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:

$A$ certain charge $Q$ is divided into two parts $q$ and $(Q-q)$. How should the charges $Q$ and $q$ be divided so that $q$ and $(Q-q)$ placed at a certain distance apart experience maximum electrostatic repulsion?

Two charged spheres separated at a distance $d$ exert a force $F$ on each other. If they are immersed in a liquid of dielectric constant $2$,then what is the force (if all conditions are same)?

Two positively charged spheres of masses $m_1$ and $m_2$ are suspended from a common point at the ceiling by identical insulating massless strings of length $l$. Charges on the two spheres are $q_1$ and $q_2$,respectively. At equilibrium,both strings make the same angle $\theta$ with the vertical. Then

Two similar tiny balls of mass $m$,each carrying charge $q$,are hung from silk threads of length $l$ as shown in the figure. These are separated by a distance $x$ and the angle $2\theta$ is small. Then,for equilibrium:

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: