The Coulombian repulsive force between two alpha particles kept at a distance of $3 \ cm$ in air is . . . . . . .

$12$ positive charges of magnitude $q$ are placed on a circle of radius $R$ in a manner that they are equally spaced. $A$ charge $Q$ is placed at the centre. If one of the charges $q$ is removed,then the force on $Q$ is

Two point charges $+q_{1}$ and $+q_{2}$ are placed a finite distance '$d$' apart. It is desired to put a third charge $q_{3}$ in between these two charges so that $q_{3}$ is in equilibrium. This is

Three charges, each of $+1 \mu C$, are placed at the corners of an equilateral triangle. If the repulsive force between any two charges is $F$, then the net force on any one charge will be:

In Millikan's oil drop experiment,an oil drop carrying a charge $Q$ is held stationary by a potential difference of $2400\,V$ between the plates. To keep a drop of half the radius stationary,the potential difference had to be made $600\,V$. What is the charge on the second drop?

Two point charges $q_1$ and $q_2$ are placed at a distance of $50 \ cm$ from each other in air and interact with a certain force. Now,these charges are placed in an oil with a relative permittivity of $5$. If the force between them remains the same,the distance between them in the oil is ........ $cm$.