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:

Two identical balls having like charges $Q_1$ and $Q_2$ are placed at a distance $r$ apart and repel each other with a force $F$. 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, $\frac{Q_1}{Q_2}$, is:

Two identical charges are separated by a distance $d$. $A$ third charge is placed on their perpendicular bisector at a distance $x$. For the third charge to experience maximum force,the value of $x$ is:

Two charges $+10 \mu C$ and $-10 \mu C$ are separated by $10 \text{ cm}$. The magnitude of the force acting on another charge $5 \mu C$ placed at the midpoint of the line joining the two charges will be: [Use $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$] (in $\text{ N}$)

The ratio of electrostatic and gravitational forces acting between an electron and a proton separated by a distance $5 \times 10^{-11} \, m$ will be (Charge on electron $= 1.6 \times 10^{-19} \, C$,mass of electron $= 9.1 \times 10^{-31} \, kg$,mass of proton $= 1.6 \times 10^{-27} \, kg$,$G = 6.7 \times 10^{-11} \, N m^2/kg^2$).

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