Applying electric field
aarondesk
aaro... at gmail.com
Thu Aug 19 17:45:03 CEST 2010
Ok, I think I'm finally understanding. Atomic units make this a little
confusing since they seem to disappear. Thank you for your patience.
Assuming only functions of z, the electric field (F) is related to the
electric potential (V) by F = -dV/dz. F units = volts/angs. V units =
eV/charge. For atomic units, the charge units drop out (=1) so it
appears to disappear and you are left with only units of energy for V.
Thus for V = (a/b)*Z = (0.5 eV/angs-(atomic units charge)*Z we get F =
0.5 eV/angs-atomic units charge. The sign gets changed on V (not
shown) since electrons are negative charge.
Convert to SI units by:
F = (0.5 eV/angs-atomic units charge)*(1 atomic unit charge/1.6E-19
C)*(1.602E-19J/eV)*(V/(J/C)) = 0.5 V/angs.
Thus the field (not potential) felt by the molecule is 0.5 V/angs.
Hopefully my math is correct!
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