TITLE: Which domain maximizes the energy of the Lebesgue measure?
QUESTION [9 upvotes]: This could be asked in more generality, but let me stick to a concrete case. 
Usually one considers a fixed domain $E \subset \mathbb{C}$ and attaches to it the equilibrium probability measure $\nu_E$, the one that minimizes the energy integral
$$
I_E(\nu) := \int_E \int_E -\log{|z-w|} \, d\nu(z) \, d\nu(w).
$$
I want to look at the opposite problem. Fix an infinite measure $\mu$ on $\mathbb{C}$, say the Lebesgue measure $dx\, dy$, and look for the measurable set $E \subset \mathbb{C}$ of unit measure $\mu(E) = 1$ for which the energy $I_E(\mu)$ is a maximum. 
Is there a literature available on this type of question? In particular, what is the answer when $\mu$ is the Lebesgue measure? I am fine with assuming $E$ is a 'nice' domain of unit area rather than a general measurable set.
The same question could be posed in Euclidean space of any dimension, or a more general Riemannian manifold. For dimension one and the Lebesgue measure on $\mathbb{R}$ the maximum is clearly attained for a unit interval, with energy
$$
\int_0^1\int_0^1 -\log{|x-y|} \, dx \, dy = 3/2.
$$

REPLY [4 votes]: That a circular disk maximizes the energy for the Lebesgue measure follows immediately from Riesz's inequality: $I(f,g,h)\le I(f^*,g^*,h^*)$,where $I(f,g,h) = \langle f, g*h\rangle$, and $f^*$ is the monotonically decreasing radial function whose superlevel sets have the same measures as those of $f$. In your case, take $f=g=1_E$ and $h=h^*=-log(|z|)$.