The dimensional formula of $k$ (Coulomb's constant) is . . . . . . .
(Take $I$ as the dimension of current.)

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

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

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 $\,N$)

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

Three identical charges are placed on three vertices of a square. If the force acting between $q_1$ and $q_2$ is $F_{12}$ and between $q_1$ and $q_3$ is $F_{13}$,then $\frac{F_{13}}{F_{12}} = $ . . . . . . .