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

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 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. Find the dielectric constant.

The capacitance of a parallel plate capacitor is 10muF . When a dielectric plate is introduced in between the plates, its potential becomes 1/4th of its original value. What is the value of the dielectric constant of the plate introduced?

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 ?

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

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?

Coulomb's force between two point charges separated by certain distance in air is F. If the charges are situated in medium at same place then Coulomb's force reduces to F//4 . The dielectric constant of medium will be