Answer:
the interaction in the protein is greater than the surface with water
\frac{F_i}{F_s} = \frac{\epsilon_s}{ \epsilon_i} \ > 1
Explanation:
The electric force for a charge is
F = [tex]\frac{1}{4\pi \epsilon} \ \frac{q^2}{r^2}[/tex]
In the exercise indicate that the charge is q and the distance r is maintained, the test charge is another
therefore if we use the index i for the dielectric constant ([tex]\epsilon_i[/tex]) in the protein
[tex]F_{i} = \frac{1}{4\pi \epsilon_i} \frac{q^2}{r^2}[/tex]
the electric force in water with dielectric constant ([tex]\epsilon_s[/tex])
[tex]F_s = \frac{1}{4\pi \epsilon_s} \frac{q^2}{r^2}[/tex]
[tex]\epsilon_i < \epsilon_s[/tex]
if we look for the relationship between these forces
[tex]\frac{F_i}{F_s} = \frac{\epsilon_s}{ \epsilon_i} \ > 1[/tex]
therefore the interaction in the protein is greater than the surface with water
