If two balls of given masses and charges suspended from the same point are released, which of the following is the correct arrangement in equilibrium?

Two positive point charges each of magnitude $10 \mathrm{C}$ are fixed at positions $A$ and $B$, at a separation $2d=6\mathrm{m}.$ A negatively charged particle of mass $m=90 \mathrm{gm}$ and charge of magnitude ${10}^{-5 }\mathrm{C}$ is revolving in a circular path of radius $4 \mathrm{m}$, in the plane perpendicular to the line $AB$ and bisecting the line $AB.$ Neglect the effect of gravity. Find the angular velocity of the particle.

Four charges are arranged at the corners of a square $ABCD$, as shown in the adjoining figure. The force on the charge kept at the centre $O$ is

Three charges are placed at the vertices of an equilateral triangle of side $a$ as shown in the following figure. The force experienced by the charge placed at the vertex $A$ in a direction normal to $BC$ is

Two small spherical balls, each carrying a charge $Q=10 \mathrm{\mu C}$are suspended by two insulating threads of equal lengths $1 \mathrm{m}$each, from a point fixed in the ceiling. It is found that in equilibrium, threads are separated by an angle $60°$between them, as shown in the figure. What is the tension in the threads? $\left(\text{ Given: }\frac{1}{\left(4\pi {\epsilon }_0\right)}=9\times {10}^9\mathrm{Nm}/{\mathrm{C}}^2\right)$

Three charges$-q_1,+q_2$ and $-q_3$ are placed as shown in the figure. The $x-$component of the force on $-q_1$ is proportional to