Physics Tutor In Kalkaji

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Very Important Question Of Electrostatics

Force between two identical charges placed at a distance r in vacuum is F. Now a slab of dielectric constant K = 4 is inserted between these two charges. The thickness of the slab is r/2. The force between the charges will now become

  1. (a) F/4 (b) F/2 © 3F/5 (d) 4F/9

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