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

The value of new coulomb force is 1.43 N.

Step-by-step 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 = (1)/(4\pi \epsilon_0) (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 = (1)/(4\pi K\epsilon_0) (q_1q_2)/(r^2)`

Take the ratio of the two forces;

`(F_v)/(F_m) = (1)/(4\pi \epsilon_0) (q_1q_2)/(r^2) \ * \ (4\pi K\epsilon_0 r^2)/(q_1q_2) = K\\\\(F_v)/(F_m) = K\\\\(10)/(F_m) = 7\\\\F_m = (10)/(7) \\\\F_m = 1.43 \ N`

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