Entropy in Conductors

In Introduction to Electrodynamics by Griffiths, it is given that

Like any other free dynamical system, the charge on the conductor will seek the configuration that minimises its potential energy. This asserts that the electrostatic energy of a solid object (with specified shape and total charge) is a minimum when that charge is spread over the surface.

He has also compared the energy of a spherical conductor(with surface charge) to solid sphere(with volume charge) and found out that the conductor has lesser energy for the same charge $Q$.

But considering the Second Law of Thermodynamics, the system should try to increase its entropy(loosely randomness) and this is possible only when the charges are distributed in the interior.

There seems like a contradiction in both statements. What is going on here?

• The Second Law doesn't say that entropy is always maximized. At constant temperature and pressure, instead, it's the Gibbs free energy $G=U-TS+PV$ that's minimized. There is essentially a tradeoff between potential energy and entropy. (In fact, free energy minimization drives some charge carriers to move inside the sphere, because the entropy increase from the first few carriers exploring the inside of the sphere is tremendous. The potential energy term, however, ensures that the vast majority of carriers remain on the surface.) – Chemomechanics Oct 29 '17 at 14:26
• Thanks for answering. From the equation, $PV$ is constant so I don't care about it. But due to the temperature dependence, does it mean that the distribution on the surface or interior depends on temperature? And is it wrong to use Boltzmann Entropy? – Alpha7200 Oct 29 '17 at 14:45
• First question: Yes, any temperature increase makes the entropy consideration more important. Second question: The $S$ is the Boltzmann entropy. – Chemomechanics Oct 29 '17 at 16:19

You're correct that the entropy should strictly be considered. However, the Second Law doesn't say that entropy is always maximized. At constant temperature and pressure, instead, it's the Gibbs free energy $G=U−TS+PV$ that's minimized. The three terms on the right indicate that it's thermodynamically preferred to (1) be in a low-energy state, (2) with a large number of macrostates (which is connected to the Boltzmann entropy $S$), (3) without doing much work to expand into the external atmosphere. As you note, the third term isn't important for this discussion.