Two point charges q1 and q2 are placed at a distance r. the electric force between them is proportional to

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 point charges Q and -3Q are placed at some distance apart. If the electric field at the location of Q is vec (E ) , the field at the location of - 3Q is

The point charges Q and -2Q are placed at some distance apart. If the electirc field at the location of Q is E , the electric field at the location of -2Q will be

Two points charges q_1 and q_2 are placed at a distance of 50 m from each each other in air, and interact with a certain force. The same charges are now put in oil whose relative permittivity is 5. If the interacting force between them is still the same, their separation now is

Two point charges +4q and +q are placed at a distance L apart. A third charge Q is so placed that all the three charges are in equilibrium. Then location. And magnitude of the third charge will be

Two point charges 2q and 8q are placed at a distance r apart. Where should a third charge -q be placed between them so that the electrical potential energy of the system is a minimum

Two free point charges +q and +4q are placed apart a distance x . A third charge is so placed that all the three charges are in equilibrium. Then

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

Two point charges +Q and -Q are separated by a certain distance. The resultant electric field is parallel to the line joining the charges at the points

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