By using Coulomb's law,define unit charge.

Two particles of equal mass $m$ and equal charge $q$ are separated by a distance of $16 \text{ cm}$. They do not experience any net force. The value of $\frac{q}{m}$ is . . . . . . (where $G$ is the universal gravitational constant and $\epsilon_0$ is the permittivity of free space).

Four charges equal to $-Q$ are placed at the four corners of a square and a charge $q$ is at its centre. If the system is in equilibrium,the value of $q$ is

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 charges $+q$ and $-3q$ are placed on the $x$-axis separated by a distance $d$. ($-3q$ is to the right of $q$). Where should a third charge $2q$ be placed such that it will not experience any net force?

$A$ charged metallic sphere $A$ is suspended by a nylon thread. Another charged metallic sphere $B$ held by an insulating handle is brought close to $A$ such that the distance between their centres is $10 \, cm$,as shown in Figure $(a)$. The resulting repulsion of $A$ is noted (for example,by shining a beam of light and measuring the deflection of its shadow on a screen). Spheres $A$ and $B$ are touched by uncharged spheres $C$ and $D$ respectively,as shown in Figure $(b)$. $C$ and $D$ are then removed and $B$ is brought closer to $A$ to a distance of $5.0 \, cm$ between their centres,as shown in Figure $(c)$. What is the expected repulsion of $A$ on the basis of Coulomb's law? Spheres $A$ and $C$ and spheres $B$ and $D$ have identical sizes. Ignore the sizes of $A$ and $B$ in comparison to the separation between their centres.