# Electrostatic energy of two opposite charges in water and in a vacuum

1. Sep 29, 2012

### JordanGo

1. The problem statement, all variables and given/known data

Compare the electrostatic energy of two opposite charges e and -e, a distance 7 angstroms apart in water at room temperature and that in vacuum (express the energy in terms of Bjerrum length)

2. Relevant equations

E = 1/(4(p$\pi$$\epsilon$D)*(-e^2)/r^2 ???

3. The attempt at a solution

First of all is this the right equation to use? If so, is the Bjerrum lenght the distance between the charges? And for a vacuum is the dialectic constant (D) just 1?

2. Sep 29, 2012

### szynkasz

Electrostatic energy is:$$E=-\frac{e^2}{4\pi \varepsilon D}$$
Bjerrum length:
$$\lambda_B=\frac{e^2}{4\pi \varepsilon k_BT}$$

3. Sep 29, 2012

### JordanGo

So, for water:
E=$\lambda$KT/80???

And for a vacuum:
$\lambda$KT???

4. Sep 30, 2012

### szynkasz

No, we have:
$$\lambda_B k_BT=\frac{e^2}{4\pi\varepsilon}$$
so put it into formula for energy

5. Sep 30, 2012

### JordanGo

I did do that, D=80 for water and D=1 for a vacuum

6. Sep 30, 2012

### szynkasz

In my formula "D" denotes distance

7. Sep 30, 2012

### JordanGo

Ok that makes sense, so D is dialectric constant and lets say r now is the distance, which is missing in the equations.