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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
$F=k\frac{|q_1{q}_2|}{r^2}$

Force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges.This is termed as the principle of superposition.
Example : Consider three charges $q_1,q_2,q_3$ each equal to q at the vertices of an equilateral triangle of side $l$. What is the force on a charge Q (with the same sign as $q$) placed at the centroid of the triangle?
Solution In the given equilateral triangle ABC of sides of length $l$, if we draw a perpendicular AD to the side BC,
$AD=ACcos30=(\sqrt{3}/2)l$ and the distance AO of the centroid O from A is $(2/3)AD=(1/\sqrt{3})l$. By symmetry AO = BO = CO.
Force $F_1$ on Q due to charge q at $A=\frac{3Qq}{4\pi {ϵ}_0{l}^2}$ along AO
Force $F_2$ on Q due to charge q at $B=\frac{3Qq}{4\pi {ϵ}_0{l}^2}$ along BO
Force $F_3$ on Q due to charge q at $C=\frac{3Qq}{4\pi {ϵ}_0{l}^2}$ along CO
Therefore, the total force on $Q=\frac{3Qq}{4\pi {ϵ}_0{l}^2}(\hat{r}-\hat{r})=0$ where $\hat{r}$ is the unit vector along OA.

$E=\frac{\sigma }{{ϵ}_0}\hat{n}$
where $\sigma$  is the surface charge density and $\hat{n}$ is a unit vector normal to the surface in the outward direction.
To derive the result, choose a pill box (a short cylinder) as the Gaussian surface about any point P on the surface. The pillbox is partly inside and partly outside the surface of the conductor. It has a small area of cross section $\delta S$ and negligible height.Just inside the surface, the electrostatic field is zero; just outside, the field is normal to the surface with magnitude E. Thus,the contribution to the total flux through the pill box comes only from the outside (circular) cross-section of the pill box. This equals $\pm E\delta S$  (positive for $\sigma >0$,negative for $\sigma <0$), since over the small area $\delta S$, $E$ may be considered constant and $E$ and $\delta S$ are parallel or antiparallel. The charge enclosed by the pill box is $\sigma \delta S$.By Gauss's law
$E\delta S=\frac{|\sigma |\delta S}{{ϵ}_0}$
$E=\frac{|\sigma |}{{ϵ}_0}$

Electric flux $\varphi$ through an area element $d\vec{S}$ is defined by
$\varphi =\vec{E}.d\vec{S}=E.dScos\theta$, which is proportional to the number of field lines cutting the area element. The angle $\theta$ here is the angle between $\vec{E}$ and $d\vec{S}$.

An electric dipole is a pair of equal and opposite point charges -q and q, separated by a distance 2a. The direction from q to -q is said to be the direction of the dipole.
$p=q\times 2a$
where $p$ is the electric dipole moment pointing from the negative charge to the positive charge.

According to Gauss's Law, flux through a closed surface is given by:
$\varphi =\int \vec{E}.\vec{d}S=\frac{q_{enclosed}}{{\epsilon }_o}$
Since electric field is uniform, it is created by a source very far from the closed surface. Or there is no charge enclosed within the closed surface. Hence, net flux through it is zero.
Note: 
This argument does not hold true for an open surface as an open surface can be arbitrarily extended to  a closed surface enclosing a non-zero charge in which case the electric flux through the surface may become non-zero.