Question

the force between two charges placed in air is 2×10^-6 N. if they are placed the same distance apart in medium whose relative permittivity 4, the force between them is​

Answer 1

Answer:

F = 2×10^-6

and F = kq1q2/r^2 F' = F/£

so, F' = 2x10^-6/4

= 0.5x10^-6 N

Answer 2

Answer:

The force between the 2 charges is $5*10^{-7}$.

Explanation:

Given is the electrostatic force between 2 charges when they are in air,

$F_{a} =\frac{Q1Q2}{4\pi E_{0}r^{2} }$

where Q₁ and Q₂ are the magnitude of charges  

E₀ = Permittivity of free space

r= Distance between charges.

$F_{a} =2*10^{-6} N$

When a medium is introduced between the charges the force between the charges changes as the E₀(Permittivity of free space) value changes.

$Force\ in\ other\ medium = \frac{F_{a} }{E_{r} }$

where $E_{r}$ is the relative permittivity of the medium.

$F_{m}=\frac{F_{a} }{E_{r} } \\F_{m}=\frac{2*10^{-6} }{4} \\F_{m}=5*10^{-7}$

Hence, The force between the 2 charges is $5*10^{-7}$