A charged drop of mass 3.2×10^12 g floats between two horizontal parallel plates maintained at potential difference of 980 V and separated between the plate is 2 cm . The number of excess or direction electron on the drop 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