Charge Distribution Costs Free Energy

Creating a Charge Distribution Costs Free Energy 

Imagine assembling a system of charged particles, one at a time. The first particle has charge ji. The electrostatic work of bringing a second particle, having charge q , from infinity into the electrostatic field of the first is 

The factor 1/2 corrects for the double counting charge j is included as contributing to the electrostatic potential experienced by i, and i is also counted as contributing to the electrostatic potential experienced by j if the sum in Equation (22.52) is over all charges i = 

If you have a continuous surface distribution of charge, with a charges per unit area on a surface 5, then the electrostatic free energy is given by a similar integral 

Example 22.7 shows how much free energy is involved in charging up two parallel planes. This is useful when considering the electrostatic repulsion between biological membranes or colloids. 

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The Self-Consistent Reaction Field (SCRF) model considers the solvent as a uniform polarizable medium with a dielectric constant of s, with the solute M placed in a suitable shaped hole in the medium. Creation of a cavity in the medium costs energy, i.e. this is a destabilization, while dispersion interactions between the solvent and solute add a stabilization (this is roughly the van der Waals energy between solvent and solute). The electric charge distribution of M will furthermore polarize the medium (induce charge moments), which in turn acts back on the molecule, thereby producing an electrostatic stabilization. The solvation (free) energy may thus be written as... 

Thanks to efficient recurrence formulae, multipole moments and multipole moment derivatives can be calculated at very high order with a low computational cost. The calculation of reaction field factors, however, may become computationally expensive at high order due to the increasing number of linear equations to be solved. Thus, in practice, the multipole moment expansion is cut off at a maximum value of f (/max), usually taken around 6. In order to get an order of magnitude of the error introduced by the truncation, let us consider Kirkwood s equations [5] for the free energy of a charge distribution of charges q, and r, in a spherical cavity of radius a ... 

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