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Sahay Sir > Question Answers > Cengage NEET by C.P Singh > Part 2 > Chapter 1 - Electrostatics > Two points charges Q1 and Q2 placed at separation d in vacuum and force acting between them is F. Now a dielectric slab of thickness d/2 and dielectric constant K=4 is placed between them. The new force between the charges will be

Two points charges Q1 and Q2 placed at separation d in vacuum and force acting between them is F. Now a dielectric slab of thickness d/2 and dielectric constant K=4 is placed between them. The new force between the charges will be

Two free charges q and 4q are placed at a distance d apart. A third charge Q is placed between them at a distance x from charges q such that the system is in equilibrium. Then