If you have 2 ions of equal but opposite charge, will the force between them be larger in a vacuum and smaller in water? Would this be because the relative permittivity of water is greater than 1 (around 80)? Also, what does this mean for the interaction energy? Would an interaction be larger in vacuum? I am trying to figure out whether a reaction is more likely to take place in the two different media between the two ions...


1 Answer 1


Yes, much larger in vacuum. The interaction is proportional to the reciprocal of the dielectric constant (relative permittivity) $1/\epsilon$ so is less in water. Think of it as an attenuation of the electric field around an ion, the larger $\epsilon$ is the more the field is attenuated and is true whatever the charges on the ions are making them attractive or repulsive. The force at distance r between charges $q_1, q_2$ is

$$F=\frac{q_1q_2}{(4\pi\epsilon_0)\cdot\epsilon r^2}$$

in SI units; $\epsilon_0 $ is the permittivity of free space. The charges are $q=ze$ where e is the charge on the electron and z the ionic valency, $\pm 1, 2 \cdots $ etc.

The interaction energy is

$$U=\frac{q_1q_2}{(4\pi\epsilon_0)\cdot\epsilon r} \qquad \mathrm{Joule}$$

The electric field around charge $q_1$ is

$$E=\frac{q_1}{(4\pi\epsilon_0)\cdot\epsilon r^2}$$

in V/m. The force acting on a second charge $q_2$ is $F=Eq_2$

Its worth plugging in some numbers for the energy for different $\epsilon$ and comparing this to thermal energy $k_BT$.

  • $\begingroup$ thank you-very informative. Just to check, if I was to calculate a force in water, would I use 80 in place of ϵ in the "ϵ0ϵ " term? Also, if then I had 2 +1 charges, and I wanted them to react, would it be best in water based on your argument of weakening the field? $\endgroup$
    – gamma1
    Commented Jun 2, 2017 at 21:07
  • $\begingroup$ yes use $80 $ for water as its the relative permittivity no units, $\epsilon_0 = 8.854\cdot 10^{-12} \pu{F\,m^{-1}}$. Yes also if charges are similar high dielectric is best. The rate const is $\ln(k)=\ln(k_0)-U/k_BT$ where $k_0$ is rate const at infinite dielectric const. You might also want to consider increasing the ionic strength of the solution, look up 'primary kinetic salt effect'. $\endgroup$
    – porphyrin
    Commented Jun 2, 2017 at 21:31
  • $\begingroup$ @porphyrin It could also be worth mentioning the reason for dielectric constant. Water molecules are polar and orient themselves to become antiparallel with the field. $\endgroup$ Commented Jun 3, 2017 at 4:15
  • 1
    $\begingroup$ @Pritt , both polar and non-polar, i.e all molecules have a dielectric constant. In non-polar ones this is $\epsilon-1 \sim n\alpha$ where n is the number density and $\alpha$ the polarisability. In polar ones additionally there is a term $\epsilon-1 \sim n\mu^2/(k_BT)$ where $\mu$ is the dipole moment. Some partial alignment takes place but this is opposed by thermal motion via the $k_BT$ term. It is too strong a statement to say that molecules become aligned, they are only ever partially so in a liquid . $\endgroup$
    – porphyrin
    Commented Jun 3, 2017 at 8:48

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