Force between two identical charges placed at a distance of r in vacuum is F. Now a slab of dielectric of dielectric constant 4 is inserted between these two charges. If the thickness of the slab is r/2, then the force between the charges will become :

Options

A. F
B. `3/5F`
C. `4/9F`
D. `F/4`

Answer

D

Explanation

From Coulomb.s law the force (F) between two charges is
`" "F=1/(4piepsilon_(0)k)q^(2)/r^(2)`
First case : k = 1
`" "F=1/(4piepsilon_(0))*q^(2)/r^(2)" …(1)"`
Second case : `" "k=4`
`" "F^(.)=1/(4piepsilon_(0) times 4)*q^(2)/r^(2)" ...(2)"`
`rArr" "F/(F.)=4`
`rArr" "F.=F/4`