought that might be it, but I wasn't sure. That makes the problem pretty easy, though, doesn't it? It would be a lot harder (and maybe not even true) if it were the capacitance to a nearby metal object...

5. Dec 7, 2015

6. Dec 7, 2015

### Staff: Mentor

You skipped listing the Relevant Equation(s)...

7. Dec 7, 2015

### Loonuh

The equations should just be:

- Gauss' law
- Surface of conductor is an equipotential, V_0
- Electric field is normal to surface conductor
- Electric field is 4*pi*sigma in CGS (sigma is charge density)
- sigma = Q/Surface Area
- Capacitance = Q/V_0

Is there anything Anything else that might be useful?

8. Dec 7, 2015

### Staff: Mentor

Please write out the equations. Just listing their names does not help you set up the equations to solve this question...

9. Dec 7, 2015

### Staff: Mentor

BTW, please check your PMs. You need to show some effort in this schoolwork thread of yours or it will be locked.

10. Dec 7, 2015

### Loonuh

Is it possible for me to still edit the thread? O don't see that option.

11. Dec 7, 2015

### Staff: Mentor

No need to edit anything Just post the Relevant Equations and your Attempt at the Solution...

12. Dec 8, 2015

### Loonuh

1. The problem statement, all variables and given/known data

"Demonstrate that the capacitance of a conductor is always smaller than or equal to that of a conductor which completely surrounds it."

2. Relevant 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:

$$\frac{Q{in}}{\epsilon_0} = \int_{S} \vec{E} \cdot d\vec{s}= -\int_{S} \nabla \varphi \cdot \frac{\nabla f}{|\nabla f|} ds\\ -\int_{S} \nabla \varphi \cdot \frac{\nabla f}{|\nabla f|}ds = -\int_{S} \frac{1}{|\nabla f|}\nabla \cdot (\varphi \nabla f)ds + \int_{S}\frac{1}{|\nabla f|}\varphi \nabla^2fds \\= -\int_{S} \frac{1}{|\nabla f|}\nabla \cdot (\varphi \nabla f)ds - \varphi_0\int_{S}\frac{1}{|\nabla f|} \nabla^2fds$$

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.