.Two similar point charges `q_(1) `and `q_(2)` are placed at a distance "r" apart in air.Now a slab of thickness half the separation between the charges is placed between the charges and the coulombs repulsive force is reduced in the ratio "9:4" .The dielectric constant "(k)" for such a slab 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 placed a certain distance apart in air. If a brass plate is introduced between them, the force between them will

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

Two point charges +q and -q are placed a distance x apart. A third charge is so placed that al the three charges are in equilibrium. Then

Two point charges+ q and-q are placed at distanced apart. What are the points at which the resultant field is parallel to the line joining the two charges ?

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 .