The product of positive charge and distance between the two charges is

The force of repulsion between two point charges is F, when these are at a distance of 1 m . Now the point charges are replaced by spheres of radii 25 cm having the charge same as that of point charges. The distance between their centres is 1 m , then compare the force of repulsion in two cases.

The force of repulsion between two point charges is F, when these are at a distance of 1 m . Now the point charges are replaced by spheres of radii 25 cm having the charge same as that of point charges. The distance between their centres is 1 m , then compare the force of repulsion in two cases.

The force of repulsion between two point charges is F, when these are at a distance of 1 m . Now the point charges are replaced by spheres of radii 25 cm having the charge same as that of point charges. The distance between their centres is 1 m , then compare the force of repulsion in two cases.

If Two charges +q and +4q are separated by a distance d and a point charge Q is placed on the line joining the above two charges and in between them such that all charges are in equilibrium. Then the charge Q and it's position are

If Two charges +q and +4p are separated by a distance .d. and a point Q is charge placed on the line joining the above two charges and in between them such that ali charges are in equilibrium. Then the charge Q and it.s position are

Two positive point charges are separated by a distance R. If the distance between the charges is reduced to R/2, what happens to the total electric potential energy of the system?

Potential energy of a system of charges is directly proportional to the product of charges and inversely to the distance between them.Prove the above statement.