Is the capacitance of a system of conductors calculated in a finite volume or over all space?

My question is motivated by the following problem.

A volume $V$ in vacuum is bounded by a surface $S$ consisting of several separate conducting surfaces $S_i$. One conductor is held at unit potential and all the other conductors are at zero potential. Show that the capacitance of the one conductor is $$ C = \epsilon_0 \int_V \left| \nabla \Phi \right|^2 d^3x $$ where $\Phi(\mathbf{x})$ is the solution for the potential.

(Source: Problem 1.17 in Jackson, 3rd ed.)

The total potential energy of the system is, $$ W = \sum_i^n \sum_j^n \frac{1}{2} C_{ij} V_i V_j = \frac{1}{2}C_{11} V_1^2 = \frac{1}{2} C $$

Since conductor $1$ is held at unit potential and all others are held at zero potential.

More generally, the potential energy of a system is, $$ W = \frac{\epsilon_0}{2} \int \left| \mathbf{E} \right|^2 d^3x $$ Equating the last two expressions yields, $$ C = \epsilon_0 \int \left| \nabla \Phi \right|^2 d^3x $$ This integral is over all space, not just in the volume $V$. In fact, if this integral simplifies to just over the volume $V$, wouldn't that imply that the field outside of the surface is $0$ everywhere (since any additional contribution to this integral is positive definite)? My intuition tells me the field is non-zero outside of the conductors because there is a potential difference between infinity (defined as $0$) and one of the conductors (at unit potential), so field lines should be directed away from the surface with unit potential and flowing towards infinity with lower (zero) potential.

My guess is that there is an issue with the self-energy contribution hidden in the energy integral over all space. If this is the case, does anyone have an intuitive explanation of why the self-energy contribution exists only outside of the volume $V$?

My questions:

  • Is the field outside of the conducting surface 0? If so, why?
  • Is there some small detail missing that forbids this approach to a solution?
  • $\begingroup$ @Angelika, the surface $S$, which encloses the volume $V$, consists of several separate conducting surfaces $S_i$. $\endgroup$ – kordon Sep 2 '15 at 4:26

Steve B is right. I just want to include a diagram. You can see that the surface $\rm S$ consisting of the conducting surfaces $\rm S_1, S_2, S_3,...., S_i$ enclosing the volume $\rm V$.
enter image description here

Suppose $\rm S_1$ is at unit potential and all the others are at zero potential. If $\rm S_1$ has $\rm Q$ amount of charge then an equal amount of negative charge is induced on the other conductors. So, the net charge on $\rm S$ is zero. The electric flux through a Gaussian surface made just outside $\rm S_3$ is therefore zero. We don't know and it doesn't matter how the charges are distributed on $\rm S_i's$. Since, electric field is perpendicular to the surface of a conductor hence the electric field outside the conducting surface $\rm S_3$ is also zero.

Like Steve B has said if it were given that the surface $\rm S$ has some non-zero net charge then we would have to calculate the capacitance of $\rm S$ which is called its self capacitance. For example, the self capacitance of a conducting sphere is calculated in this way: First the capacitance of a spherical capacitor is calculated then in the limit of the radius of the outer plate going to infinity the capacitance of the inner conducting sphere is obtained.

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Yes, the field outside the conducting surface is 0. If the net charge inside the conducting surface is 0, then the fields outside the surface are zero. It's basically the Faraday Cage effect.

And the net charge inside the conducting surface is 0, because that's the definition of a capacitor.

You might ask, "Um, well, but what if the net charge is not zero?" In that case, there would be two relevant capacitors in the question:

  • Capacitor #1 is the capacitor that Jackson is asking you about,
  • Capacitor #2 is the capacitor where one plate is the surface S, and the other is infinity.

Saying "S carries a nonzero net charge" is exactly the same as saying "We have charged up Capacitor #2."

Since the question is specifically about the capacitance of Capacitor #1, you should assume that other capacitors like Capacitor #2 are not charged up.

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  • $\begingroup$ "If the net charge inside the conducting surface is 0, then the fields outside the surface are zero." I must be missing something, Gauss' Law says that the Electric field flux through the surface is zero, not that the field is zero. Could you elaborate more? $\endgroup$ – kordon Aug 31 '15 at 22:13
  • $\begingroup$ Electric field is perpendicular to the surface of a conductor hence it is also zero. $\endgroup$ – Tea is life Sep 2 '15 at 17:12

According to Griffith, the integral over the volume, extending to beyond the volume, and effectively into all space, is just mathematical convenience- $\mid E\mid^2$ decreases rapidly as distance increases $(\propto\frac{1}{r^4})$. So, if we take all of space, the contributions of the integral beyond the volume we're concerned with is negligible, but that doesn't mean that the field is $0$.

So yes, there is a field beyond $V$; we just ignore its effects.

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