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

An electron is moving around the nucleus of a hydrogen atom in a circular orbit of radius $r$. The Coulomb force $\overrightarrow{F}$ between the two is (where $K = \frac{1}{4\pi\varepsilon_0}$):

Point charges $+4q, -q$ and $+4q$ are kept on the $x$-axis at points $x = 0, x = a$ and $x = 2a$ respectively. Then:

Two identical conducting spheres $P$ and $S$ with charge $Q$ on each, repel each other with a force $16 \,N$. $A$ third identical uncharged conducting sphere $R$ is successively brought in contact with the two spheres. The new force of repulsion between $P$ and $S$ is: (in $\,N$)

Two charges $q_1 = +6q$ and $q_2 = -3q$ are placed as shown in the figure. $A$ proton is placed on the $x$-axis away from $q_2$. To keep the proton in equilibrium,the distance between $q_1$ and the proton is:

For the regular pentagon system shown in the figure,find the net electrostatic force on the charge $q_0$ placed at the center.