According to the quantisation of charge
q = ne
Here q = 1C. So the number of electrons in 1 coulomb of charge is
n = \(\frac { q }{ e } =\frac { 1C }{ 1.6\times 10^{ -19 } } \) = 6.25 x 10¹⁸ electrons

According to the superposition principle, the total electrostatic force on charge q₁ is the vector sum of the forces due to the other charges,
\(\vec { { F }_{ 1 }^{ tot } } =\bar { { F }_{ 12 } } +\bar { { F }_{ 13 } } +\bar { F_{ 14 } } \)
The following diagram shows the direction of each force on the charge q₁.

The charges q₂ and q₄ are equi-distant from q₁. As a result the strengths (magnitude) of the forces \(\vec { { F }_{ 12 } } \) and \(\vec { { F }_{ 14 } } \) are the same even though their directions are different. Therefore the vectors representing these two forces are drawn with equal lengths. But the charge q₃ is located farther compared to q₂ and q₄. Since the strength of the electrostatic force decreases as distance increases, the strength of the force \(\vec { { F }_{ 13 } } \) is lesser than that of forces \(\vec { { F }_{ 12 } } \) and \(\vec { { F }_{ 14 } } \). Hence the vector representing the force \(\vec { { F }_{ 13 } } \) is drawn with smaller length compared to that for forces \(\vec { { F }_{ 12 } } \) and \(\vec { { F }_{ 14 } } \).
From the figure, r₂₁ =\(\sqrt { 2 } \) m = r₄₁ and r₃₁ = 2m
The magnitudes of the forces are given by
F₁₃ = \(\frac { kq^{ 2 } }{ r_{ 31 }^{ 2 } } =\frac { 9\times 10^{ 9 }\times 10^{ -12 } }{ 4 } \)
F₁₃  = 2.25 x 10^-3 N
F₁₂ = \(\frac { kq^{ 2 } }{ r_{ 31 }^{ 2 } } ={ F }_{ 14 }=\frac { 9\times 10^{ 9 }\times 10^{ -12 } }{ 2 } \)
= 4.5 x 10^-3N
From the figure, the angle θ = 45⁰. In terms of the components, we have
\(\vec { { F }_{ 12 } } ={ F }_{ 12 }cos\theta \hat { i } -{ F }_{ 12 }sin\theta \hat { j } \)
= 4.5 x 10^-3 x \(\frac { 1 }{ \sqrt { 2 } } \hat { i-4.5\times { 10 }^{ -3 }\times \frac { 1 }{ \sqrt { 2 } } \hat { j } } \)
 \(\vec { { F }_{ 13 } } =F_{ 13 }\hat { i } \) = 2.25 x 10^-3 N\(\hat { i } \)
\(\vec { { F }_{ 14 } } ={ F }_{ 14 }cos\theta \hat { i } +{ F }_{ 14 }sin\theta \hat { j } \)
= 4.5 x 10^-3 x \(\frac { 1 }{ \sqrt { 2 } } \hat { i+4.5\times { 10 }^{ -3 }\times \frac { 1 }{ \sqrt { 2 } } \hat { j } } \)
Then the total force on q₁ is,
\(\vec { { F }_{ 1 }^{ tot } }={ (F }_{ 12 }cos\theta \hat { i } -{ F }_{ 12 }sin\theta \hat { j } )+{ F }_{ 13 }\hat { i } +{ (F }_{ 14 }cos\theta \hat { i } +{ F }_{ 14 }sin\theta \hat { j } )\)
\(\vec { { F }_{ 1 }^{ tot } } =({ F }_{ 12 }cos\theta +F_{13}+{ F }_{ 14 }cos\theta )\hat { i } +(-{ F }_{ 12 }sin\theta +{ F }_{ 14 }sin\theta )\)\(\hat { j } \)
Since F₁₂ = F₁₄, the j ^th component is zero.
Hence we have
\(\vec { { F }_{ 1 }^{ tot } } =({ F }_{ 12 }cos\theta +F_{13}+{ F }_{ 14 }cos\theta )\hat { i } \)
substituting the values in the above equation,
\(\left( \frac { 4.5 }{ \sqrt { 2 } } +2.25+\frac { 4.5 }{ \sqrt { 2 } } \right) \times10^{-3}\hat { i }=(4.5\sqrt { 2 } +2.25)\times 10^{-3}\hat { i } \)
\(\vec { { F }_{ 1 }^{ tot } } \) = 8.61 x 10^-3 N\(\hat { i } \)
The resultant force is along the positive x-axis.

The proton and the electron attract each other. The magnitude of the electrostatic force between these two particles is given by
\(F_e=\frac { ke^{ 2 } }{ { r }^{ 2 } } =\frac { 9\times 10^{ 9 }\times (1.6\times 10^{ -19 })^{ 2 } }{ (5.3\times 10^{ -11 })^{ 2 } } \)
=\(\frac { 9\times 2.56 }{ 28.09 } \) x 10^-7 = 8.2 x 10^-8 N
The gravitational force between the proton and the electron is attractive. The magnitude of the gravitational force between these particles is
F_G = \(\frac { G{ m }_{ e }{ m }_{ p } }{ { r }^{ 2 } } \)
= \(\frac { 6.67\times 10^{ -11 }\times 9.1\times 10^{ -31 }\times 1.6\times 10^{ -27 } }{ (5.3\times 10^{ -11 })^{ 2 } } \)
= \(\frac { 97.11 }{ 28.09 } \) x 10^-47 = 3.4 x 10^-47N
The ratio of the two forces \(\frac { { F }_{ e } }{ F_{ G } } =\frac { 8.2\times 10^{ -8 } }{ 3.4\times 10^{ -47 } } \)
= 2.41 x 10³⁹
Note that F_e ≈ 10³⁹ F_G
The electrostatic force between a proton and an electron is enormously greater than the gravitational force between them. Thus the gravitational force is negligible when compared with the electrostatic force in many situations such as for small size objects and in the atomic domain. This is the reason why a charged comb attracts an uncharged piece of paper with greater force even though the piece of paper is attracted downward by the Earth. This given figure is shown in below.

