Force between two identical charges placed at a distance of `r` in vacuum is `F`.Now a slab of dielectric constant 4 is inserted between these two charges . If the thickness of the slab is `r//2`, then the force between the charges will becomes

One brass plate is inserted between two charges. The force between two charges will be

The potential enery of a charged parallel plate capacitor is U_(0) . If a slab of dielectric constant K is inserted between the plates, then the new potential energy will be

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

" If two charges are placed at a distance of "5cm" .If a brass sheet is placed between them.The force between two charges will "