Question

two point charges 20 microcoulomb and 10 microcoulomb are separated by 0.05 m in free space find the force between them also calculate the force when a dielectric medium of dielectric constant 3is introduced between them​

Answer 1

these are the required answers

Answer 2

Explanation:

It is given that,

Charge $q_1=20\ \mu C=20\times 10^{-6}\ C$

Charge $q_2=10\ \mu C=10\times 10^{-6}\ C$

The charges are separated by 0.05 m, i.e. d = 0.05 m

The force between the charges is given by :

$F=k\dfrac{q_1q_2}{d^2}$

$F=9\times 10^9\times \dfrac{20\times 10^{-6}\times 10\times 10^{-6}}{(0.05)^2}$

F = 720 N

Let F' is the force when the dielectric medium is introduced. The relation between the force in vacuum, force in a medium and dielectric constant is given by :

$\dfrac{F'}{F}=\epsilon_r$

$F'=\epsilon_r\times F$

$F'=3\times 720$

F' = 2160 N

Hence, when the dielectric medium is introduced the force increases and it is equal to 2160 N.