Question

on increasing the permittivity of the medium the value of Coulomb's constant



(a)decrease

(b)increase
(c)remains constant
(d)none of these ​

Answer 1

Ans is may be A

dear ........

Answer 2

answer : option (a) decrease

explanation : according to Coulomb's law,

force is directly proportional to product of charges and inversely proportional to square of separation between them.

i.e., $F\propto\frac{q_1q_2}{r^2}$

$\implies F=\frac{kq_1q_2}{r^2}$

here, k is proportionality constant known as Coulomb's constant. it depends on permittivity of medium. i.e., $\kappa=\frac{1}{4\pi\epsilon}$

in this expression, $\epsilon$ is permittivity of medium. so, Coulomb's constant is inversely proportional to permittivity of medium.

so, when we increase permittivity of medium, the value of Coulomb's constant decreases.