The ratio of the force between two small spheres (with constant charges) `F_(1)` in air and `F_(2)` in a medium of dielectric constant k is respectively..

The ratio of the force between two small conducting spheres of equal charge in (a) a medium of dielectric constant 2, and (b) air is respectively

The ratio of the forces between two charges placed at a certain distance apart in air and at half of the distance a part in medium of dielectric 'k' is А) k B) k/3 C) k/2 D) K/4

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

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

The number of tubes of force originating from a point charge of 17.7xx10^(-8)C placed in dielectric medium of dielectric constant 4 are

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?

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