Two point charges placed at a distance r in air exert a force f on each other. The value of distance R at which they experience force 4F when placed in a medium of dielectric constant K = 16 is :

Two point charges placed at a certain distance r in air exert a force F on each other. Then the distance r at which these charges will exert the same force in a medium of dielectric constnat K is given by

Two point charges placed at a distance r in air experience a certain force. Then the distance at which they will experience the same force in a medium of dielectric constant K is

Two charges exert a force of 10 N on each other when separated by a distance 0.2 m in air. When they are placed in another medium of dielectric constant K = 4 , and separated by distance R. they exert same force. The distance R equats to

Two charge spheres separated at a distance d exert a force F on each other. If they are immersed in a liquid of dielectric constant K=2, then the force (if all conditions are same) is

Two charges of equal magnitudes and at a distance r exert a force F on each other. If the charges are halved and distance between them is doubled, then the new force acting on each charge is

Two charge when kept at a distance of 1m apart vacuum have a some force of repulsion .if the force of repulsion between these two charges be same ,whenplaced in an oil of dielectric constant 4 the distance of separation is

Two point charges are kept in air with certain separation between them.The force between them is F_(1) if half of the space between the charges is filled with a medium of dielectric constant 4 and the force between them is F_(2) if (1)/(3)rd of the space between the charges is filled with dielectric of dielectric constant 9.Then (F_(1))/(F_(2)) is

Force between two point charges in air at certain distance is 200N. What will force when they are kept in a medium of dielectric constant 4, at the same distance?