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

`F^(1)=1/(4piepsi_(0))(q_(1)q_(2))/((r-t+tsqrtk)^(2))`
`F/F^(1)=([r-(r/2)+(r/2)sqrt4]^(2))/r^(2)`
= `((3r)/2)^(2)/r^(2)`
= `(9/4r^(2))/r^(2)`
`F/F^(1)=9/4`
`F^(1)=4/9F`