Minimal surface which divides a convex body into two regions of equal volume 
Question.  Given a convex body $\Omega$, what is the shape of a surface $\Gamma$ of minimal area which divides $\Omega$  into two regions of equal volume?

Background/motivation.
A 2D version of the question was posed by Michael Goldberg in Monthly: find the shortest curve which divides a convex quadrilateral into two equal areas. In the latter case the solution is given by a circular arc perpendicular to two sides of the quadrilateral (or just a segment of a straight line in  degenerate situations).  This follows from the observation that a circular sector is the shortest curve which, together with two sides of an angle, encloses a fixed area.
Goldberg himself offered a physically intuitive solution to the problem:

The curve may be considered as a restraining
member under tension produced by internal fluid pressure in the restricted area.
The ends of the curve are free to slide along the sides. Hence, the curve must be
normal to two sides of the quadrilateral. Furthermore, since the fluid pressure
is uniform, the curve must take the form of a circular arc.

The same approach suggests that in the general multidimensional case the solution is given by a spherical "cap" that intersects $\partial \Omega$ orthogonally. Now, assuming that the intuition is correct, is there a simple formal proof of this conjecture? 
Edit. As Sergei Ivanov points out the minimal surface in question need not be a spherical cap  for dimensions $>2$.
A modified question: is there always at least one solution to the problem? Should one impose any smoothness conditions on $\partial\Omega$ to guarantee existence?
 A: The conjecture as you state it is false. The variational argument (that can be interpreted in terms of fluid pressure if you like) shows that the surface has constant mean curvature and is orthogonal to the boundary. In dimension 2, this means that it is a circular arc, but in higher dimensions there are many more examples of constant mean curvature surfaces.
A small perturbation of the standard ball can produce a convex body such that no sphere is orthogonal to its boundary. For such a body, the surface in question just cannot be a spherical cap.
A: This is a classical problem on which we know the existence, thanks to the Geometric Measure Theory. In space dimension $n$, the solution is a hypersurface which is smooth of constant curvature away from a possible singular set of dimension at most $n-7$. For instance, if $n=3$, it is smooth up to the boundary, and it meets $\partial\Omega$ orthogonally.
If you relax the equal volume condition, and ask for a hypersurface of smaller area that divides $\Omega$ into two pieces of volumes $\epsilon\Omega$ and $(1-\epsilon)\Omega$, then the surface tends to a boundary point $p$ where the curvature has a critical point. See for instance the discussion in the recent PhD Thesis of Paul Laurain (ENS de Lyon).