Electrostatic attraction between a comb and pieces of papers

If the two spheres are neutral, the angle between them will be 0^o when hanged vertically. Since they are positively charged spheres, there will be a repulsive force between them and they will be at equilibrium with each other at an angle of 30° with the vertical. At equilibrium, each charge experiences zero net force in each direction. We can draw a free-body diagram for one of the charged spheres and apply Newton’s second law for both vertical and horizontal directions.
The free-body diagram is shown below

In the x-direction, the acceleration of the charged sphere is zero.
Using Newton’s second law \((\vec { { F }_{ tot }= } m\vec { a } )\), we have
T sinθ\(\hat { i } \) - F_e\(\hat { i } \) =0
T sinθ = F_e    ......(1)
Here T is the tension acting on the charge due to the string and F_e is the electrostatic force between the two charges.
In the y-direction also, the net acceleration experienced by the charge is zero
Tcosθ\(\hat { j } \) - mg\(\hat { j } \) = 0
Tcosθ = mg   ..(2)
By dividing equation (1) by equation (2),
tanθ = \(\frac { { F }_{ e } }{ mg } \)          .....(3)
Since they are equally charged, the magnitude of the electrostatic force is
\({ F }_{ e }=k\frac { { q }^{ 2 } }{ { r }^{ 2 } } \) where k=\(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \)
Here r = 2a = 2Lsinθ. By substituting these values in equation (3),
tanθ = k\(\frac { { q }^{ 2 } }{ mg(2Lsin\theta )^{ 2 } } \)       ..........(4)
Rearranging the equation (4) to get q
q = 2 Lsinθ\(\\ \sqrt { \frac { mgtan\theta }{ k } } \)
= 2 x 0.1 x sin 30^o x \(\sqrt { \frac { 10^{ -3 }\times 10\times { tan30 }^{ 0 } }{ 9\times 10^{ 9 } } } \)
q = 8.01 x 10^-8C = 80.1 nC

(a) q₁ = +2 μC, q₂ = +3 μC, and r = 1m. Both are positive charges. so the force will be repulsive.
Force experienced by the charge q₂ due to q₁ is given by
\(\vec { { F }_{ 21 } } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \hat { r_{ 12 } } \)
Here \(\hat { r_{ 12 } } \) is the unit vector from q₁ to q₂. Since q₂ is located on the right of q₁, we have
\(\hat { r_{ 12 } } =\hat { i } \), and \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } }=9 \times10^9\) so that
\(\vec { { F }_{ 21 } } =\frac { 9\times 10^{ 9 }\times 2\times 10^{ -6 }\times 3\times 10^{ -6 } }{ 1^{ 2 } } \hat { i } \)
= 54 x 10^-3 N\(\hat { i } \) 
According to Newton’s third law, the force experienced by the charge q₁ due to q₂ is \(\vec { { F }_{ 12 } } =-\vec { { F }_{ 21 } } \) Therefore,
\(\vec { F_{ 12 } } \)= 54 x 10^-3 N\(\hat { i } \)
The directions of \(\vec { { F }_{ 21 } } \) and \(\vec { { F }_{ 12 } } \)  are shown in the figure (case (b)).
(b) q₁ = +2 μC, q₂ = –3 μC, and r = 1m. They are unlike charges. So the force will be attractive.
Force experienced by the charge q₂ due to q₁ is given by
\( \vec{F}_{21} =\frac{9 \times 10^{9} \times\left(2 \times 10^{-6}\right) \times\left(-3 \times 10^{-6}\right)}{1^{2}} \hat{r}_{12} \)
\(=-54 \times 10^{-3} \mathrm{~N} \hat{i}\left(\mathrm{Using} \hat{r}_{12}=\hat{i}\right)\)
The charge q₂ will experience an attractive force towards q₁ which is in the negative x direction.
According to Newton’s third law, the force experienced by the charge q₁ due to q₂ is \(\vec{F}_{12}=-\vec{F}_{21}\) Therefore,
\(\vec{F}_{12}=54 \times 10^{-3} \widehat{i} \mathrm{~N}\)
The directions of \(\vec{F}_{21} \text { and } \vec{F}_{12}\) are shown in the figure (case (b)).
(c) If these two charges are kept inside the water, then the force experienced by q₂ due to q₁
\(\vec { { F }_{ 21 }^{ W } } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \hat { r_{ 12 } } \)
since ε = ε_rε₀,
we have \(\vec { { F }_{ 21 }^{ W } } =\frac { 1 }{ 4\pi { { \varepsilon }_{ r }\varepsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \hat { r_{ 12 } } =\frac { \vec { F_{ 21 } } }{ { \varepsilon }_{ r } } \)
Therefore,
\(\vec { { F }_{ 21 }^{ W } } =\frac { 54\times { 10 }^{ -3 }N }{ 80 } \hat { i } \) = -0.675 x 10^-3 N\(\hat { i } \)