Two similar point charges q1 and q2 are placed at a distance r apart in air when a dielectric slab of thickness t= r/2 introduced between the charges,the coulomb repulsive force is reduced in the ratio 9:4 .the dielectric constant of slab is? Share with your friends Share 3 Sakshi Kanwar answered this Dear Student When slab of thickness t is inserted than force isF'=Kq1q2(r-t+tK)2F=Kq1q2r2F'/F=9/4r2(r-t+K)2=94Taking underoot thanandt=r/2K=r24Regards -15 View Full Answer
