Whose result the whole electrostatic is ?

Two point charges $Q_1, Q_2$ are fixed at $x = 0$ and $x = a$. Assuming that field strength is positive in the direction coinciding with the positive direction of $x$, then, which following option will be correct ?

As shown in the figure, a particle A of mass $2\,m$ and carrying charge $q$ is connected by a light rigid rod of length $L$ to another particle $B$ of mass $m$ and carrying charge $-q.$ The system is placed in an electric field $\vec E$ . The electric force on a charge $q$ in an electric field $\vec E$ is $\vec F = q \vec E $ . After the system settles into equilibrium, one particle is given a small push in the transverse direction so that the rod makes a small angle $\theta_0$ with the electric field. Find maximum tension in the rod.

Two particles ${A}$ and ${B}$ having charges $20\, \mu {C}$ and $-5\, \mu {C}$ respectively are held fixed with a separation of $5\, {cm}$. At what position a third charged particle should be placed so that it does not experience a net electric force?

Suppose a uniformly charged wall provides a uniform electric field of $2 \times 10^4 \mathrm{~N} / \mathrm{C}$ normally. A charged particle of mass $2 \mathrm{~g}$ being suspended through a silk thread of length $20 \mathrm{~cm}$ and remain stayed at a distance of $10 \mathrm{~cm}$ from the wall. Then the charge on the particle will be $\frac{1}{\sqrt{\mathrm{x}}} \ \mu \mathrm{C}$ where $\mathrm{x}=$ ____________.  use $g=10 \mathrm{~m} / \mathrm{s}^2$ ]

Two point charges $Q$ and $-3Q$ are placed at some distance apart. If the electric field at the location of $Q$ is $E$ then at the locality of $ - 3Q$, it is