Answer:
The value of new coulomb force is 1.43 N.
Explanation:
Given;
Coulomb's force in vacuum (air), [tex]F_v[/tex] = 10 N
dielectric constant, K = 7
The Coulomb's force between two charges separated by a distance r in a vacuum is given as;
[tex]F_v = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2}[/tex]
The Coulomb's force between two charges separated by a distance r in a medium with dielectric constant is given as
[tex]F_m = \frac{1}{4\pi K\epsilon_0} \frac{q_1q_2}{r^2}[/tex]
Take the ratio of the two forces;
[tex]\frac{F_v}{F_m} = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2} \ \times \ \frac{4\pi K\epsilon_0 r^2}{q_1q_2} = K\\\\\frac{F_v}{F_m} = K\\\\\frac{10}{F_m} = 7\\\\F_m = \frac{10}{7} \\\\F_m = 1.43 \ N[/tex]
Therefore, the value of new coulomb force is 1.43 N.
