Two charges are in vacuum at a distance d apart. The force between them is F. If a medium of dielectric constant 4 is introduced between them, the force will now be

Explanation

When a medium of dielectric constant 4 is introduced between the charges, the force between them decreases by a factor of the dielectric constant (ϵr ). Therefore, the force will now be F/ϵr =F/4 . Relationship between permittivity and dielectric constant: The permittivity of the medium (ε) is related to the permittivity of free space (ε₀) by the dielectric constant (εr) as:ε = ε₀ * εrSubstituting and simplifying: In this case, εr = 4 (given dielectric constant). Substituting this into the equation for F':F' = k * (q1 * q2) / ((ε₀ * εr) * d^2)Since k, q1, q2, and d remain constant, we can simplify:F' = F / εrwhere F is the original force in vacuum.Therefore, the force between the charges is reduced by a factor of the dielectric constant (εr), which is 4 in this case. So, the new force (F') becomes:F' = F / 4Incorrect options: (a), (b), and (c) These options do not correctly represent the change in force due to the introduction of the dielectric medium.