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 points charges Q_(1) and Q_(2) 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 similar point charges q_1 and q_2 are placed at a distance r apart in the air. The force between them is F_1 . A dielectric slab of thickness t(ltr) and dielectric constant K is placed between the charges . Then the force between the same charge . Then the fore between the same charges is F_2 . The ratio is

Two similar point charges q_1 and q_2 are placed at a distance r apart in the air. The force between them is F_1 . A dielectric slab of thickness t(ltr) and dielectric constant K is placed between the charges . Then the force between the same charge . Then the fore between the same charges is F_2 . The ratio is

Two charges are at distance (d) apart in air Coulomb force between them is F. If a dielectric material of dielectric constant (K) is placed between them, the coulomb force now becomes.

The force between two charges in free space is F. If a dielectric medium of dielectric constant K is placed the force between two charges is…………

Two charges are placed in vacuum at a distance d apart. The force between them is F. if a medium of dielectric constant 2 is introduced between them, the force will now be

A parallel plate capacitor, each with plate area A and separation d is charged to a potential difference V. The battery used to charge it is then disconnected. A dielectric slab if thickness d and dielectric constant K is now placed between the plates. what change, if any will take place in electric field intensity between the plates.

A parallel plate capacitor, each with plate area A and separation d is charged to a potential difference V. The battery used to charge it is then disconnected. A dielectric slab if thickness d and dielectric constant K is now placed between the plates. what change, if any will take place in charge on plates?

A parallel plate capacitor, each with plate area A and separation d is charged to a potential difference V. The battery used to charge it is then disconnected. A dielectric slab if thickness d and dielectric constant K is now placed between the plates. what change, if any will take place in capacitance?