The electric field intensity at a point in vacuum is equal to

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 $g=10 \ m/s^2$)

$A$ metal sphere of radius $R \ cm$ is charged with $4 \pi \mu C$ and is situated in air. If $\sigma$ is the surface charge density and $E$ is the electric intensity at a distance $r$ from the centre of the sphere,then $r$ is equal to ($\epsilon_{0}$ is the permittivity of free space).

$A$ point charge $+q$ is placed at the origin. $A$ second point charge $+9q$ is placed at $(d, 0, 0)$ in a Cartesian coordinate system. The point in between them where the electric field vanishes is:

Two point charges $a$ and $b$,whose magnitudes are the same,are positioned at a certain distance from each other with $a$ at the origin. $A$ graph is drawn between the electric field strength $E$ at points between $a$ and $b$ and the distance $x$ from $a$. $E$ is taken as positive if it is along the line joining $a$ to $b$. From the graph,it can be decided that:

Write the equation for the electric field produced by a system of $n$ charges.