Two point charges placed at a distance of `0.20m` in air repel each other with a certain force.When a dielectric slab of thickness 0.08m is introduced in between the charges,the force of interaction is half of its previous value.The value of dielectric constant of slab is ,,,..

Two charges are separated by a , distance 1m .When a dielectric slab of thickness 40cm is introduced between , them the force becomes half its previous value. The dielectric constant of medium is

The capacitance of a capacitor becomes 7/6 times its original value if a dielectric slab of thickness t=(2)/(3) is introduced between its plates, where d is the separation between its plates, what is the dielectric constant of the slab?

Two charges 4 cm apart repel each other with a force of 2xx10^(-2) N in air. When a dielectric slab of thickness 1 cm is placed between them, The force is reduced to 1.5xx10^(-2) N. Calculate the permittivity of the dielectric.

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

Capacitance of a capacitor becomes 7/6 times of its original value if a dielectric slab of thickess t=2/4d is introduced in between the plates, d is the separation between the plates. The dielectric constant of the dielectric slab is

Two point charges are placed at separation d in vacuum and the force between them is F . Now a dielectric slab of thickness t = d//3 and dielectric constant K is placed between the charges and the force becomes 9F//25 . Find the value of K .

Two charges are placed at a distance apart if a glas slab is placed between them force between them will

The plates in a parallel plates capacitor are seperated by a distance d with air as the medium between the plates. In order to increase the capacity by 66% a dielectric slab of thickness (3d)/5 is introduced between the plates. What is the dielectric constant of the dielectric slab ?

Two charges placed in air repel each other by a force of 10^(-4)N . When oil is introduced between the charges, then the force becomes 2.5 xx 10^(-5)N The dielectric constant of oil is

Two point charges +4muC and -10 muC are placed 10 cm apart in air. A dielectric slab of large length and breadth but of thickeness 5 cm is placed between them. Calculate the force of attraction between the charges, if the relative permittivity of dielectric is 9.