Electric field inside a "Rossky cavity" in uniformly polarized water

TL;DR

This study uses numerical simulations to show that the electric field inside a non-polar solute in polarized water deviates from classical predictions, aligning instead with Lorentz cavity theory, impacting dielectric properties.

Contribution

It reveals that the interface of polar liquids carries minimal surface charge, challenging standard Maxwell boundary conditions and proposing a revised electrostatic boundary condition model.

Findings

Electric cavity field approaches Lorentz prediction with increasing solute size.

Standard Maxwell electrostatics fails to describe the interface polarization.

Differences in cavity fields significantly affect dielectric constant and mixture free energy.

Abstract

Electric field produced inside a solute by a uniformly polarized liquid is strongly affected by dipolar polarization of the liquid at the interface. We show, by numerical simulations, that the electric "cavity" field inside a hydrated non-polar solute does not follow the predictions of standard Maxwell's electrostatics of dielectrics. Instead, the field inside the solute tends, with increasing solute size, to the limit predicted by the Lorentz virtual cavity. The standard paradigm fails because of its reliance on the surface charge density at the dielectric interface determined by the boundary conditions of the Maxwell dielectric. The interface of a polar liquid instead carries a preferential in-plane orientation of the surface dipoles thus producing virtually no surface charge. The resulting boundary conditions for electrostatic problems differ from the traditional recipes, affecting the microscopic and macroscopic fields based on them. We show that relatively small differences in cavity fields propagate into significant differences in the dielectric constant of an ideal mixture. The slope of the dielectric increment of the mixture versus the solute concentration depends strongly on which polarization scenario at the interface is realized. A much steeper slope found in the case of Lorentz polarization also implies a higher free energy penalty for polarizing such mixtures.