When air is replaced by a dielectric medium of constant $k$, the maximum force of attraction between two charges separated by a distance

Consider the charges $q, q$, and $-q$ placed at the vertices of an equilateral triangle, as shown in Figure. What is the force on each charge?

A charge $+q$ is situated at a distance $d$ away from both the sides of a grounded conducting $L$ shaped sheet as shown in the figure.The force acting on the charge $+q$ is 

Two identical balls having like charges and placed at a certain distance apart repel each other with a certain force. They are brought in contact and then moved apart to a distance equal to half their initial separation. The force of repulsion between them increases $4.5$ times in comparison with the initial value. The ratio of the initial charges of the balls is

Two identical conducting spheres having unequal positive charges $q_1$ and $q_2$ separated by distance $r$. If they are made to touch each other and then separated again to the same distance, the electrostatic force between them in this case will be :-

A point charge $q_1=4 q_0$ is placed at origin. Another point charge $q_2=-q_0$ is placed at $x =12\,cm$. Charge of proton is $q_0$. The proton is placed on $x$-axis so that the electrostatic force on the proton in zero. In this situation, the position of the proton from the origin is $..........cm$.