The electrostatic force between two point charges kept at a distance d apart, in a medium ε_r = 6, is 0.3 N. The force between them at the same separation in vacuum is

The electrostatic force between two point charges is given by the equation F = (1/4πε) * (q1*q2)/r^2, where F is the force, ε is the permittivity of the medium, q1 and q2 are the charges, and r is the distance between them. In this question, the force is given as 0.3 N in a medium with εr = 6. To find the force in vacuum, we can use the fact that εr = ε/ε0, where ε0 is the permittivity of vacuum. Rearranging the equation, we get F_vacuum = (ε0/εr) * F = (1/6) * 0.3 N = 0.05 N. Therefore, the force between the charges at the same separation in vacuum is 0.05 N, which is closest to 0.5 N.