Coulomb measured the force between two point charges and found that it varied inversely as the square of the distance between the charges and was directly proportional to the product of the magnitude of the two charges and acted along the line joining the two charges. Thus, if two point charges q1, q2 are separated by a distance r in vacuum, the magnitude of the force (F) between them is given by
$F=k\frac{|q_1{q}_2|}{r^2}$

The field lines carry information about the direction of electric field at different points in space. Having drawn a certain set of field lines, the relative density (i.e., closeness) of the field lines at different points indicates the relative strength of the electric field at those points. The field lines are crowded where the field is strong and are spaced apart where it is weak.
While electric flux is the measure of the flow of the electric field through a given area. It is proportional to the number of electric field lines passing normally through a  perpendicular surface.

Magnitude of electric field strength is higher where density of electric field lines in space is more. The attached figure shows the electric field lines due to a positive point charge. Its field lines are directed radially outwards from the point charge and the dotted lines show the locus of points with same magnitude of electric field.
Since, $E\propto \frac{1}{r^2}$ for a point charge, it is larger near the charge. This behaviour is consistent in the electric field lines of point charge.

Example:
A rod of length $l$ with uniform charge per unit length $\lambda$ is placed at a distance $d$ from origin along the x-axis. A similar rod is placed at the same distance along Y-axis. Determine the magnitude of net electric field intensity at the origin.
Solution:
Let's first find Electric field due to the rod on x-axis, say $E_1$
Consider a small length element $dx$, distance $x$ away from origin.
Field  due to this $d{E}_1=\frac{k\lambda dx}{x^2}$
Integrating from $x=d$ to $x=d+l$
Gives$E_1=\frac{\lambda l}{4\pi {ϵ}_{0}d(d+l)}$ along -x direction
Now, $E_2$is along the -y direction and same in magnitude.
Thus,
$|E|=\sqrt{2}|E_1|=\frac{\sqrt{2}\lambda l}{4\pi {ϵ}_{0}d(d+l)}$

$E=\frac{kQx}{{\left(x^2+R^2\right)}^{\frac{3}{2}}}$
where $Q=2\pi \lambda R$
$R$ is the radius of the ring
$\lambda$ is the charge density
$x$ is the distance from the centre of the ring along the axis

Example:
A non-uniformly charged sphere of radius $R$ has a charge density $\rho ={\rho }_o(r/R)$ where ${\rho }_o$ is constant and $r$ is the distance from the center of the sphere. Find the electric field inside and outside the sphere.
Solution:
Electric field lines will be directed radially outwards as the charge is distributed radially uniformly.
For $0<r<R$,
      Charge enclosed in an infinitesimally small spherical shell of thickness $dr$ at a distance $r$ is:
      $d{Q}_{enclosed}=\rho (r)dV$
      $d{Q}_{enclosed}={\rho }_o\frac{r}{R}4\pi r^{2}dr$
      Charge enclosed upto distance $r$ is:
      $Q=\int d{Q}_{enclosed}={\int }_0^r{\rho }_o\frac{4\pi r^3}{R}dr$
      $Q=\frac{{\rho }_o\pi r^4}{R}$                 
      Applying gauss law at a distance $r$, 
      $E\times 4\pi r^2=\frac{{\rho }_o\pi r^4}{R}$
      $E=\frac{{\rho }_o{r}^2}{4R}$
For $r>R$,
      Charge enclosed is: 
      $Q={\rho }_o\pi R^3$
      Electric field outside is hence given by:
      $E=\frac{{\rho }_o{R}^3}{4r^2}$

Example:
A particle of mass $M$ and charge $q$ is at rest at the mid point between two other fixed similar charges each of magnitude $Q$ placed a distance $2d$ apart. The system is collinear as shown in figure. The particle is now displaced by a small amount $x(x<<d)$ along the line joining the two charges and is left to itself. It will now oscillate about the mean position with what time period?(${\epsilon }_0=$ permittivity of free space)
Solution:
Restoring force on displacement of $x$
$F=K\left[\frac{Qq}{{\left(d-x\right)}^2}-\frac{Qq}{{\left(d+x\right)}^2}\right]$
$=KQq\left[\frac{1}{{\left(d-x\right)}^2}-\frac{1}{{\left(d+x\right)}^2}\right]$
$=KQq\left[\frac{4dx}{{\left(d^2-x^2\right)}^2}\right]$
$=KQq\left[\frac{4dx}{d^4}\right](d>>x)$
$\Rightarrow F=KQq\left[\frac{4x}{d^3}\right]$
Acceleration $a=\frac{F}{m}=\frac{4KQqx}{{Md}^3}$
or ${\omega }^2=\frac{4KQq}{{Md}^3}$
$\therefore T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{{Md}^3}{4KQq}}$
$=2\sqrt{\frac{{{\pi }^{3}Md}^3{\epsilon }_0}{Qq}}$

