What is called electric field intensity? Write its $SI$ unit.

Two point charges $A$ and $B$ of magnitude $+8 \times 10^{-6}\,C$ and $-8 \times 10^{-6}\,C$ respectively are placed at a distance $d$ apart. The electric field at the middle point $O$ between the charges is $6.4 \times 10^{4}\,NC^{-1}$. The distance $d$ between the point charges $A$ and $B$ is ............ $m$. (in $.0$)

$A$ charge produces an electric field of $1\, N/C$ at a point distant $0.1\, m$ from it. The magnitude of the charge is:

Two point charges $-q$ and $+q/2$ are situated at the origin $(0, 0, 0)$ and at the point $(a, 0, 0)$ respectively. The point along the $X$-axis where the electric field vanishes is:

$A$ thin half ring of radius $35 \text{ cm}$ is uniformly charged with a total charge of $Q$ coulomb. If the magnitude of the electric field at the centre of the half ring is $100 \text{ V/m}$,then the value of $Q$ is . . . . . . $\text{nC}$. ($\epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2$ and $\pi = 3.14$)

An infinite number of electric charges, each equal to $5 \text{ nC}$ (magnitude), are placed along the $X$-axis at $x = 1 \text{ cm}, x = 2 \text{ cm}, x = 4 \text{ cm}, x = 8 \text{ cm}, \dots$ and so on. In this setup, if the consecutive charges have opposite signs, then the electric field in $\text{N/C}$ at $x = 0$ is: $\left(\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2\right)$