The correct option is C r8
When charges are placed in air, the electric force between them is:
Fe=14πϵ0q1q2r2 .......(1)
However, when charge is placed in medium,
F′e=14πϵq1q2r2
Where ϵ=ϵ0×ϵr is called permittivity of medium.
So, when charge placed in medium,
4F=14πϵrϵ0q1q2R2 ……(2)
From eqs.(1) and (2),
4=r2ϵrR2 (∵ϵr=16)
⇒R2=r264
∴ R=r8
Hence, option (c) is the correct answer.
Why this question ?Bottom line: When charges are placed at same separation (r)in a medium of dielectric constant ϵr the force betweenthem decreases.i.e. F′=Fϵr