content Text

$E=k\sigma 2\pi [1-\frac{z}{\sqrt{z^2+R^2}}]$
where $k=\frac{1}{4\pi {ϵ}_0}$ and $\sigma$ is the surface charge density

The Electric Flux Density (D) is related to the Electric Field (E) by:
$D=ϵE$ -----[Equation 1]
In Equation [1], $ϵ$ is the permittivity of the medium (material) where we are measuring the fields.
The Electric Field is equal to the force per unit charge:
$E=\frac{q_1}{4\pi ϵR^2}$------[Equation 2]
Then the Electric Flux Density is:
$D=ϵE=\frac{q_1}{4\pi R^2}$--------[Equation 3]
From Equation [3], the Electric Flux Density is very similar to the Electric Field, but does not depend on the material in which we are measuring (that is, it does not depend on the permittivity. Note that the $D$ field is a vector field, which means that at every point in space it has a magnitude and direction.
The Electric Flux Density has units of Coulombs per meter squared [$C/m^2$].

Example:
An electric field given by $\vec{E}=4\hat{i}-3(y^2+2)\hat{j}$ passes Gaussian cube of side 1m placed with one corner is at origin such that its sides represents x, y and z axes. Find the magnitude of net charge enclosed within the cube.
Solution:
The flux passing through the surfaces parallel to x-y plane is zero as their is no z component of E.
The flux passing through the surfaces parallel to y-z plane is zero as the x component of E is a constant. Flux going in goes out, so no net flux.
The flux passing through the surface parallel to x-z plane is given by,
${\varphi }_1=-6$  and ${\varphi }_2=-9$
${\varphi }_{net}=-9+6=-3$
Magnitude of charge enclosed $=|{\varphi }_{net}|\times {ϵ}_0=3{ϵ}_0$

Total charge on A is $Q_1$ and on B is $Q_2$ which remains constant by law of conservation of charge.
From gauss's law, total charge on the inner regions is zero. Hence, equal and opposite charges are induced on the inner surface. 
Net electric field at P is zero.
$\frac{Q_1-q}{2A{\epsilon }_o}-\frac{q}{2A{\epsilon }_o}+\frac{q}{2A{\epsilon }_o}-\frac{Q_2+q}{2A{\epsilon }_o}$
$q=\frac{Q_1-Q_2}{2}$

A small electric dipole having dipole moment $\vec{p}$ is placed along X-axis, as shown in the figure. A semi-infinite uniformly charged di-electric thin rod  placed along x axis, with one end coinciding with origin. If linear charge density rod is $+\lambda$ and distance of dipole from rod is ${}^{\prime }{a}^{\prime }$, then calculate the electric force acting on dipole.
Electric field at A due to dipole,   $E_A=\frac{2KP}{r^3}$            where $K=\frac{1}{4\pi {ϵ}_o}$
$E_A=\frac{2KP}{(x+a{)}^3}$
Force at A due to dipole,   $F_A=q{E}_A=(\lambda$ $dx)$ $\times \frac{2KP}{(x+a{)}^3}$
$F_A=\frac{2\lambda KP}{(x+a{)}^3}dx$
Total force exerted due to dipole on the rod,    $F_D={\int }_0^{\infty }\frac{2\lambda KP}{(x+a{)}^3}dx$
$F_D=2\lambda KP\times \frac{1}{-2(x+a{)}^2}{|}_0^{\infty }$
  $⟹F_D=\frac{\lambda KP}{a^2}$
Now force exerted by rod on dipole is equal and opposite to $F_D$  (Newton's law)
$\overrightarrow{F_R}=-F_D=\frac{-\lambda P}{4\pi {ϵ}_o{a}^2}$