Why Coulomb’s law is associated with Newton’s $3^{rd}$ law ?

The magnitude of electric force on $2\, \mu \,C$ charge placed at the centre $O$ of two  equilateral triangles each of side $10 \,cm$, as shown in figure is $P$. If charge $A, B, C, D, E$ and $F$ are $2\, \mu \,C, 2\, \mu \,C, 2\, \mu \,C,-2\, \mu \,C, -2\, \mu \,C, -2\, \mu \,C$   respectively, then $P$ is :.....$N$

Four identical pendulums are made by attaching a small ball of mass $100 \,g$ on a $20 \,cm$ long thread and suspended from the same point. Now, each ball is given charge $Q$, so that balls move away from each other with each thread making an angle of $45^{\circ}$ from the vertical. The value of $Q$ is close to ..............$\mu C$ $\left(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9\right.$ in $SI$ units $)$

Figure represents a crystal unit of cesium chloride, $\mathrm{CsCl}$. The cesium atoms, represented by open circles are situated at the corners of a cube of side $0.40\,\mathrm{nm}$, whereas a $\mathrm{Cl}$ atom is situated at the centre of the cube. The $\mathrm{Cs}$ atoms are deficient in one electron while the $\mathrm{Cl}$ atom carries an excess electron.

$(i)$ What is the net electric field on the $\mathrm{Cl}$ atom due to eight $\mathrm{Cs}$ atoms ?

$(ii)$ Suppose that the $\mathrm{Cs}$ atom at the corner $A$ is missing. What is the net force now on the $\mathrm{Cl}$ atom due to seven remaining $\mathrm{Cs}$ atoms ?

Two equal charges of magnitude $Q$ each are placed at a distance $d$ apart. Their electrostatic energy is $E$. A third charge $-Q / 2$ is brought midway between these two charges. The electrostatic energy of the system is now