In free space the force between two charges separated by constant distance is 9 dyn. When a charges and place in a medium of dielectric constant k, the forces between them becomes 4 dyn,. Then find the value of k

Explanation

Let the two charges are `q_(1)` and `q_(2)` and they are separated by a distance r.
From Coulomb's law `F_(1)=(q_(1)q_(2))/(4 pi epsilon r^(2))`
[ `epsilon=` permitivity of the medium]
and `F_(2)=(q_(1)q_(2))/(4pi epsilon_(0)r^(2))` [`epsilon_(0)=` permitivity of free space]
`:.(F_(2))/(F_(1))=(epsilon)/(epsilon_(0))=k` or `k=9/4=2.25`