Sol. It can be proved that if equal amounts of charge are carried by F and F* then the electrostatic energy of the configuration belonging to F is lower.

Let us start from the new surface F*. Imagine that the charges on it are 'fastened' to the surface, and that the surface is then hammered in such a way as to displace the charges perpendicular to the original surface. The charges have then moved in the direction of the force acting on them. As the surface was originally an equipotential, the field direction was perpendicular to the surface. Thus, the energy of the system decreases in the course of the deformation. (An outward force acts on the surface charges, regardless of their signs – a field directed outwards emanates from the positive charges, while an inwardly directed field is produced by the negative ones.)

The new surface will not be equipotential, but, if the fixed charges are 'released', they migrate, warming the metal up a little whilst doing so. The electrostatic energy of the system is therefore lower in the new equilibrium position.

This process can be repeated until surface F* is transformed into surface F. The electrostatic energy decreases all the time, while the total charge Q on the metal does not change. Since the electrostatic energy depends on the capacitance, C as Q 2 /2C, the capacitance of the surface F has to be greater than that of the surface F*.