Question

the force between two point charge separated by air is 4N. When separated by a medium of relative permittivity 2,the force between them becomes​

Answer 1

Answer:

2N

Explanation:

F = 1/(4π k∈₀)  q₁q₂/r²

      When in air, k∈₀ has its original value for air.

F = 1/(4π∈)  q₁q₂/r²

       when, k = 2.

F' = 1/(4π 2∈₀)  q₁q₂/r²

  = 1/2 * 1/(4π∈)  q₁q₂/r²

  = 1/2 * F

  = 1/2 * 4N    = 2N

Answer 2

Answer:

Force between the 2 charges becomes 2N.

Explanation:

Given that the force between 2 point charges separated by the air is 4N.

The permittivity of the air is considered as 1.

So accordingto coulomb's law of force,

$F=\frac{Q_{1} Q_{2} }{4\pi E_{0}r^{2} }$

When the same system is transferred to a medium of relative permittivity

of 2 ($E_{r}=2$) .

The Coulombic Force realation becomes

$F_{medium}=\frac{Q_{1} Q_{2} }{4\pi E_{0}E_{r} r^{2} }$

$F_{medium}=\frac{Q_{1} Q_{2} }{4\pi E_{0}2r^{2} }$

$F_{medium}=\frac{F}{2}=\frac{4}{2} =2N$

Hence, The force between the two charges changes from 4N to 2N.