Question

Two charges are kept at a distance in air what should be the relative permittivity of the medium in which the two charges should be kept at the same distance so that they experience half of the force which they experienced in air?

Answer 1

Given:

Two charges are kept at a distance in air and a medium.

To find:

Relative permittivity of the medium such that the net force experienced in medium will be half as that when kept in air.

Calculation:

Let relative permittivity of medium be k.

In air, the charges experience force f.

$\therefore f = \dfrac{1}{4\pi \epsilon_{0} } \bigg \{ \dfrac{ {q}^{2} }{ {d}^{2} } \bigg \}$

In the medium , they experienced force f/2.

$\therefore \dfrac{f}{2} = \dfrac{1}{4\pi \epsilon_{0} k} \bigg \{ \dfrac{ {q}^{2} }{ {d}^{2} } \bigg \}$

Dividing the two Equations:

$\therefore \: \dfrac{1}{2} = \dfrac{1}{k}$

$= > \: k = 2$

So, final answer is:

$\boxed{ \red{ \sf{relative \: permittivity = 2}}}$