An α-particle of mass 6.4×10^−27 kg and charge 3.2×10^−19C is situated in a uniform electric field of 1.6×10^5Vm^−1. The velocity of the particle at the end of 2×10^−2m path when it starts from rest is Force between two identical charges placed at a distance of r in vacuum is F. Now a slab of dielectric constant K = 4 is inserted between these two charges . If the thickness of the slab is r/2 then the force between the charges will becomes November 30, 2020 Category: Cengage JEE Mains Physics by B.M Sharma , Chapter 17 - Electric Charge and Field , Facebook Messenger WhatsApp Share this:TwitterFacebook Related