The force of repulsion between two identical positive charge when kept with a separation r in air is F. Half the gap between the two charges is filled by a dielectric slab of dielectric constant = 4 Then the new force of repulsion between those two charges become

Options

A. `F/3`
B. `F/2`
C. `F/4`
D. `(4F)/9`

Answer

D

Explanation

Let q be each charge .
According to Coloumb's law, the force of repulsion between the two charges when kept in air is
`F=1/(4piepsilon_0)q^2/r^2`
When half of the gap between the two charges is filled with the dielectric slab dielectric constant K(=4) , the new force of repulsion between them becomes
`F'=1/(4piepsilon_0)q^2/(((r-t)+tsqrtK)^2)`
(where t is the thickness of the slab)
`=1/(4piepsilon_0)q^2/(((r-r/2)+r/2sqrt4)^2)(as " t "=r/2,K=4)`
`=1/(4piepsilon_0)q^2/((r/2+r)^2)=1/(4piepsilon_0)q^2/((3/2r)^2)`
`=4/9(1/(4piepsilon_0)q^2/r^2)=(4F)/9`