# [lammps-users] How can i set dielectric constant of solvent in LAMMPS?

Hi, Alex,
I means, for each kind of solvent, it has dielectric constant,e.g. 80.4 for water at 20oC.
I set real as unit in my simulation, and i like to set the dielectric constant of water in my simulation as 80.4.
Does that mean i just set dielectric as 80.4?
Thanks.
Houyang

If you look at the Coulombic energy formula on this doc page:

doc/pair_lj.html

the dielectric constant as used in LAMMPS is just a unitless
scale factor on that term, 1 by default. The dielectric command
lets you change the scale factor to another (unitless) value.

I don't know what that means precisely in terms of physical
permittivities, like 80 for water, esp since those values are
temperature-dependent. But I think if you use a standard
water force-field, like CHARMM, TIP3P, etc, and the default
value of 1, you will get standard MD water behavior.

Steve

Hello, Steve,
Thanks so much!
I got it!
If we write the eq. (E=Cqiqj/(epsilon*r)), we should tell other what value does C has,
can you tell me what value does C has here for unit of “real” or “LJ”?

Thanks & Best wishes.
Houyang

C is just the conversion factor to units of energy (Kcal/mole)
when multiplying 2 charges and dividing by distance.

in update.cpp for real units:
C = force->qqr2e = 332.06371;

which is calculated as follows:

E = c (q^2 / r)
c = force->qqr2e = converts q^2/r to energy

q^2/r is e^2/Ang
eng is Kcal/mole

q^2/r (cgs) = q^2/r (E8 Ang / cm) (1.60219E-19 Coul / e)^2
(statCoul / 3.3356E-10 Coul)^2
E (cgs) = E (4.184E10 erg / Kcal) (mole / 6.022E23 atoms)
c = 332.0697

Steve

Hello, Steve,
I got it.
Steve, Alex, thanks so much again!
Houyang