What is the direction of electric field intensity?

The unit of intensity of electric field is

$A$ ring of charge with radius $R = 0.5\, m$ having a gap of length $l = 0.02\, m$,carries a total charge of $Q = +1\, C$. The electric field at the centre 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 $\pi = 3.14$)

Suppose a uniformly charged wall provides a uniform electric field of $2 \times 10^4 \ N/C$ normally. $A$ charged particle of mass $2 \ g$ is suspended by a silk thread of length $20 \ cm$ and remains at a distance of $10 \ cm$ from the wall. Then the charge on the particle will be $\frac{1}{\sqrt{x}} \ \mu C$ where $x=$ . . . . . . . (Use $g=10 \ m/s^2$)

Four equal charges of value $+Q$ are placed at four vertices of a regular hexagon of side $a$. By suitably choosing the vertices,what can be the maximum possible magnitude of the electric field at the centre of the hexagon?