Two charges `q_1` and `q_2` are placed in vacuum at a distance d and the force acting between them is F . If a medium of dielectric constant 4 is introduced around them, the force now 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 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.

Two charges +q and -q are placed at distance r. Match the following two columns when distance between them is changed to r'.

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

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 point charges A and B, having charges +Q and -Q respectively, are placed at certain distance apart and force acting between them is F, if 25% charge of A is transferred to B, then force between the charges becomes:

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 .