Question

The Coulomb force between two charges q1 and q2 at separation r in the air is 10N. If half of the separation is filled with medium of dielectric constant 7, what will be the value of new coulomb force?

Answer 1

Answer:

The value of new coulomb force is 1.43 N.

Explanation:

Given;

Coulomb's force in vacuum (air), $F_v$ = 10 N

dielectric constant, K = 7

The Coulomb's force between two charges separated by a distance r in a vacuum is given as;

$F_v = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2}$

The Coulomb's force between two charges separated by a distance r in a medium with dielectric constant is given as

$F_m = \frac{1}{4\pi K\epsilon_0} \frac{q_1q_2}{r^2}$

Take the ratio of the two forces;

$\frac{F_v}{F_m} = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2} \ \times \ \frac{4\pi K\epsilon_0 r^2}{q_1q_2} = K\\\\\frac{F_v}{F_m} = K\\\\\frac{10}{F_m} = 7\\\\F_m = \frac{10}{7} \\\\F_m = 1.43 \ N$

Therefore, the value of new coulomb force is 1.43 N.