What is the direction of electric field intensity?

Consider the following statements about electric field intensity and electric potential.
$A$. The electric field intensity due to a charged spherical shell is inversely proportional to the square of its distance from the center for points outside the shell.
$B$. The electric potential due to a point charge is inversely proportional to the distance between the charge and the point.

$A$ deuteron and an $\alpha$-particle are placed $1\,\mathring{A}$ apart in air. The magnitude of the intensity of the electric field due to the deuteron at the position of the $\alpha$-particle is:

Find the electric field at point $P$ (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length $L$ carrying a charge $Q.$ The distance of the point $P$ from the centre of the rod is $a = \frac{\sqrt{3}}{2} L$.

$A$ circular wire loop of radius $1 \, cm$ carries a total charge $1 \times 10^{-6} \, C$ distributed uniformly over its length. If $0.01 \%$ of its length (circumference) is cut off, then the electric field at the centre of the loop due to the remaining wire is
$(\text{Take} \, \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \, \text{SI unit})$

Two non-conducting solid spheres of radii $R$ and $2R$,having uniform volume charge densities $\rho_1$ and $\rho_2$ respectively,touch each other. The net electric field at a distance $2R$ from the centre of the smaller sphere,along the line joining the centres of the spheres,is zero. The ratio $\frac{\rho_1}{\rho_2}$ can be;
$(A) -4$ $(B) -\frac{32}{25}$ $(C) \frac{32}{25}$ $(D) 4$