There are two charges $+1 \ \mu C$ and $+5 \ \mu C$. The ratio of the forces acting on them will be

Two free positive charges $4q$ and $q$ are at a distance $l$ apart. What charge $Q$ is needed to achieve equilibrium for the entire system and where should it be placed from charge $q$?

Two spherical conductors $B$ and $C$ having equal radii and carrying equal charges $Q$ repel each other with a force $F$ when kept apart at a distance $r$. $A$ third spherical conductor having the same radius as that of $B$ but uncharged is brought in contact with $B$,then brought in contact with $C$,and finally removed away from both. The new force of repulsion between $B$ and $C$ is

Two point charges of equal magnitude but opposite sign are placed at a certain distance. The force between them is $F$. If $75\%$ of the charge from one is transferred to the other,what will be the new force between them?

$5$ charges each of magnitude $10^{-5} \,C$ and mass $1 \,kg$ are placed (fixed) symmetrically about a movable central charge of magnitude $5 \times 10^{-5} \,C$ and mass $0.5 \,kg$ as shown in the figure. The charge at $P_1$ is removed. The acceleration of the central charge is [Given,$O P_1=O P_2=O P_3=O P_4=O P_5=1 \,m, \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \,N \cdot m^2/C^2$]

Consider three charges $q_{1}, q_{2}, q_{3}$ each equal to $q$ at the vertices of an equilateral triangle of side $l$. What is the force on a charge $Q$ (with the same sign as $q$) placed at the centroid of the triangle,as shown in Figure?