$F$ is the force between two identical charged particles placed at a distance $Y$ from each other. If the distance between the charges is reduced to half the previous distance,then the force between them becomes:

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}} = $ . . . . . . .

In a right-angled triangle $ABC$,$AB = 3 \ cm$,$BC = 4 \ cm$,and $\angle ABC = \frac{\pi}{2}$. Charges of $+15 \ e.s.u.$,$+12 \ e.s.u.$,and $-20 \ e.s.u.$ are placed at $A$,$B$,and $C$ respectively. What is the net force on the charge at $B$ in $dynes$?

Electric charges of $1 \mu C, -1 \mu C$ and $2 \mu C$ are placed in air at the corners $A, B$ and $C$ respectively of an equilateral triangle $ABC$ having a side length of $10 \ cm$. The resultant force on the charge at $C$ is: (in $N$)

Why is the Coulombian force called a two-body force?

Two point charges $q_1$ and $q_2$ are placed at a distance of $50 \ cm$ from each other in air and interact with a certain force. Now,these charges are placed in an oil with a relative permittivity of $5$. If the force between them remains the same,the distance between them in the oil is ........ $cm$.