If the distance between two equal point charges is doubled,what would happen to the force between them?

Three charged particles of each mass $0.1 \,g$ and charge $q$ are suspended from a common rigid point by insulated massless threads of each $1 \,m$ long. If the three particles are in equilibrium and are located at the corners of an equilateral triangle of side $3 \,cm$, the charge $q$ on each particle is . . . . . . $nC$. (The angle made by the line joining the centroid of the triangle and the point of suspension with the vertical is very small). (Acceleration due to gravity $= 10 \,ms^{-2}$ and $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \,Nm^2 C^{-2}$)

$A$ charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium if $q$ is equal to:

Two particles of equal mass $m$ and charge $q$ are placed at a distance of $16 \, cm$. They do not experience any net force. The value of $\frac{q}{m}$ is

Two identical non-conducting solid spheres of same mass and charge are suspended in air from a common point by two non-conducting,massless strings of same length. At equilibrium,the angle between the strings is $\alpha$. The spheres are now immersed in a dielectric liquid of density $\rho_l = 800 \ kg \ m^{-3}$ and dielectric constant $K = 21$. If the angle between the strings remains the same after the immersion,then
$(A)$ electric force between the spheres remains unchanged
$(B)$ electric force between the spheres reduces
$(C)$ mass density of the spheres is $840 \ kg \ m^{-3}$
$(D)$ the tension in the strings holding the spheres remains unchanged

Three charges $4q, Q$ and $q$ are in a straight line along the $x$-axis at the positions $x=0 \ m$,$x=\frac{1}{2} \ m$,and $x=1 \ m$ respectively. The resultant force on $q$ will be zero if $Q$ is equal to: