Question

calculate distance at which the interaction energy between two opposite elementary charges is equal to thermal energy (kbt) at 300k

1. vaccum
2.in water (e=80)

Answer 1

$$F = k{|q_1q_2| \over r^2}$$

At 300K ktb = 0.0259 eV

$$k = {1 \over 4\pi \epsilon}$$

$$r^2 = {k|q_1q_2| \over 0.0259} = {|q_1q_2| \over 0.0259*4\pi \epsilon}$$

a)

$$r^2 = {|1.6*10^{-19}*(-1.6*10^{-19})| \over 0.325304*0.0885*10^{-11}}$$